I tried to fix my game and made it worse

[@bd648]

Man this feels like the monty hall problem but backwards. Intuiting that there is actually a different probabilty when there isnt. Fun explanation

[@bencebarabas3555]

For anyone who doesn't know the monty hall problem: It's a puzzle where the player is presented with 3 doors, one of them having a prize behind, the other 2 having nothing. The players is asked to pick a door, then after they did, a door that a; has no prize and b; wasn't picked by the player, is revealed to have no prize. Then the player is asked whether they would like to change their pick to the other unrevealed door. Now probably you would think "why the hell should that make a difference regarding my chances to get the prize, it's 50/50" but it actually isn't. If you picked the correct door initially, then changing your pick would (obviously) mean *you don't get the prize.* You initially choosing the correct door is *1/3.* If you picked a wrong door initially, then changing your pick would (since you chose 1 wrong door and the other is already opened, leaving the correct door as the other option) mean *you get the prize.* You initially choosing a wrong door is *2/3.*

[@tristanridley1601]

If you know the total number of tiles, then just like Monty Hall you do gain information near the end, and get better probabilities than chance.

[@jakethebeest2376]

@@tristanridley1601 You don't though... The proof in the second half of the video states that regardless of what strategy you use you have the same chances of winning if you used a random one. Which you don't gain any information. for monty hill the reason why you gain information in the first place is because the contest owner gave information by always opening the incorrect door and allowing you to switch. with this problem the only information you "gain" is if you are guaranteed to lose or not (or if you already lost), but the random player doesn't get this information but will still have the same chances of losing or not.

[@ValkyRiver]

If all non-winning outcomes are considered equivalent, then there is no probability difference between having the choice or not. But that’s only for the “winning” case. Since the game ends as soon as FOX is immediately spelled, this means if the game does in a loss, it could be anywhere between the 3rd and 16th tile drawn, so that could be a score from 3 to 16 (and winning would be a number 17 or more). Which one feels like the better outcome - losing immediately on the third tile, filling up less than 1/3 of the board, or losing on the last tile? IF (that’s an assumption) you consider the next best outcomes after winning to be filling as much of the board as possible, then having the choice actually makes a difference. For example, if you had the choice, you can guarantee a score of at least 9 by placing the 9 tiles in a formation that doesn’t contain a three-in-a-line. This is a slightly different version of the game where you have the incentive to try to place as many tiles, instead of going all in on high roller.

[@LoreSolver4]

I knew how it worked because of mario party, weirdly enough. There is a series of vidoes called "identifying luck" that goes in depth on each mario party game, and there is a minigame in mario party 2 called bowser's big blast that is random, and in it he explains that even though there is an order of when each player goes to pick a switch, the position doesn't matter since the chances are the same.

[@chizzicle]

so technnically as the win rate is the same regardless of the strategy, high roller is the best strategy as you go through games faster and thus in theory land on a winning game faster on average too timewise

[@Thavleifrim]

for a machine yes, for a human random is going to be faster than thinking about it in any form.

[@_puzzled]

you ever watch those sorting algorithm vids? this reminds me of that. sometimes a strategy is better because it uses less cycles while other times a strategy is better because the cycles go by faster.

[@qwerty_qwerty]

@@Thavleifrim not really

[@asheep7797]

⁠@@Thavleifrim Just to add on to what you were saying, random here just means not changing your placements based on what letter has been placed. So, it can be said that this is equivalent to just putting the tiles on the board, row by row. Then, you can eliminate any thinking.

[@KasabianFan44]

@@Thavleifrim I would argue that the risky strategy would be faster for a human to start with, when the risky squares are obvious. At some point, when you have to start properly thinking about the risk, that’s when random would be faster

[@marimbaguy715]

6:41 As a Numberphile subscriber I approve of the brown paper

[@almightyhydra]

4:20 you made a vertical fox but then kept going

[@Systox25]

Proof he is human

[@personwhoplaysvideogames351]

i was just about to comment that

[@lilmac194]

I saw that too

[@tgwnn]

your chances are much higher than 2.3% if you have a >0% chance of missing foxes

[@Mac_Omegaly]

As a high roller centerist player, I move to all the center spots first thereby ending early in many cases but then playing safe with the remaining 6 moves.

[@Mac_Omegaly]

There is a question about how to make the game more interesting. •Add more tiles: This would drastically change the static percentage of possible choices... I think? 😅 Having two tiles you didn't have to place to win would allow for more player agency. A. Add either two more O's or one more F&X. (Increases the chances of danger spaces but gives the player less calculated risk. As there's almost always going to be danger somewhere.) B. Have two wild "Non Fox" letters like "B" (box) (fob) (Bob) with bonus points for making the bonus word. Bonus points could reduce the number of previous losses or there could be some other methods for scoring. •Allow a "once per game" hint for the letter of the next tile you place. With bonus points for winning without using the hint. •Bonus round: add either to the top & bottom, or left and right 8 more empty tiles to fill up to not find the fox, with each additional tile in bonus round adding to the final score.

[@radical_rat]

A version with actual strategy: Board is empty to start with, and you can fill spaces in any order you choose. With each tile placed, alternate between looking at it before placement and placing it face down without looking at all. Reveal all tiles once the board is full. The idea is that you use the known tiles to try and create the board with the highest chance of success win the hidden tiles are revealed, but you can still never be 100% sure.

[@ValkyRiver]

Another version: just don’t consider all losing states as equivalent. If all non-winning outcomes are considered equivalent, then there is no probability difference between having the choice or not. But that’s only for the “winning” case. Since the game ends as soon as FOX is immediately spelled, this means if the game does in a loss, it could be anywhere between the 3rd and 16th tile drawn, so that could be a score from 3 to 16 (and winning would be a number 17 or more). Which one feels like the better outcome - losing immediately on the third tile, filling up less than 1/3 of the board, or losing on the last tile? IF (that’s an assumption) you consider the next best outcomes after winning to be filling as much of the board as possible, then having the choice actually makes a difference. For example, if you had the choice, you can guarantee a score of at least 9 by placing the 9 tiles in a formation that doesn’t contain a three-in-a-line. This is a slightly different version of the game where you have the incentive to try to place as many tiles, instead of going all in on high roller.

[@Imperial_Squid]

There's no strategy to this or any other version of the game. So long as the win condition is placing _every_ tile on the board, the number of winning arrangements out of all possible arrangements _remains the same._ You could put them all down face up in order, face up randomly, face up with your choice, face down in order, face down randomly, face down with your choice, mixed face up/down, etc etc etc. *_ALL OF THESE VERSIONS_* have the same chance of winning the game. The only thing that changes is how soon you'll find out if you lose.

[@ValkyRiver]

@@Imperial_Squid What about a version of this game with multiple end states (and not just a “win” and “lose”)? For example, 1 point for every tile placed without making FOX, and the final score is the amount of tiles placed before either winning or making FOX?

[@Imperial_Squid]

​@@ValkyRiveryou can use the graphs at 5:19 to give you an idea, If you score 1 point per turn before making a fox, that's directly equivalent to scoring X points for losing on turn X of the original game. As such, we're looking for strategies with taller bars on later turns. Comparing between the shy, high roller and random strategies, shy would do best in your 1 point per tile version. (Though I don't doubt you could finesse the strategy with some better heuristics to push that average score even higher)

[@damnerd]

​@@Imperial_Squidwoah, I'm pretty sure that if I can look at every tile before placing them I win 100% of the time, so the number of winning arrangements is not the only factor involved here.

[@NStripleseven]

A more intuitive argument for the absence of strategy in this game is if you imagine laying out the face-down tiles in a row before playing, or numbering them 1-12 or similar. Then, whatever your move (let’s say you want to play in the third unoccupied square, the top right), you choose the tile out of the lineup corresponding to that square (so tile #3). This means that the final layout of tiles, while still using the same random distribution as normal and so changing nothing about the game, is entirely predetermined. If tiles 1,2,3 contain F,O,X, you’re going to lose no matter what order you place them in.

[@AyeletZuckerAz]

And the earlier version had the squares lined up with the tiles randomized, so it's just reversing which one is "predetermined", which changes nothing

[@tgwnn]

Yeah that's the argument that convinced me a few videos ago!

[@670839245]

you thought you introduced an element of skill, but you introduced an illusion of choice. (It's still a valid thing in game design though.)

[@R1ver7th]

one way to think about it is that your strategy really just determines what order you place the tiles in. framed through this lens where you're basically just discovering if your board is a loser or a winner then high roller is the best strategy because itll allow you to cycle through games the fastest

[@ampisbadatthis]

okay, hear me out, you get 5 Os, 5 Fs, 5 Xs, and 1 wild card. When you pull the wildcard, you immediately have to decide what it represents. Most of the time, it's completely random, but every so often, it becomes a tense game of odds.

[@radiol]

I suggested in one of sorths the wild card can be a fox hole. Like a handicup for a player so you have one spot where you won't have fox.

[@limonlx7182]

​@@radiol So like.. an empty tile? Wouldn't that result in the same thing as mentioned in the video though? It's still a completely random chance that the empty tile will end up on any possible square. The only thing that would change is that it would be slightly more likely to win, but it still would lack any meaningful player input. The suggestion in the comment above is different though, as it's giving the player an actual choice that can impact the game.

[@JollyTVance]

hear*

[@ampisbadatthis]

to fix players calculating the best option for the wildcard or fox hole, you could just have a 1 minute timer, or a 10 second timer that resets everytime you place a tile.

[@ampisbadatthis]

@@limonlx7182 I realise that you weren't replying to me, but I want to defend that other commentor. My version is the same as theirs, in theory. There's always one or more best possible options for which letter to pick, obviously if the wild card lands in the corner, pick O, if it lands at a point where O or X would make fox, pick F, etc. but in practice, radiols "fox hole" (as long as the square isn't preselected at the start of the game) becomes a completely different kind of problem to the problem in the video, placing the fox hole is an "optimal stopping" problem. You need to not only calculate the chance of you getting a fox on that exact tile but also the statistical probability that you could get a fox on any other tile in the future. My version only has you calculate the chance of losing due to the exact tile that the wild card lands on. eg, if the board was F X F F X O O O F O - - - - - - it might feel that the optimal play would be to play the fox hole on the next tile because there's three "FOs" pointing at it, but in reality, the chance of that tile losing is 3/5, but the tile at the end of the row has a 4/5 chance of losing (remember that this version of the game has 15 playable tiles, not 16.) In my proposed "wild card" game, however, the board presented isn't any trickier than the normal game, since the optimal strategy is obviously to pick O if it lands on either tile in the third row, or to pick X if it lands in the bottom row, without even needing to calculate the odds of losing. That specific board is an extreme example, but a similar logic would have to be used regardless of where you place the fox hole, except for, obviously, the last tile in the game. There's a wikipedia page dedicated to "optimal stopping" algorithms and problems if that interests you.

[@darkZarchon]

The easiest way to make the strategies work in the way you were envisioning is to increase the number of tiles. Adding, say, two of each letter will create situations where the probability is zero because the board is full before it gets to choice 13. That said, there are already many common games with the illusion of choice but no actual choice. There are also many other ways to create a different kind of strategy. So what you decide is truly up to you.

[@vytah]

You're still blindly filling a square with the same number of random tiles. Since you don't have an influence on which tiles are left over, it's essentially the same, just with a different letter distribution. It's as if you decided before the game which two tiles to toss out.

[@scottclowe]

SUGGESTION: If you want to improve the game so there is some player agency, make it so the tiles are divided into two groups of tiles at the start of the game. The frequency of each letter in each group should be known to the player at the start of the game. On their turn, the player chooses which square to place the tile in, and which group to draw from. A tile is randomly selected from that group and placed at that location. The set of tiles for the two groups at the start of the game do not need to be the same size or have the same distribution as each other at the start of the game. Proof that this version of the game has strategy: initialise the two groups as 1: 6 O and 2: 5 F, 5 X. If the player chooses randomly from groups 1 and 2 with probability proportional to the number of tiles remaining in the group, they have the same success rate as in the original game. If they chose to use group 1 for a group of tiles around the edge of the board, they have a 100% success rate. Hence player agency is meaningful. The optimal initial distribution and count of tiles between the two groups to make for a fun game remains a question of game design. As a game designer myself, I would expect two groups that start similar but a little different would be good. e.g. Group 1: 2 F, 3 O, 2 X; group 2: 3 F, 3 O, 3 X. The distributions would change over the course of the game, causing the player to adjust which they draw from and when. However this initial distribution may be too similar for the two groups for the game to have sufficient strategy. One would need to playtest and simulate to verify. From a UI perspective, place one group either side of the board (left and right) to make the groups clear. You could even call them the left and right group instead of 1 and 2.

[@Nachiebree]

"Is it even a game with no player input and complete randomness?" Yes. Candy Land exists.

[@quintopia]

I mean, by the broadest definition they're games, but they are definitely not strategy games, and one could argue they aren't particularly fun either. It's definitely not the category of game you typically want to be playing.

[@trunkulent]

Candyland does exist and, crucially, _isn't a game!_ Yes I will die on this completely pointless hill.

[@quintopia]

@trunkulent i feel kinda the same. But the academic field of games studies would disagree and i can't deny that their reasons for doing so are good.

[@yoavhaklai6819]

wow these long form videos are so well made! you should make more of those, they're so interesting!

[@ValkyRiver]

If all non-winning outcomes are considered equivalent, then there is no probability difference between having the choice or not. But that’s only for the “winning” case. Since the game ends as soon as FOX is immediately spelled, this means if the game does in a loss, it could be anywhere between the 3rd and 16th tile drawn, so that could be a score from 3 to 16 (and winning would be a number 17 or more). Which one feels like the better outcome - losing immediately on the third tile, filling up less than 1/3 of the board, or losing on the last tile? IF (that’s an assumption) you consider the next best outcomes after winning to be filling as much of the board as possible, then having the choice actually makes a difference. For example, if you had the choice, you can guarantee a score of at least 9 by placing the 9 tiles in a formation that doesn’t contain a three-in-a-line. This is a slightly different version of the game where you have the incentive to try to place as many tiles, instead of going all in on high roller.

[@thehexagon_yt]

Losing immediately is better, IMHO, because you don't waste unnecessary time. Also the idea is pretty simple - harder start, easier ending. You're not avoiding the danger, you face and overcome it for greater future. After all, if the result is inevitable, what's the point of delaying it?

[@UltraLuigi2401]

4:22 made a fox in the top right but didn't notice

[@ilexdiapason]

exactly what i was about to say

[@galkain7685]

Maybe next time put the o's from top right to bottom left, that for sure will change your luck and give you the win

[@max7heki11er]

ah yes the classic illusion of free will, most games are like this where the choices you make do little to nothing, yet if you remove the choices people don't like not having the pointless choice

[@johannschiel6734]

The easiest explanation: in the end, all tiles are randomly placed and you decide in the end, if the game is lost or won. The order, in which the tiles where placed doesn't matter at all for this question.

[@goosewiz]

The best strategy is to count how much letters of all types left unused. Take one then, assuming that you picked the most possible variant, place it on square, where you won't lose immediately. Also you can analyze whether you will create dangerous tiles by placing new (again, assuming that you picked most possible letter)

[@victorvirgili4447]

the whole ending of this video is talking about how that doesn’t change your odds, with mathematical proof to back it. the real way to improve this game is to give the player info on what the tiles are, even if very little

[@Dash123456789Brawl]

@@victorvirgili4447As far as I could tell, the mathematical breakdown didn’t address that the player can identify different probabilities through an inherent understanding that there’s a limited number of each tile, and what their strategy accordingly. For example, with the version of the game that starts with 4 O-tiles already placed, that leaves only 2 O-tiles remaining. A player can recognize when there are no more O-tiles to place, and it seems likely that they can improve their odds of winning by playing around that fact. It would happen far less frequently, but the same applies to running out of F-tiles and X-tiles. Perhaps those were all covered by the explanation, and I failed to recognize it as such…

[@coltynstone-lamontagne]

This was a really good video! Very interesting how unintuitive and not random it ended up being

[@GoldenSandslash15]

On the website version, could you add a setting that lets you toggle whether or not it displays the "risk number" for the shy/high-roller strategies?

[@jan_harald]

I think a viable improvement would be that you can either place the tile blindly, OR flip it over and see what it is BUT that will require the board to be spun a random amount of 90° changes, and before you place it, you can't see the board, if you try to place where another piece is you'll just have to retry... downside: that requires either an assistant, or a virtual game, while somewhat possible, it'd be really difficult to play it alone upside: you can have actual strategies, random rotation will mean you can't actually make sure to place the piece safely (the board has 4 possible rotated positions, and if you get it wrong you can just as well accidentally spell the fox), but it does give you general control, like do you try to place it more towards center or more towards edge

[@jaykebird2go]

I wonder how this might change if some portion of the shuffled tiles you know beforehand, and some portion you don't... like for example, of the 12 tiles, you flip over 4 of them right away before putting them on the board, and then you leave the remaining 8 unflipped (and thus, still random). Probably would increase the odds of winning, but leaving some tiles unflipped would still have that randomness factor.

[@bot24032]

As a mathematics-near person, after about 2 days of watching you do that, i got a weird feeling that while you can change the time you stay alive, you can't change the chance if winning but until you did the video with all tiles pre-placed, i couldn't find an explanation. When I saw that (and especially when you mentioned it's the same game), I was like "THAT's what I've been looking for!!"

[@tristanridley1601]

What about the games where your last few tiles are mostly x, o, or f? You gain information.

[@francescocostanzo8225]

That near doing heavy lifting

[@Imperial_Squid]

​@@tristanridley1601you gain information, not luck. If I need to roll a dice and beat a 4 to win, it does me no good to know I've got a dice with 1s, 2s and 3s on every face. Knowledge of the outcomes doesn't matter if you can't affect them.

[@bot24032]

@@tristanridley1601 again, place all the times on the board from the befinning but they will originally be flipped face down. you can flip them one by one in whatever order you want to. this game has the exact same probabilities as the original, and clearly your choices don't decide whether you make a fox eventually. they do, however, decide when you see it

[@tgwnn]

When you write "it it's picked" on the brown paper at 7:10, you should write specifically that you mean square X, not tile Y. I got distracted for a moment and misunderstood the sentence on the paper (actually it's a more natural reading to assume "it" refers to the subject in the previous sentence).

[@blockshift758]

The background editing is top tier i love it. One thing i wanna say ~high roller's~ random's background was too saturated for me. Not as much on 5:25 forsome reason.

[@Magikarpador]

you should make a version of this for "bat" but the goal is to find the bat and you find it every time and bats are cute and everyone is happy

[@CaptainSpaceCat17]

that's really interesting. i suppose to make player choice have agency, you'd have to introduce some amount of information about the hidden tiles before they're chosen

[@neilk.3398]

Wow what a great video, can’t believe KZbin is free.

[@JuliuszCovers]

Suggestion for making it more of a game of skill (I hope - I haven't made the calculations): first look at the tile, then place it. However, you can only place it in a field that's orthogonally adjacent to the tile you placed last. If there is no such space, you can place it anywhere. Not sure how easy or difficult it would make the game.

[@TheOne_6]

i would call high-roller the daredevil

[@our_lord_and_savior]

I think you should add an extra feature that does add player choice into the equation. First scratch, give the player some rabbits. They distract the fox by allowing a player to swap the just played tile with an adjacent already placed tile. Might make the win rate too easy giving too many rabbits but also possible scaling difficulty

[@kinyutaka]

New idea: do not find the fox 3d

[@tristanridley1601]

I was disappointed when my intuitive strategy wasn't used. Both the shy and high roller strategy prevented you from winning by tile counting. If you play to gain information and avoid a couple risky locations until younhave few tiles left, you will still have a lot of random losses but the times where your last tiles are uneven will let you play with much better odds.

[@Anarchodemsyak]

Tile counting still does not change your odds.

[@LachyDachySachy]

You just described the shy strategy

[@stellasdoesstuff]

There needs to be a "calculated risk" group: I will choose the highest risk option of the subset of squares with risk < x% The result will probably still be the same, but I want to see it!

[@AlphaSquadZero]

I think the way you can make the choices meaningful is by having whoever can fill the board in the fewest moves including the number of moves used for failed boards wins.

[@TheLmb10000]

I think you should add some special one-time-use moves to the game in order to make it more strategic. Something like the ability to swap two adjacent tiles, reroll one of the already placed tiles or reveal one tile in the hand. Such rules can bring a little bit of control over the game and make it less random.

[@TheLukeskywalker2]

I just went to your website to do the o variant and I got it first try.

[@shelbygreene]

As you explained the random strategy, the sixth tile you placed was a loss immediately. Sadly, didn't make it as far as you thought 😢

[@rileyzook-moran4012]

Why not just make it a 2 player 1v1 game? Have it so the tiles are still picked at random, but let the player look at the tile before placing. Then the goal could be to box your opponent in like a lot of classic board games a la connect 4, tic tac toe, etc. I'm sure there's still problems and I haven't thought at all about possible dominant strategies but it would give the players more agency than these last iterations

[@PLMMJ]

Maybe you should try to make a board game based on finding as many cows as possible to tie into your other book?

[@deadbeatonthemooneatingkfp8500]

This should be game design 101 :)

[@trigun1081]

I actually played this with my niece as a two player game. First one to spell fox loses.

[@thehexagon_yt]

Now that's a version that has strategy. Kinda like buffed tic-tac-toe though, but anyway. I actually made a bot to play with the same way during the first version series.

[@victorvirgili4447]

@@thehexagon_ytI’d say it’s more like a compressed 4 in a row

[@qladinado]

What about a more uno-esque multiplayer version? Each turn someone plays a tile from their hand, and if you play the word fox, you lose. This version could also have special tiles where you can remove, swap or maybe even push tiles into adjacent squares.

[@noworldrecords8924]

I think that the best strategy is the high roller since even if it doesn't increase your chances of winning, it will make you win in less time on average since most failed attempts will be shorter

[@edene5493]

i have a strategy that i think gets it down somewhere between 1/20 and 1/30, because the issue with these strategies is early game tiles are still largely random and they aren’t allowed to plan ahead

[@edene5493]

because there’s a bag of f, o, and x, even if an individual tile will always end up somewhere 1 in 12, the question of whether an f or x would go somewhere actually has an impact, as long as stuff like that is considered from the start

[@vispreni]

Oh man I have a follow-up vid in the works where I think I managed to prove that having free placement DOES influence the outcome, you beat me to it by a couple days 😅 this is going to be fun

[@ThelsdeKwant]

You think? If you work out the details, you should realize that free placement doesn't do anything to influence the win chance. In the end, you're placing one of 12 tiles at random in each of 12 locations. The order in which you do this doesn't matter. Each location is going to have one of the 12 tiles at random, regardless of whether you place there first or last.

[@1992coen]

By pure chance I beat the game on your website on first try 😂

[@byeguyssry]

Me trying to figure out why any strategy could work

[@manas21899]

I just went with the high-roller method and nailed it on my first try

[@nillanapier1161]

My prediction is random will do the best, high roller is an obviously bad strategy, as you are consciously trying to make the word fox, and while shy might look different, it's actually that exact same strategy, but in reverse, you are always maximizing the chance the next move will be lethal

[@cloudyblueskye]

On the random example it actually lost on the sixth placement (top right corner going down)

[@patches2654]

I just tried it for fun on your website and got it in 4 attempts

[@Yonkage-ik5qb]

I don't have any ideas for how to make this more strategic, but I do think it would be funny if on your attempts to not find the FOX, your tally marks for attempt number started going down the right side of the page and then wrapped around the bottom and then up the left side, as it may be dozens more attempts, and you're rapidly running out of room. Of course, this will probably mean you immediately win the game...

[@ExEvMusic]

Just tried it, 11 attempts, and funnily enough it was a random strategy 😂

[@FilmscoreMetaler]

You can technically quadruple your win chances if you don't rotate the f when placing it down. ɟox != fox and so on.

[@uiinpui]

I love when that happens

[@toolebukk]

1:40 you may as well only start with those two in the center. Although this will make a difference probability-wise, it should feel more "fair" as it were.

[@YayDragons]

What about shy but placing in the center tiles as much as possible early on so as to give more opportunity to use strategic methods

[@quintopia]

Can you prove that high roller is the optimal strategy for maximizing number of wins per tile placed? (Which is to say, minimizing tiles played per game)

[@NikkiTheViolist]

Nah, this result is very intuitive: whether you're going high roller or low roller or random, it's all random in the end because you're not making any choices at all you're just picking one random tile and placing it in one random place (where the random place is either defined by you or by the current turn), nothing's changed I am curious, tho, if taking into account the probability of fail state would change anything compared to the simple number of fail tiles (since the current High Roller and Low Roller strategies simply depend on whether the fail is an 'x' or an 'f' and not the proportion of 'f's and 'x's remaining in your deck. It's possible that the math all cancels out in the end but...) well, anyway, my strategy is still going for losing faster than losing slower since then I can play more often and win more times

[@kmag2084]

This Video is so great

[@raphaelbell2791]

would you be interested in changing the rules to allow multiple players? I'm a TA and I think this would be fun to play with kids/teens

[@AlexCheddarUK]

I've been playing 2 player with my partner and am going to start posting Shorts about it soon :)

[@raphaelbell2791]

@@AlexCheddarUK yay!

[@stillbuyvhs]

What if you gave players points based on the number of tiles left over? Game ends when the board is full, or when someone spells FOX. Lowest score wins.

[@cuberomer]

As a mathematician, it was just so obvious to me that it doesn't matter that I thought you were pretending you had a choice as a joke. The first half of this video drove me absolutely crazy.

[@_puzzled]

some other comments have suggested that the average length of the game could be a factor in deciding the best strategy. it wasnt until the second half of the video where he stated that he was looking for the higher win rate.

[@cuberomer]

@@_puzzled I mean he was doing one attempt per day, so I knew the length of the game didn't affect him. Optimizing for shorter lengths is interesting though.

[@kman6004]

When he implemented the "strategy" last time, I was kind of confused because I was instantly like "But... that doesn't change anything...?" Glad he was able to figure that part out!

[@thehexagon_yt]

I remember someone in comments under one of his videos stated that there is a strategy and that they work with probabilities like everyday. I'm not even a mathematician, but I figured out there's no strategy even before this version of the game appeared. They deleted that comment though after being proven wrong by many people.

[@plazmi1]

Maybe you can explain me if this is so obvious, as i am missing something. We assume that with preplaced tiles every slot has a propability of being an X for example equal to 5/12. It is clearly not the case with certain strategies, as there exist a strategy where we place a tile in this spot only if w propability is greater than 5/12. This will drag the propability up. Such strategies change the distribution of letters among slots, so why is it obvious that they do not change the propability of winning?

[@bizdaretz8679]

9:32 There is a winning outcome there Actually there's two of them... And going from left to right, top to bottom... F, O, X, X, X OR O, F, X, X, X Soo.... I don't know math to be fair... sooo that COULD still be 1/42, buuut my brain is just saying "No, there's 2 winning outcomes"

[@cianchetto4]

1/42 from the start without having anything placed

[@oboo1225]

2:50 it doesn't really matter, but you missed the possibility of an O making a diagonal fox

[@salkinds]

I've found a strategy that wins 2.64% of the time, according to my simulations - a VERY slight improvement (only 1.1x better), but it shows that this game does have some strategy! Here's how it works: Each turn, play your tile in a space where it's possible to make a fox, but only if the letter that is most likely to be drawn would NOT create a fox if placed there. If there is no such space or a tie for the letter most likely to be drawn, then play in a random safe space. If there are no safe spaces, then play randomly. This strategy has an edge over the ones discussed in the video because it leverages the fact that we know what tiles are still left to be drawn. Once a potential space to create a fox appears, you know that you MUST play on that space eventually during the game. Therefore, you should play on those dangerous spaces when the "killer letter" is a minority of the tiles left to be drawn, and avoid playing on them when the killer letter is more likely to be drawn than any other letter. Games with constantly changing odds are often about changing your strategy based on opportunities that arise, and that's what we're doing here. All in all, I think your game has more strategy than you concluded here, even if just a little bit. I think there's a strategy out there that can win even more, this is the only one that I tried myself! Thanks for the video and for creating an interesting problem :)

[@asheep7797]

How many simulations did you do?

[@salkinds]

@@asheep7797 I ran 50 million, just like in the video! To verify that my simulator is working right, I also ran 50 million simulations using the Random strategy which resulted in a win rate of 2.38%, matching the results of Alex's simulator :)

[@scottclowe]

I think you need to run more simulations or there is a bug in your code. Since you have to fill the whole board with tiles, and each tile is selected with a uniform random distribution over the remaining tiles, all strategies have the same success rate on this game.The proof is basically just the chain rule for probabilities.

[@ThelsdeKwant]

This tactic matters not. If X is the tile you don't want to place at a certain location, and you wait for X to be a minority tile, there's a good chance you end up with only Xs in the pot near the end of the game, and you have to place it there regardless. In the end, you're placing a random tile in each of the 12 locations. The order in which you place the tiles don't matter. Each location is going to have one of the 12 tiles at random in it.

[@azai.mp4]

​@@ThelsdeKwant Just because each tile is equally likely to end up in each square, doesn't mean that every configuration of all the tiles is equally likely. If you spin a color wheel real fast, every hue may be equally likely to go anywhere, but that doesn't mean you should expect all the hues on the wheel to get completely scrambled. In other words, variables can be uniformly distributed, but not be independent. That is something which must be argued separately. (It could be true in this case, but OP's experiment says it's not.) I'm not sure if complete independence of the probabilities is necessary, but it would be sufficient for the proof.

[@RoderickEtheria]

There's no skill in laying random tiles without having information. The best you can do is wager on which tile will come up.

[@rafa57games]

I think that this could be like playing cards. There's no one way to play it. You can play de randon version, the pvp version, the new version or any other way you want it

[@Flyce_9998]

I don't care if it doesn't make a real difference, the illusion of choice makes it more fun

[@abdijabarkhalif]

what if you flip over 1 tile before starting

[@lijath]

You should score the game based on how many tiles you placed.

[@TheOne_6]

oops

[@chgl708]

4:22 saw the fox

[@GuardingDarkness]

This assumes you use the same strategy the entire game, and not mix strategies. A lot of player agency is picking strategies. What if I go for an early high roller strat, and switch to shy if I dodge that first bullet?

[@cianchetto4]

it literally changes nothing as he explained in the calculations. Your "changing strategy" is still a strategy, and he showed that ANY strategy has 1/42 probability, cause you are simply creating an implicit order of the tiles where it doesn't matter when you pick a tile, it's always as likely

[@feva6774]

The only way I can see this game being fun is if it were a 2 player game and you tried to make the other person lose, and you obviously need to look at the tile before placing it. Maybe add a rule that new tiles have to touch at least one other tile

[@Doauwed]

Matpat time

[@temmie5764]

Took me 140 attempts when I tried, the tallys went far off the screen

[@broor]

Welcome to vegas

[@mathguy37]

ah yes, the illusion of choice how does this change if we, for example, add a guaranteed f? (both choosing it whenever or at a certain position order work interestingly) also just got a silly game on the regular mode where the Os went like this: ?OOO ???? ???? OOO? meaning with just that it was impossible to lose

[@phyphor]

You've claimed to graph "win rate over time" but it's "win rate over multiple games" which is important for certain strategies. If the number of games you can play is limited by, for example, how much time you have then it becomes interesting whether you play 100 games with a 1% chance of winning each game (where you will win once about 3/4 of the time) Vs 1 game with a 66% chance of winning.

[@FilmscoreMetaler]

Nothing you did affected any probabilities so I expected no changes and was not surprised at all. :|

[@aoay]

What is the probability of winning if you don't pre-place the diagonal O's?

[@Imperial_Squid]

That was the first version of the game. Chances of winning were 12.65% or about 1 in 8. He did another video going over the stats of that version if you're curious.

[@thehexagon_yt]

The exact ratio was 4559 in 36036 which is close enough to 1 in 8 as was mentioned by person before me.

[@sarvagyajagatram]

On the website I got it in my first try LOL

[@samoilis2276]

I find it interresting that this felt counterintuitive to you, because to me, it felt very obvious and I was getting a little annoyed any time someone critiqued your strategy in the comments

[@nimiugn]

Yea and lots of comments arguing there's a strategy to increase the win rate, which you can't convince them to think otherwise lol

[@Tahgtahv]

But ... 1 to 200 cows is a total of 20,100 cows. Am I missing something here?

[@shavranotheferanox7809]

I hate maths, why do the 11 cancel out when you multiply 11 over 12 by 1 over 11?

[@PLake-q9w]

What if in one game you switch between all three strategies

[@thehexagon_yt]

Still no result. As long as you see the letters after you place them, there won't be any strategy involved no matter how you play.

[@chgl708]

12 attempts to win

[@NYKevin100]

How about this version? Fill in the diagonal with O's, flip and shuffle the remaining tiles (so they are all face-down and randomized), then repeatedly do the following: 1. Select a random tile. 2. If there are an even number of tiles on the board, place the tile wherever you like, and then flip it over (so that it is face-up). 3. If there are an odd number of tiles on the board, flip the tile first, and then place it wherever you like. 4. Do not find the fox.

[@Imperial_Squid]

Still the same win rate. So long as the win condition is that you need to place _every_ tile to win, there is _no strategy_ that affects your win rate, the only thing you can change is when you find it you lost. Think of it this way, to win the game you need to place every tile without making a fox. There are only so many arrangements where this is the case. So your win rate is the number of winning arrangements of the tiles divided by the number of total arrangements. So long as every tile needs to be placed, this number does not change. Using different strategies only changes how quickly you find out if you lost, but if you kept playing after you lost and put down every tile, that arrangement was always going to lose.

[@NYKevin100]

@@Imperial_Squid If you read my suggestion more carefully, you'll realize that it includes flipping some tiles before they are placed, which allows you to influence the outcome (for example, you could put O's in the remaining corners, F's and X's next to each other, etc.).

[@Imperial_Squid]

@NYKevin100 I read it, it doesn't change the numbers. You could choose where to place every tile or none of them, so long as you pick up tiles from the pile at random, you can have as much or little strategy as you like when placing them, it *_does not_* affect the win rate.

[@kemcolian2001]

@@Imperial_Squid I disagree. Take the most extreme example, where you flip over the tile every time before placing it, then it's trivially easy to win the game every time (Alex Cheddar demonstrated this in his previous video about DNFTF probabilities). You can make a similar argument for a game where you flip every single tile before placing except one, which you have to place at random. then again It's way easier to win the game with a little bit of strategy (no matter whether the unseen tile is at the end, the beginning, or at some point in the middle) Of course at the extreme opposite end, where you don't look at any of the tiles before placing, it's 100% random. so based on this I think it's reasonable to conclude that either: the effectiveness of strategy slowly decreases and randomness slowly increases as you go from one extreme to the other; or eventually you reach a "critical mass" of unseen tiles that makes it so that strategy doesn't matter any more and randomness completely takes over.

[@Imperial_Squid]

@@kemcolian2001 As I said "so long as you pick up tiles from the pile at random", if you turn the tiles over and pick specific ones, _that's not random anymore_

[@G.Aaron.Fisher]

This is a joke, right? You're playing up toward a big reveal about why the order doesn't matter? Part of me wants to watch to make sure. But it's going to be devastating if I keep watching and that never gets addressed.

[@AlexCheddarUK]

Keep watching my friend 👀

[@thehexagon_yt]

I'm a little bit disappointed that only approximate chance of winning was stated in the video. The exact ratio is 1 in 42.

[@PLMMJ]

Maybe you should try to make a board game based on finding as many cows as possible to tie into your other book?