How Games Twist Probability - Extra Credits Gaming

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Probability! A cornerstone for the design of any game. It can build excitement and suspense but, it can also confuse and upset players. So how do you keep probability and players' choices in games meaningful without bogging them down with the specifics? Let's enter the psychology of gaming and how you can twist and obscure probability to build anticipation and emotion for your next game design.
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#ExtraCredits #Gaming #GameDesign

[@extrahistory]

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[@denislamesch8155]

The Monty Hall Problem is a famous and often missunderstood problem. In one way, the intuition, that switching makes no sens because it's now simply a 50/50 chance between your original pick and the remaining door is corect. But there is one thing that is often overlooked, that changes the ods. Wenn one of the doors is revealed, it is always a goat(or turd in this video). This means that the host never reveals the main prize. This makes sens since if he did, switching would be meaningless. But it also means, that since the revealed door is not random and we are ignoring the cases where the main price would have been revealed, the ods change. Now it's like instead of a 50/50, you should see it as the original 1/3 but you and the host get a pick, meaning it becommes said 2/3 or 66%.

[@fantalandia4273]

Thanks to YOU!!!

[@badcactus819]

Fire Emblem actually has done interesting things with their RNG. Due to the permadeath mechanic the games have historically had, the RNG can feel particularly punishing. As a result, starting about halfway through the series, the developers started using a system the community dubbed "true hit," which is essentially just averaging 2 random numbers on every roll. Crucially, the games that use this system still display the chance as if only 1 number was being rolled. This results in displayed hit values over 50% being more likely to hit than their displayed chance, with the inverse for values under 50. Ironically, this actually lines up better with human perceptions of probability, making the game feel less punishing

[@KnucklesEki]

Now I'm curious if tabletop RPGs like D&D has you roll two d20 on tests and take the average of the two. How more or less balanced would their combat be?

[@badcactus819]

@@KnucklesEki considering how much it would nerf crits I am not sure it would work as well. Critical hits in FE are completely seperate from the hit roll, having their own role (which iirc is not affected by the true hit systems in some of the games). In ttrpgs, nat 20s would be exceptionally rare. FE doesnt have this issue

[@MoiMagnus1er]

@@KnucklesEki That's equivalent (up to slight variation due to rounding) to the "roll 2d10 and sum", which is used by some tables. Though this is not the most popular variant like that, the most popular variant like that is "roll 3d6 and sum". It gives a widly different gameplay, and while I don't like its effect on combats and other "chaotic" situation, but it works reasonably well for skill checks or any situation where you'd expect an expert to almost never fail.

[@tijnvannelmen5832]

I thought True Hit was the name for the original formula, with the new one being '2RN' or '2RNG'

[@antenna_prolly]

Oh, that explains that option you can set for matches in Fire Emblem Tactics Online.

[@Throgan]

"Clarity when you want to make them think; obfuscation when you want to make them feel" explains so much of the world right now, well beyond video games.

[@drosberg3680]

A very interesting wrinkle in the Monte Hall problem that I've never seen explored is that if Monte were truly a random actor making decisions blindly then switching and staying would indeed have equal probabilities of winning. If Monte were to reveal doors at random, then there would be three equally likely possibilities: 1/3rd chance that the prize is behind the door you picked, 1/3rd chance that the prize is behind the door Monte offers as a switch, and 1/3rd chance that Monte accidentally reveals the door with the prize before offering the switch. This means that both switching and staying win 1/3rd of the time, and 1/3rd of the time you lose with no chance to switch. However, in the actual problem Monte is not acting randomly and has perfect information, and he will never allow that last outcome to occur. As a result you get 1/3rd chance that staying wins and a 2/3rds chance that switching wins, because Monte will always pick the situation where switching wins over the one where you just lose without getting to switch at all. I think that's a fascinating probability lesson, as without knowing Monte's behavior the two situations are indistinguishable. If Monte revealed a dud by random chance then it's indeed a 50/50, but if Monte has perfect knowledge and used it to ensure he revealed a dud before offering the switch then the odds are 33/66. There are actually many tellings of the Monte Hall problem that aren't explicit about this, so the 50/50 interpretation is completely valid in those cases. It shows a lot of the nuance in statistics, that two situations that look identical can in fact be very different!

[@StringsOfTheHarp]

I haven't done the math since you said you haven't, but I'm pretty sure if you chose random actors like that it turns out to be a 25% to 75% outcome, since the 33% chance of picking the prize makes it an 100% chance for the randomly opened door to become a dud. This is a common issue in statistics IIRC, because people 'stop testing when a positive result is reached' So overall while it might look like a 1/3 chance for the randomly opened door to be the prize, the first 33% chance completely cancels out the probability of the next rolls, since both doors become a 0%% chance and not a 50/50 If door A has the prize and you picked A at random, there is no other 33% or 50/50 between B and C; the area of operation becomes invalid. Lets say you were shooting a gun at a target and you had a specific number of shots to get a bullseye. Because people who 'succeed' don't take the rest of their shots; the statistics look more like a staircase than a bell curve if we see if are counting if a person shot each bullet or not on a graph. If it is truly random while testing for a positive result the distribution ends up changed. The monte hall problem is specifically geared on the change of condition vs innate bias. The fact the third party reveals a dud is part of the required problem, changing that removes the problem all together and becomes what is basically a coin flip. The attachment people have to their initial choice even if they are aware of the solution; as well as how people operate and think is a major part of it. Most of these statistical philosophy things rely on the participants having 'a choice' rather than 'random chance'

[@michaelmurphy2112]

I don't think in that scenario it was actually Monty picking the door. There's probably someone in the directors booth, who knows where the prize is, telling Monty which door to reveal. Monty likely didn't know where the prize was either, as that could allow someone to gather some information about the location.

[@flatulentdragon]

I have abandoned many games with character development trees because I had no idea what effect the various choices would have, and I was worried about choosing the "wrong" one. Stuff like: "Do you want Necro or Gia?" The categories were completely obscure.

[@LamanKnight]

This is exactly why I've always gone to PokemonDB before deciding which moves to let my Pokémon keep, or to a wiki that lets me know what the upgrade paths are for weapons in Monster Hunter. I have no way of knowing, on my own, which path is better. It makes me tense to think that I could be moving away from the build I actually want, without even knowing it. I like being able to make informed decisions instead.

[@KillahMate]

@@LamanKnight It's not even about knowing which path is 'better'. In any well-designed game (single-player at least, so you don't have to worry about the 'meta') both paths should be more or less equally 'good', equally powerful and feasible to play. But you don't know which one _you_ would like better and you can't know without consulting a wiki.

[@1BlueYoshi]

Terry Cavanagh, the designer of Dicey Dungeons, has mentioned that he decided to only use 6-sided dice in Dicey Dungeons because people tend to be able to better estimate probabilities when using them compared to using other dice. At the very least, it’s definitely easier to intuitively understand how those dice rolls effect gameplay then something like a weapon in an RPG having a +1.5% crit chance or something like that

[@KristofDE]

Great to see some Dicey Dungeons love, it's a great game!

[@tohitarmorclass0846]

I swear that game is rigged, I've put a fair number of hours into it and the sheer number of times probability screws me is just wild

[@dagroth123]

an ode to the D10, whos statistics are so plain they couldn't be confused. alternatively my favorite, the d12: a d6 with more options.

[@HebaruSan]

If you ever feel baffled by the Monty Hall problem, remember that it's a relatively small effect with just 3 doors. "You should always SWITCH" does NOT mean "switching will always lead to a WIN", but I think we sometimes hear it that way. With 3 doors, you'll still lose a lot by switching and win sometimes by not switching, it's just that a completely rational probability computer would recognize that switching always has an edge.

[@ICountFrom0]

I find most intersting the paradox with the 2 door version. Though I disagree with the montey hall problem being statistics. It's psycology too, as you have to assume an all knowing host that wouldn't reveal the key as door 3. They aint just blindly revealing door 3, they're revealing a GOAT, and your taking information from the position of that goat.

[@wisesquirrel4986]

I just don't understand what logical leap they are trying to make here. Whether or not the door is revealed is meaningless for two reasons. First, you have no way of knowing which of the 3 doors is correct, therefore neither has better odds of being the correct answer (god knows that the definition of "probability" is, but one thing is for sure, neither answer is more correct than the other when it comes to choosing the prize). Second, revealing that one of the doors you didn't pick was a dud only increase the odds of the door you picked being correct (so why would a person feel more nervous? They have no reason to feel more nervous. The only reason they feel more nervous is because the people organizing the 'Game Show' are trying to bluff you). I never really liked probability. I'm good with math, NOT probability. Probability never made sense. It probably works, but I'm incompatible with it. This 'Game Show' problem is smoke and mirrors to me.

[@Leivve]

Can someone explain how it improves your odds? Cause when you take a step back and examine all the parts, it was ALWAYS a 50/50, just disguised as a 1/3. You pick an option and then a wrong option is removed, so really you were always picking between a wrong and a right option; with a "fake" third option that never actually existed, but you were lead to think it does.

[@ragingfred]

Fun fact: choosing whether or not to switch at random will give you a 50/50 chance of winning. This seems like it should be identical to choosing a door at random of the two remaining.

[@ArchvileHunter]

@@Leivve Imagine your strategy is to stay, not switch. The only scenario where you win by staying, is if your first choice was correct. In a game with three doors, the chance that you initially picked the door with the prize behind it is 1/3. The host's actions after your pick do not change this probability at all. And since the sum of all probabilities is always 1, the door left closed by the host must have a 2/3 chance of containing the prize. When the two doors remain, there is indeed one correct, and one incorrect choice, but just because there are two choices does NOT mean the odds are 50/50.

[@leetri]

Xcom commits a big sin when it comes to RNG: it rounds numbers up without telling the player. I was playing the final level of the game with my veteran squad, when a Cyberdisc appeared out of the fog. It flies up to one of my injured soldiers and will instantly kill them as soon as it becomes the aliens' turn again, so I have to kill it in a single turn. I spend a lot of time thinking how to optimize my damage so I can deal with the other aliens also attacking me. According to the game, my sniper has a 100% chance to hit the Cyberdisc so I plan around that. I take my 100% chance shot... and it misses. A guaranteed 100% chance shot missed. Years later, I find out that the game rounds numbers up, so in actuality I had like a 99,99% chance to hit. Still sucks to miss a shot like that, but I wouldn't have been nearly as upset if it didn't present it as a guaranteed hit. They should've just rounded it down to 99% instead.

[@barryfraser831]

EU4 does the opposite with sieges, they always make the shown chance of success smaller than the actual chance, when its 50/50 it displays it as 49. That way it doesn't feel as bad when you fail twice in a row.

[@JarieSuicune]

Heh, a similar effect happened in the original Red/Green/Blue Pokemon games, where most attacks have at least a 1/256 chance of missing without any modifiers (other than Swift which always hits). Always frustrating to have that critical last-chance attack that shouldn't miss suddenly whiff.

[@nathanaelsallhageriksson1719]

@@barryfraser831 It's worth pointing out that this is because eu4 sieges work in incriments of 7. If you pass a 21% chance without a critical success it will increase to a 28%, and so on. That's why 50/50 is displayed as 49%, eu4 never shows 50% on a siege. Also interesstingly with eu4 is that the reason why you almost always win a siege before it has a 60% chance or higher is cause by the time you get to 56% you have had at least 4 chances at succeeding. Pretty neat.

[@UnreasonableOpinions]

It's reasonable to round out numbers given there are so few 99.XX chances in the game that it will almost never come up, and the odds that this scenario is also one significant to the outcome of the round is vanishingly small. Annoying an inevitable few players out of millions is less of a problem than introducing decimal places to the probability calculation; not because of issues with engine or display but simply because that doubles down on the far more common problem of players mentally rounding out numbers.

[@Daemonworks]

@@nathanaelsallhageriksson1719 That said, when luck isn't with you, the sieges can tun stupidly long - and those are the ones the player is going to remember, not the ones where they managed to land 7%. This is probably why CK3 uses a siege system where there's a base level of progress gained automatically with RNG mostly serving to shorten the timer. The final result is still driven by RNG, but the worst case scenario is capped, so you don't get the aggravation of those really bad runs, and they just feel better.

[@pseudopod]

Critical hits is something you need to be careful with as a designer. Diablo 3 for example lets you stack both bonuses to critical damage and critical hit change to the point that doing anything else is substantially suboptimal.

[@YourFunkiness]

Thanks for talking about probability in games. Next, can you explain how I missed a 100% shot in XCOM?

[@extrahistory]

[@Professionalmoneymakerr]

@@extrahistory what an amazing and cursed emote

[@ASpaceOstrich]

It was rounded up and wasn't actually 100%

[@aerowalker3]

That example with 1,000 doors really helped me understand the math. I’ve been told about the probabilities behind the Monty Hall problem, but it was never intuitive until that example. Thanks!

[@extrahistory]

Thanks! It's hard to get your mind around it without the pictures and context. We're really glad to have helped!

[@lancerguy3667]

Yeah, the 1000 door example is what I use when teaching the Monty Hall problem to children, as well. The issue with the 3-door example is that it deals with such a small sample size that it can seem completely arbitrary to someone hearing about it for the first time, since it directly contradicts what feels like common sense. This leaves them feeling like they don't really get 'why' it works like that. 1000 doors makes it a lot easier for a layman to appreciate the effect, even though the mathmatical principle is the same regardless.

[@TheSirmousavi]

We had a big debate about the Monty Hall with a friend who insisted the probabilities are 50/50. We finally coded the problem and simulated 10 million rounds. But even before running the simulation his confidence (that had not budged an inch during the day) was shaken just by seeing the code. Basically if you code it you have to: 1) randomly pick an item from the list [1,2,3] and store it as prize_door. 2) randomly pick an item from the list [1,2,3] and store it as player_choice. 3) remove prize_door and player_choice from the list [1,2,3] and then pick a random item from the remaining list and store it as opened_door. 4) if prize_door == player_choice increment a counter 5) run the whole thing a large number of time and at the end divide the counter by the number of runs. The result will converge to 1/3. Now notice that you can remove step 3 as it has no bearing on what we are counting in step 4. And the result will be the same. **The host's action is irrelevant to the chances of your initial choice being the winning choice. Nothing he does changes the chances of that door when you picked it out of 3 doors being the winning door. It cannot magically jump up to 50%.**

[@Stirdix]

@@TheSirmousavi I really like that explanation - very straightforward.

[@TheSirmousavi]

@@Stirdix 🙏

[@SharurFoF]

I think there's an important insight missing in your description of the Monty Hall problem. If I didn't already understand it, I don't know that I would have been convinced that your argument was valid. Your example with 1000 doors came close, but then I felt like it didn't follow through. It's not that you'd have the choice between 999 and the initial door (which doesn't feel as obviously analogous to the three door situation), it's that you would open 998 dud doors. Which should make the informational asymmetry between your initial pick and the other unopened door much clearer. The odds of you happening to pick the right door are much lower than the odds that it's just another dud door, and now all the other duds are eliminated and the remaining door is the prize door.

[@StompinPaul]

The Monty Hall problem is an interesting thing. I'm glad you mentioned in the beginning the need for it to be a constant that one of the other doors opens after a choice, that ends up being one of the stickier parts, and the nature of the whole thing can change. Supposedly at one point the man himself was told about the problem and asked about it, and his answer to whether someone should swap or not was that it actually came down to the host and contestant trying to read each other. He might choose to open the other door or not depending on what the chosen door actually contained, or maybe just to mess with the contestant. There are some games that can get great value from obfuscation. However, those tend to do so knowingly, so it's still very important to know what is and isn't important to distance from understanding, and how to do so.

[@googamp32]

I'm sorry. Did you say that I can win goats in this game?

[@LuckyFlanker13]

Lord I hate the monty hall problem

[@bobross547]

I would much prefer a goat .

[@quietone610]

One time, Monty paid a total of $1000 for three different costumes, and then--later--promised a contestant a prize "worth a thousand dollars." Which, of course, was those costumes.

[@extrahistory]

@@bobross547

[@extrahistory]

Math is complex but a goat... now that's a statement!

[@tomfeng5645]

There's two other on-average situations with crit in games: 1) Crit as an effect on damage formula In some games, dmg is calculated in lines with total dmg - defensive modifier, so crits are meant to penetrate high defense (relative to character dmg) better. Another way is like in pokemon, where crits ignore defensive buffs. This is generally meant as a design that limits the effectiveness of "walling" with defensive stats. 2) Crit as a stacking effect Many games have both crit chance and crit damage, which means stacking both effects multiply on themselves. For a simple example, imagine no base chance/effect at all - on average, +50% crit dmg and +50% crit chance is +25% dmg, while +100% crit dmg and +100% crit chance is +100% dmg! With careful control, this can favor crit for a pure-damage set of modifiers (favoring pure-dmg builds too) while hybrid builds would prefer a straight attack/dmg boost instead.

[@D3Vlicious]

Second can clearly be seen in Monster Hunter where the meta for the past few games has revolved around getting 100% crit chance (or Affinity, as it's known in game) and then maxing out crit damage as much as possible. No matter how many changes Capcom makes to the system, the fact that you can build for 100% crit means that crit is always going to be meta-something both the in-game tools to test builds for damage and player-made tools continue to verify.

[@misternoodle1236]

Fundamentally, I disagree with the solution most people think is correct for the Monty Hall Problem. I understand the math involved and have some background in Probability, however, the Monty Hall Problem is not one problem, but two. In the first, you basically are given a 3-sided die with words "heads," "tails," and "body." You are then told to guess what the roll will be. After that, you find out the call didn't matter. The 3 sided die is then taken away from you and you are given a coin to call heads or tails. This creates an entirely new and independent event from the first with even odds heads vs tails (let's just ignore the fact that most coins aren't EXACTLY 50/50 due to imbalanced weighting of each side and physics). This does not mean you always switch or always don't switch. What it means is you instead assess the situation and realize "I am now making a new decision between two things." Not recognizing that fact makes you run amok of the "Gambler's Fallacy" and you are making your decisions based on the wrong reasons. That is how I have always seen the problem.

[@misternoodle1236]

tl:dr - I think the Monty Hall Problem is a dumb problem on a fundamental level.

[@qwerty11111122]

It's a useful problem though. It has a very simple premise and execution, people's gut tells them to do one thing (staying) and the outcomes show a clear advantage in doing the opposite. This discrepancy is interesting.

[@garyermann]

Unfortunately, you are interpreting the problem incorrectly. One of the prerequisites of the Monty Hall Problem is that before any door is guessed, all three doors have an equal chance of being correct. What your example suggests is that the correct answer was only ever going to be between two doors, which simply isn't the problem.

[@Purble]

FINALLY found somebody in here who also thinks the Monty Hall problem is ridiculous

[@Jstar300]

@@garyermann But if they always reveal one of the duds from the doors you didn't pick then ask if you'd like to switch or stay, hasn't it now become a choice between two doors?

[@MrDylan2125]

Jokes on you. I'd take the revealed door because I want the goat.

[@bluedarkyugi]

Ironically my introduction to the monty hall problem was zero time dilemma with 10 doors instead of three

[@driver2469]

This channel has gotten me through so so many hard times please please keep making videos

[@brockmckelvey7327]

0:50 My years of watching VSauce have paid off. I'm gonna switch, Matt! (Here's hoping I get a goat. I think they're adorable and good company)

[@extrahistory]

We fully support your goat decision!

[@Nevict]

My kind of game: A game where the complexity does not come from the single elements it is made of, but from their interactions with eachother.

[@g3heathen209]

I remember the mtyhbusters episode on the "monty hall paradox". Worth a watch if you find it.

[@DarkBloodbane]

I have watched this Monty Hall problem in Brain Games years ago but never bother to think about it deeper cause random is random, you either get or don't. However, the last part about human perception is interesting cause this is what's special about this problem.

[@okaminodin4321]

If you want a game that's all about probability, the Mounty Hall problem and stuff like that, I highly recommend checking out Zero Escape: Zero Time Dilemma! (And the previous two games, to understand the context)

[@leakingamps2050]

I think the most intuitive way to view this for me is to make it a 50/50 choice between all the doors that got opened and your first choice, and the other choice. The new choice gets the whole 50% assigned to that half, but your first choice only gets the 50%/the number of doors on that side. This doesn't give accurate probabilities (you'd have to do some math to the two percentages to get those, since the open doors don't actually take up probability space), but it's pretty intuitive for me. I.e. if you have 6 doors, and you pick one, and 4 duds are revealed, you have a 50 chance for the new one (50 chance for a group of one door), and a 50/5 (50 chance for a group of 5 doors) = 10 chance of the first pick. Doing the math to make it actual percentages gives (50/60 or) an 84% chance that switching is the correct option, and a (10/60 or) a 16% chance that staying is correct.

[@TheKingOfApples100]

the thing about crits in video game is that often they are multipliers and stacking many multiplying effects gets more powerful fast. Ie you have 10 dmg and a +50% bonus, adding a +10 damage two times gives you 45 dmg, but adding two x2 effects, 10 x2=20 x2=40x1.5= 60 Total damage. Even tho adding 10's looks like you are getting the same thing looking at base dmg each +10 is double your starting damage adding ends up being worse

[@Lilith_Harbinger]

That's right. You'll see that in almost any game where "crit builds" are possible, that they are more optimal. When you can increase both your crit chance and crit damage, they are multiplicative and strictly better than upgrading your attack. The example Extra Credits discuss looks like it is only possible to improve the damage of critical hits. Even then, in most games enemies don't die in 1 hit and the argument about excess damage doesn't work.

[@KristofDE]

There's one thing that could potentially be happening under the hood in a video game which puts the maths of this into question: is that "switch or not" question asked always, or sometimes? And what are the conditions for that? If you add an extra step there, maybe you are offered to switch most often when you DID pick the right door? But if you balance this subtly enough, there's barely any way of tellling other than looking into the game's code.

[@geckoo9190]

So... if I understand well, there are three possibilities, by staying you can only pick possibility 1, but by switching you are simultaneously picking possibilities two and three, since one of them was already revealed... If I have ever said that I don't know how something works is now. I understand that you have 1/3 that your initial pick would be correct, then your probability rises to 50%, but by staying with your original pick, wouldn't it also rise to 50%?, I mean the whole difference between one possibility, another and even certainty is to know the outcomes. I think that the only way to know for sure is to run this experiment a lot of times to see if it makes any difference on the final outcome.

[@safe-keeper1042]

I've seen an example of the opening problem applied to bathrooms at a festival. Once you've tried a stall, should you go with it or keep opening stalls until you've found a better one? The answer is yes. They had to lay out all the various possibilities for me to be convinced (like in this vid) :p , but it makes so much sense once you get it.

[@JMcMillen]

The festival bathroom problem is an example of the 37% rule. With that, you evaluate and reject the first 37% of possibilities. After that you pick the first thing you get that's better than anything you have previously seen. It's actually been found to work pretty reliably with tons of different things when you have to accept/reject a series of 'things' one at a time.

[@slycoop1997]

I'm not sure that I understand. In the example with three doors, door number three is opened regardless of whether the player chooses to switch right? If the player chooses to stay, he has effectively opened doors one and three. If he chooses to switch, it's doors two and three. Two out of the three doors are opened in both circumstances. I understand that having more doors opened gives the player a better chance of winning, but I don't understand how switching accomplishes this.

[@RockLee0034]

It doesn't. I'm half convinced this video is itself an example of how humans are bad at math. The psychological dilemma is nothing more than a farce because no matter what you choose the show can ALWAYS show you a bad door. While the probability becomes 50% once one door is opened at the beginning you always had a 33% chance that will always be reduced to a 50% chance by the show itself.

[@technologicalMayhem]

I honestly really dislike it when people go for the 100 door variant of the problem as I feel it doesn't really explain what the choice of switching or not switching affects. Imagine that the doors were made out of glass and you could clearly see what is behind them. Of course now it's easy to just pick the prize. If you were to switch you would change to a goat instead and lose, so of course you wouldn't do that. But if you had to switch, what door would you pick? One with a goat behind it, since the other goat will be opened switching always results in you getting the prize. If you won't switch you need to select the winning door first to win. If you are planning on switching you need to pick a goat door to get the prize. So when the doors aren't see-through you want to use the switching strategy as it's more likely that you picked a goat than the prize.

[@anderskorsback4104]

The key to understanding the Monty Hall problem is that, at the point you get to change your choice, the two doors are not equivalent, the situation is not symmetric with regards to them. The one you first picked was, at the time, equivalent with the other two. After the elimination of one of the wrong doors though, the symmetry breaks down. The third door is a survivor of this elimination, unlike the first door. Which means that the third door is more likely to be the right one, since it always survives the elimination.

[@NinjaNanya]

Dear God this explanation was so much better than when I learned it in college.

[@SkywardShoe]

This was my second favorite goat-themes game show in the past year, right after the one in Psychonauts 2.

[@ASLTheatre]

This video is great! OG views of GameTheory would recognize this same concept being taught but I like this video better.

[@extrahistory]

Thanks for the compliment!

[@UnreasonableOpinions]

XCOM is a great example of this effect in action - even though the game has been unpacked completely and the exact RNG algorithm has been solved, even though the game offers strict RNG and the only exception is actually to BENEFIT the player on lower than the hard difficulties, people will die on the hill that the game is cheating. It's simply that most players mentally round out percentages - everyone has an innate point where they round out a percentage to 'yes', 'even', or 'no'. 80% feels close enough to 'yes' that even though people accept it missing ,a few misses in a row feels like a cheat even though it's statistically very likely over the course of a campaign. The role of perception is clear - because it's your small, elite squad versus a large, weaker group of aliens in an action-economy game, you the player have a small number of high percentage chances to hit or kill, versus the enemy's large number of low percentage chances to hit or kill. When the difference between perceived odds and actual odds comes up, it will usually be the player missing and the enemy hitting, and because these are very memorable, they take on outsize significance.

[@christianhumer3084]

2:36 While it is correct, that you have 2/3 probability when switching, you cant increase your actual chance of winning (unless the open door is the prize, we can ignore that, since you would be either blind or an idiot if you dont choose the open prize). When there is leftover behind the open door, you have a 50/50 Chance to choose the correct door, since you cant say what is behind the other 2 doors for sure from the open one. 5:15 Paradox: HAHAHAHA.....Ha.... *NO*

[@HexAF8B9]

That is what i was thinking too. If I am always shown the Bad Doors and never if I missed the Prize, how does this Matter if I want to win. I don't want to open 999 Doors, I want to open the winning door.

[@saxor96]

@@HexAF8B9 You don't have to open 999 doors. A better explanation would be... You choose one out of 100, that's a 1% chance of winning. The host then opens 998 doors, all of them duds. The prize doesn't move. Will you keep that door, or switch?

[@HexAF8B9]

@@saxor96 so the Doors the Host will open could also contain the Prize? Only then I think the Percents adds up. Since I have won nothing if he 100% will not reveal the Prize. This would not change my Odds.

[@saxor96]

@@HexAF8B9 no, the host always opens a door without prize. So the choice always ends up in this: Option 1: You got it right the first time. In a 3 door contest, that would be only a 33% chance. In this scenario, if you switch, you lose. (33% of winning by staying) Option 2: You chose any of the wrong doors. In a 3 door problem, this is 66% likely to happen. If you switch in this scenario, you're guaranteed to win (so, 66% of winning by switching). The thing you're letting to chance is guessing in which one of the two options you're in. You have a 66% chance of getting it wrong the first time, so the most plausible scenario suggests that you switch so you rectify your wrong, original selection.

[@HexAF8B9]

@@saxor96 I do not get how my Odds to win get better if the Additional Doors I get for Free are bad anyway. I get the Percent Stuff of 3 Doors each 33% and 2 Represent 66%. But if is is guaranteed that the right door will never be the one I get for free before choosing to switch, at which Points do my Odds increase that the Door I have chosen is the winning Door. Sure, I had 33% of picking the right one, but how does having a bad one eliminated raise my odds of winning? At the End of the Day the Door that I could switch too did not have 66% of being the Winning Door at the Time of choosing. I do not get 2 Doors by switching if the one I additionally get as Information is not the Winner for certain. In either Scenario I will see 66% of the Contents behind the Doors. Switching in a 50/50 Scenario which will be created after I have chosen does not benefit me, or does it? I get hung up on the 100% a Bad Door gets eliminated. How does that impact my Odds of being in the right Scenario if I switch?

[@noyes4968]

I first discovered this channel by history know im even more interested in another sections great video as usual

[@extrahistory]

Thank you!!!

[@TheCreepypro]

always love hearing about the monty hall problem and probability

[@blaster915]

Was waiting for this subject to be covered! Never know how to get it right. Thanks for this!

[@extrahistory]

Thank you for watching!

[@impcirca1988]

When you choose to "stick" you are not sticking with the original probabilities - you're making a fresh choice between 1 of 2 doors, regardless of which you choose. It's not staying with the same decision made previously and retaining the previous chance of success, it is a new decision with new probabilities either way.

[@amwoodco3049]

Not quite. I'll tell you why, but we have to change the game a little using a die. I roll a d3 and ask you to guess what I rolled. You guess one. I tell you that I didn't roll a three. Stop here! What, if anything, at this point, changes the odds that a d3 rolled a one? Nothing, that's what. The odds are still 1 in 3 to roll a one, which is your guess. All I did is say it isn't a three. The odds don't change to 1 in 2 that the die rolled a one, they remain 1 in 3, so it's a 2 in 3 chance that it's not the value you initially guessed. That's where people mess up the Monty Hall paradox; they incorrectly assume the second choice is equal because it's the last choice, but the reveal of the dud has stacked the odds.

[@impcirca1988]

@@amwoodco3049 after you say you didn't roll a three, you ask me to stick with one or change to two. The odds that I rolled a 1 are the same. Yes. The odds that I choose the right option out of the remaining two is 1/2. Because I am choosing one of two options. One over two is the same as 50%. After the reveal, the second choice is made out of two options, not three. The relevant percentage isn't the chance of rolling a 1. The relevant percentage is the chance of choosing correctly after the reveal.

[@sillum]

It should be noted that the door example only works if you know that the door that is opened is intended to always be a dud. If the door opened is random, then there is 2/3 chance that your are in a world where you picked the correct option, and a 1/3 chance where you exist in a world where you didn't pick the correct option. This makes the probability work back to be 50:50

[@RickJaeger]

In the parameters of the Monty Hall problem, it is always a dud that is opened, yes. Wouldn't be much fun as a game show if they opened the door and revealed the prize you could no longer have.

[@Alphasoldier]

If the correct door can be randomly opened, you won't have a chance/reason to switch, so this is entirely a moot point.

[@arandomdumbass8542]

Tf2 crits are always a nice thing when they come in clutch

[@garyermann]

To me, the easiest way to hammer home to someone how the Monty Hall problem works is to use a deck of cards. Tell them to randomly pick a single card and if it's Ace of Spades, they win. Then, take the rest of the cards and start eliminating every single card in the deck that isn't the Ace of Spades until you have one card remaining and ask them if they want to keep their card, or switch with the one in your hand. People realize very quickly how obvious it is that the last card in your hand is almost definitely the Ace of Spades. It has the benefit of taking the same concept as the Monty Hall problem to an extreme, while still using something people readily have available and understand intuitively.

[@ASpaceOstrich]

I still don't see it. If they picked the Ace initially your actions would be exactly the same as if they hadn't. Eliminating more wrong options doesn't make it make more sense to me. I don't see how the initial choice is affecting this second choice.

[@garyermann]

​@@ASpaceOstrich Think of it like a game between two people where one person gets a single card, and the other person gets to keep the entire rest of the deck. Who are you going to guess will win? Naturally, the person with the deck is almost always going to win in that scenario. Better yet, if you have a deck available, try it out yourself. It takes maybe a minute each time, and you'll see how the probability works within the first one or two tries.

[@AnonPer]

I am the ANTITHESIS of this example: My school class (I believe it was middle-school, but it has been a long time) presented this, and AFTER switching when one of the bad doors was revealed, I found that I had originally picked the correct door. I hated myself all day afterwards because of this.

[@samuelmason2703]

I've used the Monty Hall problem as a hook for a presentation a couple of times, always a fun way to mess with people's intuitions

[@antenna_prolly]

Close, but not quite. In your scenario, the 1/3 chance only applies at the outset, because your guess was *one of three, two of which* are wrong; when a different option is removed, it's *no longer* the same problem or probability, because _now_ your _original guess_ is *one of two, one of which* is wrong. But if you're talking about a scenario in which a host or code can manipulate the results, then you might be right, but it's no longer a straight math problem, it's a guide to Las Vegas.

[@antenna_prolly]

Another way to explain it: if all three options are equal from the start, and if according to you, opening door 2 after door 3 is revealed is like opening both doors 2 & 3, then re-choosing door 1 after door 3 is revealed is like opening both doors 1 & 3. Same addition/multiplication of odds, same probability of outcomes. Be logically consistent.

[@rosscalhoun3389]

The Monty Haul problem has been explained to me before, and it still makes no sense to me.

[@petertrudelljr]

goddamned monty haul problem. I STILL can't wrap my head around it.

[@timothyjaydyning3247]

A gygaxian dungeon is like the world's most fucked up game show. Behind door number one: INSTANT DEATH! Behind door number 2: A magic crown! Behind door number 3: ten pounds of sugar being guarded by six giant KILLER BEES!

[@bronsoncarder2491]

I wonder if the probability isn't actually slightly better than 66.6%, because they don't randomly open a door, they specifically open one of the wrong doors. Well, I guess that's not a change of probability though, huh? That's just a change in our ability to measure the probability. We have more information, and can adjust our assumptions... This was a really interesting video, made me think. Love it. lol

[@lshanksy1]

Ngl I've always disliked crits in games, I really love games that have stagger mechanics where by playing well you make the enemy vulnerable to follow up damage. Or to a lesser extent simply targeting thier weakness, eg. holy damage against undead.

[@JohnnyEscopeta]

- EC: You should always switch - Me: Why? - EC: Idk man

[@tylerward4386]

The 3 door problem is a worthless question. Both probability equations are identical if you redo the first equation with the new information (1 door being eliminated). Meaning switch or not your odds are identical. If you go based on common belief then all you did was make your original equation wrong since you didn't correctly account for the entire equation (lacking an accurate variable) And just to stave off some of the useless counterpoints. I'm assuming that you are given another free choice (which you are) and can freely choose the same door again (instead of only being able to choose the other, which you can). This is just to keep the conceptual statistics to a minimum.

[@jocaleb0236]

Ah crockets, my favorite

[@AbyssalManta]

If critical hits don't entirely shake up the direction a battle is going (both ways!), you're doing it wrong. The Persona series got that.

[@brianphelps2415]

I knew the right answer, having seen the problem in a couple other places but I'll be honest it never actually made sense to me until watching this video. Thanks!

[@legocitykilldozer]

So dexterity/agility builds are always the smart choice

[@Lack_Of_Interest]

"Nat 20 baby!" -Marcy Wu

[@alli_mode]

The CURRENT "let's make a deal" doesn't let you switch anymore unless it's cash for some reason. But, all of the prizes are pretty great now. (Unless it's gym equipment. It's nice gym equipment, but who wants gym equipment?)

[@fence03]

This sounds like a Ted-Ed explanation of a riddle

[@Ggdivhjkjl]

Wonder what the game mentioned at the start was?

[@NoName-hg6cc]

I think there is a movie where Kevin Spacey act as a professor of an university and a brilliant student answer correctly

[@joshdavis3743]

21

[@shawnheatherly]

Because probability can feel exciting right until you lose.

[@MM126.90]

0:50 switch, Mythbusters tested it and switching had a 70% higher chance to win

[@jameshalladay9149]

the remixed intro is good I'll say that.

[@neepungupta5289]

I know this a strange probability concept for folks. I recommend for anyone who finds this hard to believe to look up Monty hall simulation programs, (or write your own if you want, it's not that hard), and see the results of 100 trials either ways. The numbers shake out practically as well, with changing having about a 66% success rate, and keeping having 33%.

[@majorfallacy5926]

or just think it through practically. There's a 100% chance the prize is in one of the 2 closed door, but the one you picked has a 1/3 chance and the open one is obviously a 0. So the other one must have a 2/3 chance unless they changed how math works

[@Ashebrethafe]

@Paul Calixte Since there are two duds, the host can _always_ reveal one you didn't pick, so doing so doesn't actually reveal any new information -- assuming the rules of the game say he _has_ to do so. Otherwise, it becomes a mind game -- you have to guess whether he's letting you switch because you picked the right door, or to make you think you did. The actual show _did_ always end with a contestant picking one of three mystery doors (which was called the "Big Deal" because it included the most valuable prize, the dollar value of which the contestant would be told in advance), but two of the doors were good (one was just better than the other), and they were never given the chance to switch -- Monty Hall just opened all the doors one at a time, saving the chosen one for last. (IIRC, most of the smaller deals _were_ mind games -- Monty would give the contestant a box and then offer to trade it for a door, or make several offers to buy it back from them.)

[@ArchvileHunter]

@Paul Calixte The chance that your initial pick is the prize is 1/3. The chance that the prize is in one of the other two is 2/3. The host reveals that there is a goat in one of the other two doors. But you already know this, since regardless of your first pick, there will always be at least one goat among the other two choices. Remember that if you stay, the only scenario where you win is if your initial pick out of three was correct. The host revealing a goat changes nothing.

[@majorfallacy5926]

@Paul Calixte disagreeing with a rigorously proven mathematical solution as if it was a random opinion is the most hubristic thing i've read in a while

[@postapocalypticnewsradio]

PANR has tuned in.

[@extrahistory]

Thanks so much PANR!

[@trevinbeattie4888]

This explanation doesn’t make sense to me. Opening one of the “dud” doors doesn’t increase the probability you would win by switching; it just reduces the possibilities. In the first example with 3 doors you just go from a 1 in 3 chance of being right to 1 in 2. If instead the game always opened door #3 regardless of whether it was a winning or dud door, then if it did reveal the prize you would instantly know you had a 0 in 2 chance (assuming you couldn’t switch to the now-open door) and if it didn’t you’re still left with a 1 in 2 chance.

[@technologicalMayhem]

They don't always open the third door. They always open a door with a dud behind it.

[@mercyloke7720]

6:22 sounds like how gambling makes people feel

[@McRusen]

The example with 1000 doors gets actually even more intuitive when you use the original rules. You pick one of the doors, 998 get revealed. Do you want to change?

[@donkosaurus]

the critical argument made partway through the video does not account for any incoming damage, or any kind of health regeneration mechanics, and specifically the boss scenario(most players will base their characters around these) in which large amounts of damage are required a direct dps mean average does not correlate with the experience of anyone who plays games

[@strikereureka345]

I literally learned the opening segment last year in science weird to think you find these random things everywhere

[@jeffbenton6183]

The joke's on Monty Hall... I actually *wanted* the goat!

[@HealyHQ]

Ha, this is literally the opening scene of the movie "21." I already knew the correct choice. B]

[@AdamSchadow]

Players are not stupid some of them will simply crunch the numbers and use their brains to crush the others forcing the rest to copy them no matter what your goal initially was.

[@JarieSuicune]

That depends on the game, it's not always an option. Sometimes some dude with a Pachirisu comes along and just wrecks your ultimate tier plan anyways.

[@joshdavis3743]

In Soviet Russia Monty Hall problem is zero percent chance to win. Only the gulag behind both doors.

[@ChroniclerC]

Oh my gods, thank you for explaining the Monty Hall paradox better! I was running off the Mythbusters explanation for YEARS, and this is *so much better*!!! (And more accurate.)

[@colinbrown248]

Make more videos like this

[@SuperParkourio]

5e D&D has an issue with probability in its challenge rating rules. Assuming a dragon that starts combat using its fire breath and has a 1/3 chance to do it again on each subsequent turn, the developers state that the challenge rating of the dragon (a measure of how difficult the monster is) should assume that the dragon doesn't recover the breath until its 4th turn and it's average damage across the first 3 rounds is therefore (breath + multiattack + multiattack) / 3. But that's actually the average damage of each round AFTER the first. The true average damage is (breath + turn 2 average + turn 3 average) / 3. Every monster with a rechargeable ability punches a little above its weight because of this probability mistake.

[@Ryu_D]

Thank you for the video.

[@I_Hate_YouTube.]

4:25 Tf2's random crits are broken. The more you kill, the more likley you are to get a random crit. It should be in the other way to balance it out. So if you are new, you have am evel lesser chance to get a kill. One of the many things that are killing tf2.

[@leetri]

Nothing is killing TF2, it's literally doing better than ever before. Back in the day the average was about 40-50 000 players per month, nowadays it's between 80-100 000 average players.

[@nickspencerfishingrodd2383]

For how long it stayed there, idk how much is "killing it". But that's probably cuz people got used to it over the years. If something like that would have been added to a modern game, it most likely wouldn't last long

[@andrewtormanen]

In the initial example you gave, you are essentially choosing either doors 1 and 3 or doors 2 and 3 so it's still a 50 50 chance of success. Especially since you as the player still don't have any new information about doors 1 and 2. In other words, the scenario at 3:08 is different from the original one and it's misleading to treat them as identical.

[@joshdavis3743]

I don't think you understand the Monty Hall problem then. I am assuming by initial example you mean I pick door one, door three gets opened and it is a dud and then I get to switch or not. If I switch it is roughly 66.6% that I win. That is why they call it the Monty Hall Problem, it appears 50/50 to most people, but is actually 66.6%. The example at 3:08 is also 66.6%.

[@OpaqueDreamer]

I have great respect for Marilyn Vos Savant, the genius (highest IQ ever recorded) who answered the Monty Hall problem. But this logic, it's just wrong, and it's not even that difficult. The argument hinges on the notion that your probability of success (1/3) does not change once one door is opened. But it obviously does. That's new information, which requires a new equation.

[@amwoodco3049]

@Jason Allow me to point out your error. Think of it as me rolling a d3 in secret and guessing what I rolled. You guess one. I tell you it's not three. Stop here! How does that information change the odds I rolled a one on a d3? The answer is it doesn't. That is the catch to the Monty Hall paradox, the new information doesn't change the odds of the initial choice, but stacks the odds in the second choice. Remember, the initial conditions set for all this is includes equal propability distribution and you aren't being tricked. Mythbusters did an episode on this and brute forced the problem, showing that switching results in twice as many wins. They, however, also went into the psychology of the problem: most people don't switch.

[@the6ofdiamonds]

I've never understood the idea of "wasted damage". If you deal more damage than needed, you rarely lose... anything. Combine that with the idea of a "crit build", and you basically flip the chances of dealing 100 vs 200 damage. If you can make it so that you consistently (let's say 75% on the stat page) deal 2x damage (200, in this case), how is that "worse" than consistently dealing 120 damage? Things with less than 100 health don't care about the surplus damage, but things with 150 do care, since the "inferior" crit build does 200 damage. Unless you limit the ability to land a crit (specific ammo requirement, special combo requirement, etc), I see no reason not to just go for overkill.

[@Yurmanator9000]

Not wasting damage is a term that comes from when you need to minimize the time in a fight. The more time a fight goers on, the more chances you have to take damage. For example, if you fight an enemy with 10hp, you need to deal 5-9 damage per hit to win in 2 hits. If your extra damage (6-9 value) comes from NOT investing in other more useful stats, that's statistically "wasted damage". Of course this scenario converges as time in a fight is lengthened, and amount of turns taken to kill your enemy is less important than simply winning, so stat investments aren't always an easily solvable problem. Crit is a hindrance when it comes at the cost of consistency, especially in a complicated scenario like most video games. But, just to present an example: Imagine a turn based RPG: you go, then your opponent goes, etc. You have 10hp and so does your opponent. Let's say you have no crit chance and deal 5 dmg per hit. your opponent deals 5 damage as well. On your first turn, you deal 5, opponent deals 5, second turn, you deal 5, opponent is dead. This took 2 turns. Now imagine you deal 3 damage but have 33% crit chance (1/3 attacks), whcih deal 3x damage each (9). Your statistical average per hit is still 5. Fight goes: you deal 3 (or 9), opponent deals 5, you deal 3 (or 9), and if the opponent is not dead (because you failed 2 crit rolls) deals 5, and you are dead instead Chance of failing 2 crit rolls is 0.66*0.66 (chance of NOT critting squared) is 4/9 or 44% In this hypothetical scenario, you die almost half of the time rather than winning, but you win in 2 hits either way. Of course like the first example, a larger hp pool smooths out the equations and dps is equivalent, and the only difference is variance in how much damage you take (or lose in the unlikely event you fail every crit roll)

[@the6ofdiamonds]

​@@Yurmanator9000 "Not wasting damage is a term that comes from when you need to minimize the time in a fight." If I take 1 turn per enemy, the fight lasts as many turns as there are enemies, it's irrelevant how much overkill is involved. If I can use an attack that hits multiple enemies and still have overkill, that's even better. Winning the fight is the end goal, the faster I achieve that, the better. The example EC gave makes it sound like players should narrow the margin as much as possible. If I'm dealing 200 to a single enemy with 5hp, I don't care about the leftovers, I do care the enemy is gone. if I'm one-shotting something, I'm literally spending the minimum amount of time possible against one enemy. I don't think "oh, this goblin has low health, let me bring out a weaker weapon", I just hit the goblin and move on. I'm not gonna fret over 195pts of damage, unless the game has some way of limiting my ability to do 200 damage. "Crit is a hindrance when it comes at the cost of consistency, especially in a complicated scenario like most video games." Technically, yes, practically, I disagree. I can probably sit here and list games that are just broken if you invest in crit (Skyrim and Oblivion come to mind, mostly because it's not random if you're sneaking, but even KOTOR2 cracks wide open, and that's basically 3ed D&D). It's like saying "don't go for a headshot" in an FPS, because of how small a target it is. Technically center mass gets you more hits, but the right combo trivializes the combat, if you can land the headshot.

[@Yurmanator9000]

@@the6ofdiamonds I agree on all points. Like i said, the effects are less extreme both ways as health pools get larger, so for certain games where the numerical advantage of crit is just SO MUCH better than consistant damage, yeah it's correct to go crit. Certain games it's correct, and certain games it isn't . I'm just explaining an example of how these two concepts came to be. Wasted damage is in relation to total stat investments, as a concept of min-maxing encounters. Crit being a detriment is again, only applicable if the game in question rewards consistancy over big numbers. It all depends on how the math of the systems work out, but you can't say that crit investment has NO downsides, even in most cases that's true.

[@elbuenohombre]

Never mind, I'm teaching Father the math. Whatever, Rosa.

[@biblicallyaccuratecockroach]

Mythbusters tested this conundrum back in their day, that was a fascinating episode. Humans are notoriously bad at probability, and once we make a decision we tend to get attached to that decision, and in such a stressful environment like a game show we're even less reliable. It takes conscious effort to step back and think. Always switch!

[@Daemonworks]

A thing that might help with Monty Haul is that the reveal and switch question don't actually do what most of us naturally think they do. So you got three doors, and one has a prize, and it's not moving. To be very clear: there's a total of three possible ways for things to be set up, so each door has a one in three chance on it's own. You pick a door, and call that door Set A. One door, 1/3 chance the prize is here. Call all the other doors Set B. There's a 2/3 chance that the prize is behind one of the two doors here. One losing door in Set B is revealed. The reveal doesn't change anything about the odds between Set A and Set B. All three locations of the prize are still possible, it's just that there's two different ways to have one prize door and one loosing door in Set B. If the prize is anywhere in there, it's the one door that's left. In terms of probability, it's as though you merged the original odds of the doors in the set. Then you're asked to either stay with your original choice or switch. And while you seem to be choosing between two doors, you're /actually/ choosing between your single door and a set of the rest of the doors. So a 2/3 chance of getting it if you switch. Also... drawings, charts, simulations and the like will help, because it's easier to see what's going down if you make it concrete.

[@danielkane8568]

The problem with the Monty Hall problem is that the actual probabilities depend on the behavior of the host. The intuition that switching is a 50-50 shot is CORRECT if the host always opens a random unchosen door (and thus sometimes reveals the prize). Even saying that the host always opens an unchosen door without a prize doesn't completely pin things down. If for example you knew that the host always opened the LOWEST NUMBERED unchosen door without a prize things would be different: If you chose 1 and the host revealed 2, there would be a 50% chance that the prize is in door 1 and a 50% chance that the prize was in door 3. If you chose 1 and the host revealed 3, there would now be a 100% chance that the prize was in door 2.

[@saxor96]

You are adding cases that aren't in the original problem. The behavior of the host (in the mathematical problem) is always open one random door without a prize. Or to put it in more perspective, it reduces the number of choices to "the first choice with 1/3" or "the other, only door". Because the prize doesn't move either at any point, it's always in its initial position.

[@codergeek42]

That was one of the clearest explanations of the Monty Hall problem I've ever seen. Great episode! 💙👍

[@kaelhound]

Just gotta say, the goats are a *great* prize

[@N1GHTw4tching]

love all youre subjects

[@extrahistory]

[@nimbulan2020]

Games often don't even use true RNG, they will increase probabilities after repeated failures since long failure streaks will give the player the impression of a lower than indicated chance of success.

[@cornRipper]

Anyone else already familiar with the Monty Hall Problem from that counting cards movie, “21”? I did, anytime I see goats behind doors now I always know what is about to be explained. 😂😂😂❤ Love you guys for going over this is also another easy way to learn such a fun idea of probability.

[@quietone610]

*hits that switch*

[@crapshoot]

"you should always switch" but what if I really like goats?

[@OmegaX9]

I’m sorry, but I’ve yet to hear a version of the core Monty Hall problem where this makes sense. By core version, I mean the one where you go from picking a door out of 3 to choosing whether or not to switch your choice when your odds become 1 in 2. Because while you will have a 50% chance of guessing correctly by switching doors after the first dud is revealed, you also have improved from a 33% to a 50% chance on the door you originally chose. The only thing that’s actually changed by opening the door is that the division of odds between the doors goes from 33/33/33 to 50/50/0. The bit about choosing 1 door out of a thousand and then being given the option to take all of the other 998 doors (which is what that example sounded like to me) doesn’t work because you still have better odds with the majority.

[@technologicalMayhem]

I think the biggest problem with most explanations I find is that the the choice to switch doesn't introduce a new roll of the dice but just changes the result by "inverting". If you picked the prize initially and switch you land on a dud. If you picked a dud initially and switch you land on the prize door. And since there are two duds and one prize you should always switch since it's more likely than not that you landed on a dud. If you don't switch you have a 1/3 chance of winning because you need to land on the prize to win. If you plan on switching, the chance is 2/3 that you land on a dud and get to the prize by switching.

[@OmegaX9]

@@technologicalMayhem even that I don’t fully agree with. When you switch you don’t get to keep your original choice, so you still have a 33% chance of being correct. It’s more of a psychological trick than anything statistical, since you know how low your chances are the idea of choosing another answer seems like the better option. It’s like taking a multiple choice test where you don’t know the material: you might initially pick a choice at random and change it later, only to learn that your first choice was the correct one. In something like the Monty Hall problem where you don’t have any prior knowledge or any way to alter probability (like knowing which door is most or least likely to hold the prize), every choice is valid and has the same chance of success no matter what you do.

[@technologicalMayhem]

​@@OmegaX9 The thing is though, when deciding to switch or not, there is nothing about it that has to do with chance. Imagine for a moment that the doors are made of glass and you can clearly see what is behind them. Now it's obvious what door to pick and when the host opens one of the doors with the goats you just decline to do the switch and win. But what if you had to switch? Well in that scenario you would want to pick one of the doors with the goats. Since the host will always open the other door with goat you haven't picked, when asked if you want to switch or not you will always remain with a goat door and the prize door. Since you picked a goat initially you now switch to the prize and win. In more simple terms, if don't switch and pick a goat, we get the goat. If pick the prize, we get the prize. If we do decide to switch: when we pick a goat, we get the prize. If we pick the prize, we get a goat. Since we know there are two doors with goats behind them and only one with the prize, we know that when we pick one of the doors it's most likely one with a goat behind it. And that's why we would always want to switch. Because if it's a goat the other goat door will be opened and we will switch to the prize. There is always the chance that we pick the prize and switch away to a goat, but the chance of picking a goat and switching to the prize is 2/3 which is better than 1/3 without switching.

[@OmegaX9]

@@technologicalMayhem okay, let me counter that hypothetical with a different hypothetical. What if we were asked to switch before the third door was opened? Would you still do it then? Because nothing really has changed, your chances of the door you picked being the correct one or not it would only be proven or disproven after all other doors were opened. It's changing the odds to make you THINK you're better off choosing the other door because your odds of choosing the right door have now risen, but people tend to forget that the odds that they were right in the first place have risen the exact same amount.

[@technologicalMayhem]

@@OmegaX9 How would switching before the door with a goat has been opened even work? Could I chose between both unopened ones? If so that would not give any higher chances of winning as that effectively if like picking again. Remember that the whole thing with opening a door and allowing changes the outcomes, not the chances. It's just that when you are planning on switching you need to pick one of the goat doors to win, of which there are two. Whilst when you don't switch you need to pick the prize door, of which there is only one.

[@kotzakozidiskosmas7206]

Can you do a video about the greek indepedence war?

[@nathanrobbins7502]

I think a better way to understand the Monty Hall problem using the 1000 door analogy is this: Imagine that you pick a door, and then 998 doors are revealed to be duds, leaving only your original choice and 1 other unopened door. Clearly, you would want to switch to the remaining unopened door rather than keep your original choice.

[@ASpaceOstrich]

Why though? Why does the extra step with the dudes result in a different probability to if they just said "pick one of these two doors?"

[@GrifoStelle]

How the montyhall problem was proven true is interesting. It was believed it didn't make a difference. It was showcased in some "did you know" section in a newspaper claiming the chances were 50/50 but a woman wrote a letter to the editor that this wasn't the case. Others wrote back to her through the newspaper arguing back with women stereotypes. She slapped down the mathematical proof in her next reply and a university (I believe) got involved to back her up and prove her right. It was a pretty cool read. Hope I remmeber the story right! It's been a while

[@danielmorton1606]

I mean it's on wikipedia. It was not showcased, it was a regular column "Ask Marilyn", Marilyn being a statistician. She did receive lots of critisms from the public including PhDs. I think a lot came down to how the problem was posed.

[@AidanRatnage]

I never got what was so bad about winning a goat.