The Poker Paradox

[@kezzyhko]

What if you don't consider the ranking when counting ways to get a hand? Like, you get a hand with the joker, which can be turned into either two pair or full house. Add +1 to the counter of both. Of course, you would always choose full house, since it is higher in the list, but that does not matter, you still theoretically can get two pair. That way we will get unambiguous number for "ways to get a hand", and we can rank hands by that number afterwards

[@markoboksic2238]

to get hand with joker that can be turned to full house, you will have from 4 leftover cards either two pairs, or 3 of a kind plus 1 different. So when you have 2 pairs + joker it is always full house. When you get 3 of a kind and 1 different you can choose between poker or full house.

[@kezzyhko]

@@markoboksic2238 > "2 pairs + joker it is always full house" Not if you turn joker into any card that is not in the leftover cards. This may not make sence for the player, but you still can do it

[@chitlitlah]

This is what I was thinking. The probabilities with a joker should be the number of ways you could possibly make that hand even if you'd probably use the joker to make something else. So a pair plus a joker would be counted for both trips and two pair. Trips plus a joker would count for both quads and a full house. The probabilities would add up to more than one, but in this case, it doesn't matter.

[@markoboksic2238]

@@kezzyhko that is nitpicking and in general not correct. You are obliged to use joker to get the best possible outcome.

[@xAtNight]

@@markoboksic2238 where was it stated that you *need* to use the best outcome? In his problem he says the joker "can be whatever you like". And thus this should be solvable with the above mentioned method.

[@f5673-t1h]

The root of the paradox is that you have two things whose definiton depends to the other. The rankings determine the probability, but the probability determine the rankings. It's the same as the paradox where you say "the next sentence is true" and "the previous sentence is false". Most paradoxes in math are from things like this where there is a "loop", like self-referrential objects, e.g in Russell's paradox. So we always avoid them.

[@alegian7934]

but when exactly is self-reference unsafe? "the function whose derivative is itself" leads to no paradox, just e^x. Have we pinpointed what causes this incosistency in math? Im intrested in formal logic but Im looking for resources to read on

[@nicholastessier8504]

Russell's paradox is a set, not an object

[@nicholastessier8504]

@@alegian7934 when it leads to a contradiction, read the SEP on formal logic

[@f5673-t1h]

@@nicholastessier8504 Sets are objects 🤦‍♂️

[@jacoblojewski8729]

@@alegian7934 That's the hard part. There's now way to know right from the get go that self-reference will lead to contradictions. You basically have to investigate each case. In your example of e^x, this is why many books don't *define* e^x as a function whose derivative is itself, but rather something completely different (one common example is as the inverse of ln(x), which in this route is defined as the integral from 1 to x of 1/t dt). That way you don't have to go through the whole usual problem with self-referential definitions: doing a follow on proof that your definition "makes sense": in the e^x case, an existence proof. Anyway, a few places to look if you want to dig in more: *The Logic Book* - Introductory level logic book, but goes into a lot on proving consistency and completeness of various logic systems. *Mathematical Logic* - Intermediate level book focusing on logic's application in math, but digs in deep again on consistency, completeness, etc. *Set Theory* (Jecht) - This one's very dense and will take a bit to get through unless you're practiced in baseline set theory proofs. But it covers (eventually) some of the techniques needed to tackle the really hard versions of proving a self-referential system is consistent and/or complete (for example, *forcing* and *models*). For something quite a bit lighter, I'd recommend reading up on Gödel's Incompleteness Theorems. There's a great video by Veritasium (called *Math's Fundamental Flaw*) that gives a good layman's explanation on how they function, and the surprising way that self-reference can pop up in a system of rules with it being less obvious that it can.

[@johnchessant3012]

This video must be just a by-product of Zach's preparation for his next high-stakes poker game against James Bond

[@zachstar]

True highest ranking is obviously when you have two primes.

[@morkovija]

I sense a comedy skit with evil protagonist explaining his plot to a hero and he doesn't understand any of it. you're welcome Zach ;)

[@strangelyrepulsive77]

it's not a byproduct it's a side effect

[@tc6818]

When playing Baccarat or poker, Bond isn't playing his hand. He plays the man sitting across the table from him.🃏♣♠♥♦

[@gabor6259]

*Daniel Negreanu

[@CauchyIntegralFormula]

I mean, the resolution is obvious, right? You just count how many hands could be treated as the given rank. So a hand with a single pair and a joker should count as both a two pair and also a three of a kind, since you could treat it as either type of hand. Instead of ordering by how many hands *will* be of a given type (which depends on the ranking of the hands), just order by how many hands *could* be of the given type. Technically, this also applies to the original ranking. A four of a kind is also two pair and a three of a kind, if you think about it. The only reason we exclude them from the count of the lower hands is because they're lower hands, but the only reason they're lower hands is because they occur more often; it's the same circular reasoning, and it only pans out without jokers because of the structure of the deck. There happens to be a self-consistent ranking without jokers, but we just got lucky, and our luck didn't hold. Instead, update the ordering rule so that self-consistency is not necessary, and just abandon that each hand has to slot into only one category

[@jdkemsley7628]

Yes, you just stick to counting, and THEN decide on rank. The paradox only occurs because of recursively using rank WHILE coming up with the numbers that decide rank

[@nocodenoblunder6672]

This would lead to a, in praxis, more likely hand to to be also more powerful than a hand that occurs less often because as soon as you as the game designer decide which hand is stronger two pair or three of a kind every player should from a game theory perspective always choose the stronger one 100% of the time since obv there is 0 incentive to choose a weaker hand in poker. Therefore, as a player, you would not care about the theoretical probability as much as you would about the actual probability because at showdown you would face that your opponent announces that his joker is obv a third 10, completing his three of kind, beating your more rare two pair of Aces and Kings.

[@CauchyIntegralFormula]

Okay. And? If that's gonna occur no matter how you rank the hands, then you obviously need some other way to rank them. Which I provided

[@B3Band]

All you really have to do is set a house rule for two pair vs three of a kind. "Natural Three of a kind beats Natural Two Pair (no jokers in hand), which beats Joker three of a kind, which beats Joker two pair" and therefore joker two pair would never be created. You could even extend it to "Natural ______ always beats Joker ________" when two players have the same hand, but one of them got it via the joker. So if you needed a joker to get a full house, but someone else got a full house naturally, they beat it regardless of the high card.

[@CauchyIntegralFormula]

@@B3Band Yes, you could double the number of hand types and re-rank everything. Though in that case, just by the numbers presented in the video, "pair + joker" is rarer than "natural two pair" (82368 hands vs 123552 hands) so your ranking is still not sorted by rarity correctly

[@brianb.6356]

There seems like there's a simple way to solve this, which is to rank hands as if you were using the joker to achieve that hand, and not the "best possible hand". That way, three of a kind is ranked higher, since 137,280 < 205,920.

[@xAtNight]

He should have specified that you always need to use the best possible hand. Without this rule your method is correct.

[@methatis3013]

@@xAtNight He did specify that...

[@freshrockpapa-e7799]

@@methatis3013 No. He said that's what a player would do, not that it's enforced by the rules.

[@phatkin]

Then you have a balancing issue, since in practice two pair would become (relatively speaking) quite rare, whereas the now comparatively common three of a kind is still worth more points.

[@tcoren1]

@@phatkin because it is harder to achieve. What is the problem with something harder to achieve being worth more points

[@lapiscarrot]

Missed opportunity to call it the Joker Poker Paradox

[@omegahaxors9-11]

Going in: "I can do anything!" Leaving: "Who keeps spinning the world around?"

[@shadowcage72]

You can resolve this by counting combinations not as their highest rank, but by what they make. Aka a full house counts as a full house, 3 of a kind, 2 pair, 1 pair, and high card. This would resolve it, as changing what you choose to score it as does not alter what it counts as

[@raynmanshorts9275]

If you've read the Super System, there's an interesting situation Mike Caro got into when a friend of his made a bad side bet involving a pair of jacks.

[@kotrob6205]

Was going to post this. This is the correct answer, and how you assess the relative probabilities correctly. The fact that you wouldn't make a two pair instead of three of a kind doesn't mean that you count. The video is therefore misleading.

[@ThePeterDislikeShow]

Or you could count natural hands vs joker hands as different classes? Some video poker machines do that.

[@adamscura9826]

Could you solve the paradox by ranking Joker + 1 Pair separately? There are 82,368 combinations of Joker + 1 Pair, so that would put it higher than Natural 2 Pair, and lower than Natural 3 of a Kind. The new ranking would be: - Natural 3 of a Kind: 54,912 - Joker + 1 Pair: 82,368 - Natural 2 Pair: 123,552

[@HaydenNK3]

I don't think it works. If you think about it, that means the joker has its own value that cannot be changed or decided by the player, doesn't ? If the joker can become any value, when you have Joker + 1 Pair, you'll change that to 3 of a Kind. Doesn't that mean you just can't change the value of the joker anymore then ? To me, it seems whatever you do, you'll still have a paradox. The only way to solve this would be to remove one of the issue (meaning unnatural 3 of kind or unnatural 2 Pair) and declare that one of those just doesn't count.

[@HaydenNK3]

I've read your comment again, and I'm wondering if my last sentance isn't what you're implying there ? It's just that the wording "ranking Joker +1 Pair separately" seems wrong to me, and I'd say something like "3 of Kind formed with a joker is considered as 1 Pair only" => meaning it's useless to change "1 Pair + Joker" to a "3 of Kind", therefore 3 of Kind still beats 2 Pair. Wouldn't be simplier like that ?

[@F_A_F123]

Easy way to solve this is to just delete 3 of a kind

[@phatkin]

this was exactly what I was thinking. Basically, threes of a kind and two pairs would be required to be "pure" (no joker), and there would be a new type of hand, "one pair + J", which would sit between one pair and two pair.

[@goldenwarrior1186]

@@HaydenNK3 Actually, there are only a select few scenarios where the joker can’t change value. Most of the time, it can

[@john_titor1]

The paradox is that it's fallacious to argue that the rankings of hands, based on probability alone, should be affected by what you, as the player, would preferentially turn your joker into. Just because most players would choose to turn their pair and joker into a 3 of a kind instead of a pair and a random leftover, does not mean that the existence of that hand goes away. If i have two 2's, a 5, a 6, and a joker, it is indeed smarter to make it into a three-of-a-kind 2's, but i can still use it to make a 2, 2, 5, 6, 7, for example, if i wanted. So in reality, there isn't actually any paradox, and it is indeed possible to objectively sort the hands based on probability.

[@redpepper74]

Right, it’s a paradox in the sense that it might not make sense intuitively, not in that it actually is impossible to solve

[@francogaray3273]

This is exactly what I thougth.

[@coug1973]

He said it in the video a couple times, that the player would want to turn his poker hand into the best possible hand. That seems like a rule in the set up that is required. Otherwise, you're correct and the joker can be make into something that doesn't help the hand. But that is changing a rule that is being utilized in this example.

[@john_titor1]

@@coug1973 But it isn't a rule. I didn't change any rules, it's simply why the "paradox" arises. We make the assumption that everyone will want to make the best possible hand, and therefor, assume the probabilities will change. But the option to be suboptimal always remained. It was simply a failure to see all the options. In fact, in order to actually end up with these paradoxical probabilities, you would have to add a rule to force what card you can turn your joker into based on your hand, which would cancel out the mutability of the probabilities in the first place meaning it is no longer paradoxical. You are always able to rank hands by probabilities, regardless of the which ruleset discussed you go with.

[@Muhahahahaz]

With your method, the new hand frequencies would just be 6 times the previous table (without jokers) But in reality, most poker games do not actually allow you to “choose” anything. Your hand automatically becomes the strongest possible hand (see Texas Hold ‘Em, for example) I believe “choice” here was primarily being used for pedagogical purposes, so we can think through what each hand is supposed to be In either case, a “paradox” still exists. You can change the rules of the math puzzle to wish it away, but there’s still a version of the puzzle where the paradox exists You don’t have to “fix” the paradox though. Paradoxes like this come up all the time in logic/mathematics, and the reason they exist is due to inconsistent use of self-reference Basically, the puzzle says “this hand is ranked higher if and only if it’s ranked lower”. It’s a badly formed statement that can never be self-consistent. These type of scenarios are important to look out for so that we can redesign things in order to avoid them

[@Furiousjp01]

5:41 love that low quality 10♦

[@edwardagarwala]

The paradox isn't caused by the joker, it is caused by the assumption that the probability of a hand should determine it's ranking. For example, in a regular deck there are 1,020 ways to get the worst hand in poker, a non-flush 7,5,4,3,2. We don't then say those hands must be better than the 3,744 full house hands.

[@danielyuan9862]

One rather interesting idea to fix this "issue" is to make the rankings of 3 of a kind and 2 pair the same. When comparing a three of a kind and a 2 pair, compare the rank of the triplet of the 3 of a kind with the higher pair in the two pair. The higher rank wins. In the case of a tie, the 3 of a kind automatically wins. Although I think the solution is just to keep the rankings the same, because as someone else commented. When counting probabilities, you should count a hand if there _exists_ a card that the joker can turn into that would create that hand. That way, it doesn't matter the ordering of the hands, the probabilities will stay the same.

[@B3Band]

All you really have to do is set a house rule for two pair vs three of a kind. "Natural Three of a kind beats Natural Two Pair (no jokers in hand), which beats Joker three of a kind, which beats Joker two pair" and therefore joker two pair would never be created. You could even extend it to "Natural ______ always beats Joker ________" when two players have the same hand, but one of them got it via the joker. So if you needed a joker to get a full house, but someone else got a full house naturally, they beat it regardless of the high card.

[@livedandletdie]

@@B3Band Well, each hand of a 3kind is different to all other 3kind, after all, an 3 of Ace is the highest hand in that case. 3 Kings is the second highest, etc. So just assuming that we can always extend it to the Jokers. And sure Ace beating all is arbitrary, but if we take the card values. We can always evaluate the Joker in relation. And it's traditionally known as a Wildcard, so I'll write it as W. AAA AA+W is a lot more likely than any Natural Three of a kind. There's only 4 ways to make AAA there's 6 ways to make AA+W There's still only 4 ways to make KKK so AA+W is lower than 222 which is the lowest Natural three of a kind. So we can order all the naturals as usual, then followed by the Wild Three of a kind. We can extend this to all the others as well, including pairs and 2 pairs and W never being the high card. We can evaluate the Wild card to have a value of 0.

[@Nosirt]

The problem with this is that in poker, there is a mathematical concept called “blockers”. With your “look at the rank of the both card and determine which wins”- the 2 pair would win more often- just due to logic. Because the 3 of a kind only has 1 rank to its name- the one 3 of a kind. But the 2 pair will have 2 different rank to compare. So you are giving the 2 pair 2 chances to win while the 3 of a kind only has 1 “application” to say it. Basically, comparing a 3 of a kind of 2, it would always lose to all 2 pairs and all other 3 of a kinds. Because to make 2 pair, you need 2233 minimum and 33>222 in rank of number. While due to blocker concept, when you have 3 of a kind, there is only 1 other number of that rank available to the other player, meaning their 2 pair will most likely be 2 different numbers rank. And if we chose an average of 3 of a kind- which is 777 ( so there is 50% chances of one pair of the 2 pairs being below 7 and above 7) this actually means- vast majority of time, when ever you compare 3 of a kind to 2 pair only based on the highest pairs value- the 2 pair is more likey to come out ahead because it just has more probability to have at least 2 pair higher rank than the rank of the 3 of a kind.

[@rsm3t]

​@@NosirtThis isn't true, because the highest hand is 3 As, which beats 2 As plus another pair. And so what if the lowest hand is also three-of-a-kind? Three -of-a-kind has the advantage in a tie, so one slight advantage and one slight disadvantage offset each other. And anyway, since we've replaced two ranks of hands with one, the relative ordering of hands within the rank is immaterial; any arbitrary scheme will do, but as humans will be comparing the hands, we'll go with something easily processed. The only requirement I see is that there is some interleave between the 3-hands and the 2x2-hands. Because if the lowest of one category beats the highest of the other, then there really are two separate ranks, and all we've changed is vocabulary.

[@Nosirt]

@@rsm3t the problem arises because the “pros” given to the 2 pair would far outweigh the “pros” of the 3 of a kind always beating when tie. In words, it’s like 1 pro and 1 con for each but mathematically , the “compare higher of the 2 pair” to the trips gives way more mathematical advantage. For example, there are 13 trips- AAA being the best (out of even the 2 pairs, given our “tie equals trip win) but the second best would be AAxx and this rank alone will have 12 combos. AAKK to AA22. This mean every trip will be the biggest of its group, but 12/13 if it’s in-group will be two pair which will beat all other trips below it. This actually means the probability of card distribution already garuntees you that there are more combos of two pair that beat trips individually than vice versa.

[@kajacx]

The math is neat, but there absolutely is a way to order the hands based on probability. You order them based on the chance to get that hand, regardless of whether you would actually choose that hand.

[@Mutual_Information]

The Joker is always wreaking havoc..

[@someknave]

My solution to this is that you include all hands that could qualify even ones that you would claim a higher category. I would do this even without wilds eg. A full house would also count as a three of a kind, two pair, a pair and high card, as it has all of those things.

[@bobfr4806]

Fun fact: If you're playing poker with a 32 cards deck, then a flush is less likely than a full house and even less likely than four of a kind !

[@jimmyh2137]

What kind of 32 cards deck?

[@8szczypiorek]

@@jimmyh2137 not sure what variant he means, but there is a quite populart Short Deck Holdem variant where you use 36 cards(6 is lowest card) and hand rankings also change there.

[@oisyn-]

@@8szczypiorekIn fact, in regular Texas Hold'em the probability of a high card (17.4%) is already lower than those of a pair (43.8%) and two pair (23.5%)

[@jdkemsley7628]

The paradox only happens because the column is misleadingly named "possibilities". Those numbers involve more than just counting possibilities. There are 40 possibilities for getting a straight flush but only 36 of those would result in the 2nd-ranked hand. So it should be labeled "possibilities of having these cards while also not having a better, already defined rank". If you're saying you want to rewrite the definitions of rank based on probability, you have to stick to just using probability. What you're doing is recursively using the definition of rank while writing that very definition.

[@firelow]

The "Royal Flush" isn't a different type of hand, it's just the best type of "Straight Flush". You wouldn't call a "Pair of Aces" a different type of hand from a "Pair".

[@firelow]

This is just me being pedantic btw, still cool idea and interesting video

[@farismusic9713]

This is an official poker hands ranking. I am not a fan either, but has to be said this way.

[@tc6818]

What if two players each have a Royal Flush in different suits? Is there a suit hierarchy?

[@farismusic9713]

@@tc6818 officially, no. If there are 2 players with the same hand strength, they split the pot (50:50). However, there are some methods that are occasionally used in unofficial games, like private or home games. For example rules of other games, or alphabetical order are some of the ones that are being used. But to not complicate further, if you are playing official poker game with no special rules, its split 50:50 or so called "chop" in poker.

[@skilz8098]

@@tc6818 Not officially but some places may rank suites. I haven't seen it too often but usually the most common rank of suits I've seen was typically Clubs than either Hearts or Diamonds then Spades, but then again this depends on location as these would typically be considered house rules.

[@batlrar]

I have an old handheld video poker game where twos are wild, that I instantly thought of when seeing the subject of the video and tried to remember how it ranked them. I thought I remembered it making the payouts for them even, but upon checking it actually does that for four of a kind and full house, interestingly. I think they would have done the same with two pairs, but instead it's not even in the rankings! Five of a kind is actually under a royal flush, but keep in mind that this is with four wild cards instead of just one. The top three ranked hands you can get, in ascending order, are a royal flush using twos, a four of a kind of all twos, and then finally a natural royal flush. The payouts are also all very low, since you're winning a lot more often on average even without the two pairs even being listed. It's very interesting to see a three of a kind be the lowest possible payout, which just returns the same amount of points it cost to play a hand in the first place!

[@matthewparker9276]

There is a solution, and that is to count the number of hand that could produce that result, even if it could also produce a higher result. This does make the probabilities sum to more than 1, since every hand has a high card, and full houses also count as both 3 of a kind and 2 of a kind, etc. but it does allow you to rank the hands consistently.

[@johnlawlor7931]

A better solution is to just allow the joker to be any card, except in cases like high card, where you can only turn the joker into a card you dont already have and still not have a pair. I think it is a mistake to say a two pair is easier to get than a three of a kind because while it isn't the best move, the joker could represent any value not just the one that gives you the best hand.

[@blacky6552]

Not a paradox, just bad logic. There are still more ways for your hand to be a two pair than a 3 of a kind. The fact that you would always want to turn your pair + joker hand into a three of a kind, doesn't mean you have to. You can still turn it into a two pair, also increasing that number of possibilities. If you want to calc the proper number of possibilities, you would need to count all transformations of the joker separately.

[@Riokaii]

the possible way to order the list is to just arbitrarily choose one and force a wildcard with one pair in hand to be used always to make 2 pair or 3 of a kind

[@zaksmith1035]

But this defeats the rule that the wildcard can be whatever you want. There are better solutions explained in the comments.

[@Pervitin_enjoyer]

It blows my mind how someone can make comprehensive math problem videos and hilarious sketches at the same time

[@stonefreak5763]

This Video (and "Riddle") is awesome, just one tiny tiny little thing: When you switch the Text "Three of a Kind" and "Two Pair", you do not switch the Cards to the exact right of it.

[@lsi3585]

I just came from class thinking I hate this shit, then I watch a zach star video and go yeah I f*ckn love math.

[@Khaim.m]

4:00 This would be better if you gave the number of "one pair plus joker" hands. Don't make me do subtraction on six digit numbers.

[@BramCohen]

It seems like this should also happen with full house vs. 4 of a kind and one pair vs. high card, at least if you add in more than one wild card. There's another interesting phenomenon that if you deal out a card at a time until either a straight or a flush is possible from any five cards on the table then a flush is more likely to come up first even though it's a higher ranking poker hand.

[@skilz8098]

The flush is higher. You need 5 out of 13 possible cards from a single suit to make a Flush. You need 5 out of 52 cards to make a straight. A Flush is always harder to get than a straight. Now a straight flush is harder to get than both that isn't a royal straight flush which is the only exception as it consists of the first double digit card all 3 face cards and the ace within a single suite. Also not shown here it is unlikely but still probable for multiple people to turn up a royal straight flush. Now depending on where you are playing some places will determine the tie breaker based on suite. I'm not 100% certain but I think the ranking of the suits only in regards to Royal Straight Flushes and no other hands is Clubs, then either Hearts or Diamonds then Spades. Some places may not enforce that and just consider it to be a draw.

[@danielyuan9862]

The last bit _is_ mindblowing. It just shows the probabilistic nature between straights (having enough cards of connecting) and flushes (having enough cards of a certain category.

[@B3Band]

@@skilz8098 I don't what what you're trying to tell us. If you deal 4567 of spades, there are 8 cards that can make a straight, and 9 cards that can make a flush. However, two of those cards satisfy both requirements and would therefore be a Straight Flush, which is a different hand. So there are 6 cards (out of 48) that can make a straight, 7 out of 48 to make a flush, and 2 out of 48 to make a Straight Flush. There are fewer ways to make a flush when given 0 cards at the start, but if you already have 4 consecutive suited cards, the straight is now less likely (given that you already have 80% of the Straight Flush).

[@skilz8098]

@@B3Band For a flush there are 4 suits each having 13 cards. You need 5 of them to make a flush. The draw is still 52 choose 5 for the hand you are dealt but the ratio of cards available to get that hand is 5:13 out of 52. The ratio for making a straight is 5:(13*4) out of 52 since a straight can be a mixture of the suits and every card has a valid place in a straight except for the Ace it has 2 locations. It can form a low straight or an Ace high straight. This spans the entire deck. To make a 5 card straight. You have 4 chances of getting each numbered card from the deck spanning the entire deck. You can have a straight from {Ace,2,3,4,5 } ... {10,Jack, Q, K, A} where there are 10 different orders of straights you can get and for each card of the set there are 4 possible cards to draw to get that number. In other words there are 10^4 - All straight flushes to make a 5 card straight out of 52 cards. When it comes to a flush not being a straight flush nor a Royal Straight Flush there are these possibilities from each suit: {A,2,3,4,6}, {A,2,3,4,7}, {A,2,3,4,8}, {A,2,3,4,9},{A,2,3,4,10} up to ...4,K and so on with the amount of permutations where none of them make a straight. Yes there are 4 suits but all of the cards must be from the same suite. So you can take the amount of permutations above and multiply it by 4. The difference between the two is that the amount of ways possible to make a straight is much higher than that of a flush. Even with added Wilds or Jokers, a Flush is still harder to get than a regular straight.

[@Nosirt]

@@skilz8098 you didn’t read the OP’s comment fully. He isn’t saying there are more flush than straight possible- he’s saying the probability to make them changes as you deal more cards. If you start with 2 cards, if both are spade, then any of the remaining 11 spades will help you get to your flush. If you start with 2 connected cards, say 76, then any 5,4,3,8,9,10 will help you get to the straight (6*4= 20 cards). So making a straight is easier as expected. But in poker you need 5 cards. So let’s deal the 4th card, with flush, there are now 10 spades than can help you and you have 3 spades out of 5 needed. With the straight, 76, no matter which card you got that was needed from 10-3 bar 76, you now have less options. If you got 8, than now a 3 would not help you. If you got a 10, than only 8 and 9 would help you. Etc. In the end- on your 4th card- if you have 4 spades, then any of the remaining 9 spades will help you reach your flush. But if you have 4 straight cards in a row- say 5678, then now only 4 and 9 would help you make a Straight, which is only 8 cards. This is the most outs for straights possible. Which is lower than a flush. So the paradox is kind of like zenos paradox- to make a 5 card straight, you must have made a 4 card straight first, and to make a 5 card flush, you must have made 4 card flush first, but even tho there are more stright possible than flushes- in that 4th moment, you should see more flush than stright because it’s more probable by 1 card more.

[@khoinotfish]

Great video!

[@lsi3585]

R u a time traveller?

[@khoinotfish]

@@lsi3585 Yes I am :))

[@elimiuzu]

it shouldn't be a paradox since the probability of getting a two pair and three of a kind shouldn't vary depending which you rank higher...sure, the probability a player plays said hand might vary but the number of ways to get each hand would still be 205920 and 137280 respectively and thus there is no paradox and nth should change

[@danielyuan9862]

@@DrDeuteron No, you'd only count it once. You count a hand in a hand ranking (once!) if it's _possible_ to create that ranking by modifying the joker in the hand (if there is one) to any card.

[@elimiuzu]

@@DrDeuteron if a hand's card's order can be rearranged to get a "different" hand then it's understood to be the same hand, no degree of subjectivity there...just like topology if you can show that one shape can become another through deformation and within the confines of the rules set by topology then they are fundamentally the same...idk if that was a particularly good choice of example but i hope you get my point~ edit: I thought of a possibly better analogy...if I have a ball and I draw a dot on it, no matter how I turn or twist the ball to rotate and move the dot's location, it's still recognised and understood as the same ball... shifting the joker around produces no meaningful difference compared to say the K's being Q's instead...if it does have a meaningful difference the the order of which K you put first in your hand should also count as a different hand...which we know everyone still recognise it as the same hand so that's debunked... but also if it did count it still wouldn't change the order since both with and without the joker you would have those extra hands and each kind of hand would have those extra hands so in the end after the dust settles the order of superiority remains unchanged, just now you have inflated numbers to deal with

[@Gagagooi]

You will more often have the possibility of getting a single pair rather than a 3 of a kind, therefore ranking a 3 of a kind higher than a single pair. You may choose to turn a single pair into a 3 of a kind by utilizing the joker card, but that shouldn’t influence how the different hands rank against each other

[@FluffyEmmy1116]

I still love how your videos are still interesting whether or not it contains aliens and/or Jigsaw

[@mr.condekua6141]

When you calculate the possible ways to get any state (such as two pairs) you use the former result without the joker. Thats not how it works since the former result is when used 5 cards but you are calculating 4 cars and then adding the possibles ways of the joker

[@marc68521]

The simple solution is to do it off of all possible ways to get to the hand. I would argue that anyone that says this doesn’t work is simply doing so because they want this to be a paradox when in actuality there is a clear number of the actual probability that a hand could be formed, not what a person is going to choose as poker never has taken that into consideration but instead the actual chances that the hand can be formed.

[@kangmoabel]

Keep up the good work! much love from Ethiopia ❤❤

[@shashipancholi]

Some food for thought: What if the joker can only be card that is currently not in play? eg not in your hand, on the board or revealed by an opponent at showdown? This removes the 5 of a kind option. Also adds a new level of jeopardy where you may have to change your hand at showdown. Eg your hand is AAAK + joker. Your opponent reveals the final A in their hand so you get downgraded from four of a kind to full house

[@Muhahahahaz]

Yeah… the “5 of a kind” really bothered me Like no, you don’t just get to make a 2nd ace of spades that doesn’t exist 😂

[@willguggn2]

That's how we sometimes spice up our Hold'em games. The wild card is the best possible card not in play.

[@johnskrb]

You simply need to add the “one pair plus joker” combos to BOTH two pair and trips. It’s accurate and actually relevant in wild card variants of open face Chinese poker where you sometimes use a wild card to make a lower hand to fit within the 3-hand hierarchy.

[@MattArseneau22]

Well, you correctly added "5-of-a-kind" at the top of the chart. The next thing to do is simply to remove "2-pair" and "high-card" from the chart, since 3-of-a-kind is better than 2-pair, we would never make a 2-pair hand (and a high-card hand automatically becomes a 1-pair hand).

[@cparkrun]

The paradox, if i'm understanding correctly, is that if we assume 3oaK to be the higher ranked hand, that will cause players to interpret all hands containing one pair+joker as 3oaK (rather than 2 pair), which flips the order of scarcity. The thing is though... so what? We should allow that scarcity order to be flipped, because that is not the order that matters. The only thing that matters is the scarcity in our *opportunity* to make a hand from a well shuffled deck, not the scarcity in outcomes post-interpretation. If 2 pair becomes scarce only *because* we choose to discard it, that tells us nothing about how difficult it is to construct that hand from a well shuffled deck, which is the measure that should be tied to ranking. It is always harder to construct 3oaK than 2 pair from a well shuffled deck, even with a joker, so that is the hand that should rank higher.

[@arya6085]

Nitpick, but I hate seeing hand rankings with royal flush at the top. It's not a new rule, it's just a word for a type of hand. It would be like putting a "wheel" between three of a kind and straight or a "broadway straight" between straight and flush.

[@lifeiswonderful22]

What I find paradoxical is the MDF from the bettor's perspective is always lower than the pot odds for the potential caller.

[@tcoren1]

The fundamental misunderstanding of the video is that it wrongly assumes the hands are ranked based on how common you see them in games (which he calls "probability"), while they're actually ranked on how difficult they are to achieve. For regular poker the two are one and the same since you have no control over your hand, but for this special poker people try to achieve the more powerful hands, making them show up in more games. Like, there are varieties of poker where you have some control over your hand, so more powerful hands are more common in games, but that's not a problem because hands are ranked based on difficulty to achieve, not on commonality in human (or AI) games

[@tcoren1]

Heck even in plain poker royal flush might be more common than a straight flash, because both are common when cheating but if you're gonna cheat might as well take it all the way and give yourself the god hand. Does that mean that straight flash should out-weight royal flash?

[@YTSparty]

A better visual representation would be using Venn diagrams. Have 2 circles that are intersected. One circle represents ways to get 2 pair, the other is ways to get 3 of a kind and the intersection is having the option of 2 pair or 3 of a kind with a joker. Assuming hands automatically become the higher ranked hand with the joker, the area in the circle that represents the higher ranked hand keeps the intersection and the other circle loses the intersection. If initially one circle is smaller than the other you rank it the highest, but that shrinks the other area, Hence the paradox.

[@gregorsmirnow6337]

I believe this is only a paradox if the game rules dictate the joker HAS to be the card that makes the best hand. If the player chooses what the Joker represents, then you can adjust the right column to be calculated differently. Instead of iterating over all hands and adding one to the possibility count of the hand type with the highest ranking that hand can make, add one to ALL hand types that are satisfied by the given hand. For example, the hand (8H, 8S, 8C, 8D, 2H) would add one to the possibility count of High Card, Pair, 2 Pair, 3 of a Kind, and 4 of a Kind. Then rank them based on those possibility counts. Despite the total possibility count exceeding the total possible # of hands, I believe the hand types' corresponding possibility counts still correlate to rarity. This means 2 Pair and 3 of a Kind would have unconditionally defined possibility counts with which to be ranked. Is there a flaw in this ranking algorithm? Let me know if I missed something!

[@montano0222]

What about the high card? There is mayn’t ways to get 4 cards that don’t match and then get a joker, to make the hand a Pair. Would that be enough to make the likelihood of getting a single pair way higher?

[@andrewbloom7694]

Im seeing a lot of comments about how it would be pointless to calculate the results without regard for the "optimal" choice, as if there are never situations where you would choose to create a lower ranked hand. I can give a few that I've PERSONALLY done, and some I have seen others try: - Teaching little kids, where you might go easy on them the first few times so they don't get frustrated while learning. - You are playing on an app or in a casino where for a certain amount of time get a reward of some sort (free drink, game token, etc) for getting a certain number of each specific hand (ie, you have gotten 99/100 2 pairs necessary for a reward, but only 80/100 3 of a kinds). Assuming the amount you bet was less than what the reward is worth (say you bet 5 quid but the drink is worth 10), AND you won't be able to get enough 3 of a kinds before the reward period ends, you'd pick the 2 pair - You deliberately threw a match for some reason (other than teaching a kid). - You are trying to be clever by masking your tell. You have good cards, you flinched, the other players saw you. But you could deliberately tank the hand by choosing the lowest ranked possibility, so the opponents think that's your tell for BAD cards. Ultimately whether these things make sense or would work doesn't even really matter, because even if they aren't wise, someone with think they are and will try it. So TLDR, I think a video where you don't assume the player would always pick the better ranked hand would be worth doing

[@arekkrolak6320]

Royal flush is a sort of straight flush, it's not like something will beat a straight flush unless it is a royal flush, likewise you could say 4 aces is a better hand than 4 of kind :)

[@birdbeakbeardneck3617]

i think what should be done is compare them both as if they were both advantageous to the game result(ie. if the players were dummies) and compare the two numbers basically ignore the intelligence of the players

[@Tker1970]

In Pai Gow poker, a joker is used as a quasi wild card-can be a 5th ace, or used to make a straight or a flush. As a result, a straight pays less than three of a kind on a side bet for having a straight or better.

[@2Skool4Kewl]

First thing that came to my mind

[@GregHuffman1987]

7:47 sierpinski triangle. Learned about that yesterday

[@matthewcooke4011]

You could take this idea to the extreme. Imagine a deck consisting of only jokers. Then if we have 5-of-a-kind as the highest ranked hand, then 100% of all hands would be [chosen to be] 5-of-a-kind, making it the most common hand and should therefore be ranked lowest. As other commenters have pointed out, the problem is the interdependency between the ranking by probability and the choice for the joker. We rank based on probability, but then the choice for the joker is based on rank, which then skews the probabilities (and the rankings).

[@logosking2848]

this is actually the result of a poorly defined ranking system, not an absence of a proper rank

[@aldrikvoldus585]

While the paradox works on paper and in theory in practical use most of not all players will still rank the hands of poker in the same order if using a wild card because the likelihood of having the same hand is still relatively the same. 1 wild card doesn’t drastically change the outcome of each hand in practical settings.

[@rattlehead999]

Three of a kind with a joker automatically becomes 4 of a kind too, and even more frequently than before full house.

[@stevenking3323]

There's also the added question of what if two players have the same hand, but one has the joker. Do they tie or should the joker have less weight? And should a natural 2 pair have greater or less value than a 3 of a kind with a joker?

[@lucusloc]

Yeah, this can be solved with one extra dimension of analysis. Since jokers are not a random card, they are a free choice, all joker hands need to be counted in a separate tally from the purely random chance hands. You can model this by adding every possible joker hand choice (52 possibilities for each joker hand) to the counts. This will have the effect of dramatically inflating the counts, but will not change the order of the hands (99% sure of this base on thought experiment, not actual numerical calculations). This will exactly model what would happen if we treat the joker as just a random extra card, which is actually what we want when we are discussing the probability of each hand. The fact that we get to *choose* the best hand in actual gameplay just means a lot of those possibilities will be ignored due to the *not* random nature of free choice. That does not mean those possibilities are impossible, someone could choose to make two pair instead of three of a kind, but it is acknowledged that we are not in a pure random chance xCy scenario anymore. To put it differently; there are still more possible two pair hands than three of a kind, it's just that you would have to be dumb to *choose to make* a two pair hand instead of a three of a kind. That is an entirely different argument than there being less possible two pair combinations in total.

[@omamba5105]

What happens when you add more possible winning hands, separating out the wild cards? For example, one winning hand is a royal flush using the joker and a separate winning hand being the royal flush without using the joker.

[@rsm3t]

The idea of a wild card is to become any card in the deck. To become that card, it must have the same power as that card.

[@ZaityGaming]

This is not a paradox. The ranking is based on the possibility of that hand to show up, not which one the player would choose. When the joker gets introduced, both two pair and three of a kinds possibility to show up increase and two pair is still more likely, therefore its still a weaker hand.

[@daviddumitru9102]

7:06 Wouldn't high card be 0 since you could always make 1 pair with the joker?

[@StrayCatInTheStreets]

Very simple to solve this if you made a game of it. 1 - The non-wildcard hands are ranked higher than their wildcard ones. 2 - No 5 of a kind, as there are only 4 kinds of cards (hearts, diamonds, spades and clubs). 3 - Wildcard has to be discarded in the flop/turn/river, wildcard is only in play the hole cards 4 - Three of a Kind Wilcard / Two Pair Wildcard are treated as the same rank and only beat the One Pairs. 1 Royal Flush, 2 Royal Wildcard Flush, 3 Straight Flush, 4 Straight Wildcard Flush, 5 Four of a Kind, 6 Wildcard Four of a Kind, 7 Full House, 8 Wildcard Full House, 9 Flush, 10 Wildcard Flush, 11 Straight 12 Wildcard Straight 13 Three of a Kind 14 Two Pair 15 Three of a Kind Wildcard / Two Pair Wildcard 16 One Pair 17 One Pair Wildcard 18 High Card

[@wqeasd7107]

Fixing the joker outcome to the optimal hand and calculating the odds creates an issue but allowing for the joker to be any of the 52 cards to calculate possibilities still gives a proper list.

[@Qermaq]

What changes if we say a natural hand beats an equal wild card hand? So Ac Kc Qc Jc 10c beats Ad Kd Qd Jd Joker, but a 10-high straight with a joker still beats a 9-high natural straight. So higher hand still always wins, but a hand with no joker beats equally-ranked hands using a joker. This might remove some incentive to upgrade one way or another. Not ever sure how to analyze this into formulas.

[@Filip.S.]

you should change perception of hands into 4 cards versions so you can have 4 of a kind -> 5 of a kind 4 for a royal -> royal flush 4 for a straight flush -> straight flush 4 for a flush -> flush 4 for a straight -> straight a trips -> full house two pair -> full house pair -> trips Nothing -> pair so the only hands that are possible are on the right, and you should reorder these by probability (adding probabilities of full houses together)

[@willguggn2]

That's only true if you're guaranteed to get the Joker. But you're not.

[@TheFinalChapters]

The paradox is easily solved by using the number of possibilities that yield each result *regardless* of higher ranks achieved. Really, this should have been done from the start, since it was only by chance that "high card" happened to have more possibilities than "one pair". But it would be silly to rank "high card" higher than "one pair", given every single instance of the latter also includes the former.

[@stephen2876]

With a Joker wildcard included there'd also be zero possibilities for a High Card hand because you'd always make a pair with your highest card of the four, with the Joker assuming the same rank.

[@maxevans5611]

There is another place where the paradox shows up. If you draw 3 of a kind + Joker + Other you can either make it 4 of a kind or a full house.

[@zaksmith1035]

Easy solution: those two hands are now of equal value. You can differentiate them by if both pairs are higher than the three of a kind the two pair wins, and if the three of a kind is higher than both pairs it wins. If the three of a kind is between the two pairs, chop the pot.

[@Liwet.]

In Poker, when you have more than 5 cards in your hand, you don't get to choose which cards play and which don't; the rules of the game dictate which cards play and will force you to use the 5 _best_ cards. Same with the Joker; you can't choose which value it takes; the rules of the game will dictate it. As shown at 4:44, the Joker would become a 5. Poker rankings don't assume a wild card is in play, only video poker payouts do. In reality, games that include a Joker only allow it to be wild if it completes a straight, a flush, or a straight flush. Otherwise it's considered an Ace, and 5-of-a-kind (only 1 possible way of getting it), beats a Royal Flush.

[@djdoc06]

Not really a paradox. It’s just not a well thought out initial question. Try THIS: it is MUCH clearer if you reframe the problem, and rank the probability of all the hands that have a wild card. [Because the “paradox” arises from lumping “natural hands” and “Wild hands” together] Instead, rank only hands that have a wild card to create probabilities, and you realize we actually have less classes of wild card hands than natural hands. We are currently taking “3 of a kind + WC” class and lumping it with “natural 4 of a kind”, but we could just as arbitrarily lump that in with “natural full house” before we figure out what the lumped group hand frequencies are. Likewise, We are currently taking “1 pair + WC” class and lumping it with “natural 3 of a kind”, but we could just as arbitrarily lump that in with “natural 2 pair” before we figure out what the lumped group hand frequencies are. It becomes very intuitive to recognize, how arbitrary the choice is of how to lump these “wild card hands” with natural hands.

[@TheAgamemnon911]

You could resolve the paradox by combining the two kinds of hand into one category. (in other words, two pairs against three of a kind results in a draw, so only highest card value decides)

[@goldenwarrior1186]

@@DrDeuteron Joker is higher than king

[@TheAgamemnon911]

@@DrDeuteron With the aces being equal, then the second highest cards decide. Those would also be aces. Then the third card decides: With the joker standing in for an ace (in this specific case, because it must be for the hand to be considered TOaK) it beats the king. Just to reiterate: The hands 'Three of a Kind', 'Two Pair' and '1 Pair + Joker' are all equal value and draws between them are only decided by who has the highest values on the board. Inelegant perhaps, but it eliminates the ambiguity of what you choose for the joker. It will always be the highest value needed to construct a hand in this combined category and the ranking paradox is avoided.

[@mulyadicahyono442]

Considering that the set of 1 pair plus joker always included in the higher group, then the natural 3 of a kind would still be higher. So the order should be the calculated order, except for the line of the 3 of a kind would be shown as natural 3 of a kind/1 pair plus joker.

[@OzixiThrill]

The problem entirely stems from a pointless exclusion of the probability space. The question isn't wether or not players will pick the "most optimum hand" (hint - no, they won't), but what possible ways there are to make a given hand. Is it possible to make two pairs with a pair and a joker? Yes. Thus excluding it from the possible outcomes because there is a higher value hand a player can chose is not exactly sound; Not to mention that it introduces a problem that doesn't even need to be there.

[@sabriath]

This is why most prefer the "pai gow" method....in that a joker can only be used to complete a straight or flush, otherwise it is an Ace only, nothing else.

[@realTornado2000]

Basic logic left the room here. The probability to create Two Pair or Three of a Kind with a Joker stays the same regardless of the ranking and how we want to use the Joker. This means Three of a Kind should always be with higher ranking because it is more difficult to get it with or without a Joker.

[@BR-lx7py]

Doesn't the same paradox exist between One Pair and High Card, considering how close their numbers were in the non-joker case?

[@dejanhaskovic5204]

Yeah but high card is guaranteed

[@Kyango]

I feel like the solution would be counting hands with a joker as a separate rank, just for those 2 ranks. i.e.... -3 of a kind (without Joker) -2 Pair (without Joker) -3 of a kind / 2 Pair (using Joker) The rest of the ranks could be played as normal, regardless of if you have a Joker or not.

[@ijustwannaleaveacommentony6511]

there's only 2/1 jokers so the probs almost remain the same, generally. you're almost never getting a joker

[@blacky7801]

4:28 this is the exact moment the logic faills. it's not about what you should, it's about what you could

[@caitsith2]

Another possibility, is to rank Joker + pair higher than natural two pair, but lower than natural 3 of a kind, since there is exactly 82,368 hands that gets a pair + joker, which can be turned into either three of a kind, or two pair.

[@riccardoscaglione9140]

The poker paradox: a paradox that doesn’t use poker rules

[@ericdculver]

Seems to me an interesting game would be a to deal out poker hands with wild card. Then have a bidding round (like in bridge) where players try to compete on who decides the ranking, Three of a Kind high, or Two Pair high. Maybe the stakes for bidding would be poker chips. Then we play poker based on the ranking chosen. Interesting because if you have one of those One Pair plus Wild card hands, you might care whether it becomes a Three of a Kind or a Two Pair based on what cards those are.

[@darkseraph2009]

As always, it is the case that paradoxes do not actually exist. They are only the effect of some sort of misunderstanding or incorrect assumption in our logic. In the case here you incorrectly assume that because one should not use the joker for a certain hand that one cannot do so. The possibilities must count everything one can do and in that case you find that three of a kind remains higher in the list then to pair regardless of the fact that one would not use the card for two pair. The fact that they can must be taken into account.

[@MrJronson]

Seems to me like the solution is for two pair and three of a king to be equal in value and depend on their high card to declare a win or draw

[@nicknamenescio]

I am not sure that logic at 5:14 fully checks out: counting the ways to get Two Pair could more reasonably involve only the deck AFTER the wild card has been changed into the best possible hand. That means Two Pair will not be chosen at all and is not part of the rankings anymore. It drops out of the set. A counterargument in the vein of "but, but why would people choose to make One Pair into Three of a Kind if they do not know what the real rankings are?" falls flat: you cannot have both simultaneously, at the same step of the process - a choice that determines the rankings and the rankings that determine the choice. That way you would enter a rather typical logical paradox like the classical set theory paradoxes. Still, such a paradox is not necessary. A paradoxical outcome sometimes simply means you got your processes not in order. In this case, it means you need to axiomatically establish a hierarchy of choices: a) you may either determine that the old rankings stay as they are and people choose according to them and ONLY AFTER such a choice you would determine what the new order is. Maybe still slightly paradoxical because such a new order might lead to different choices, but well, that is what our order of choice was for. That is the order that I have chosen in the first sentence above, rather arbitrarily, of course. I could have also chosen b): b) Or you may determine that you first see how many possible ways there are to reach a specific order, like Two Pair or Three of a Kind, and if more than one choice is possible, let all choices count for the total. Then make the rankings, creating a kind of singular pathway for people for those cases where before there was more than one option. All in all, the process might not be mathematically fully satisfying or completely elegant, but it gets the job done and avoids deliberately jumping into any unnecessary pitfalls. Those will only swallow you whole if you insist on the impossible.

[@adissentingopinion848]

With the invention of Balantro, a roguelike poker hand making game, the introduction of wild face cards, flushes with 4 cards, and so on, this suddenly becomes an applied problem. How much should you upgrade the two pair vs the three of a kind payouts if you obtain this wild card in your deck, including downtime where you do not have it and when you are allowed to mulligan your hand?

[@hongxiawang2049]

I'm surprised and not surprised that balatro was mentioned

[@penciljazz]

You didn't include the odds of getting a joker into the equation.

[@darkdudironaji]

If there are 54,912 ways to get a 3 of a kind, but 137,280 with a wild added, that means there are 82,368 ways to get a 3 of a kind with a wild. So just add "3 of a kind with wild" in between 3 of a kind and two pair. Rankings fixed. I'll take my Fields Medal now.

[@cliftonchurch6039]

Seems like a One Pair plus Joker should be ranked as its own hand, between a Natural Two Pair and a Natural Three of a Kind.

[@nightterror8775]

Now just imagine the probability when you play poker and get to swap out cards. I’d guess the possible combinations would rise to about 10^20

[@4400marko]

That's why poker is played without a joker...

[@VVeZoX]

By this theory, with a joker, there are zero possibilities of a 'high card' hand.

[@ParrotParrot]

The solution is to rank all hands with a joker lower or separately from hands without a joker. So that means having a pair and a joker now only beats a pair and lower.

[@nosteinnogate7305]

Why add "5 of a kind" as a new category but not e. g. "super royal flush" (10-2 suited) or "super straight"?

[@sillysad3198]

it's amazing how many comments solved this paradox correctly. now i am happy.

This just means the given way to compare hands isn't useful. What should be done is not force the Joker to make the hand the best possible, that is with a pair in hand a Joker should be a way to make both a three of a kind and a double pair. In this way we would be able to compare their probabilities and thus decide which combination is best. (Even though this would, in practice, make it so players would only choose to make the Joker so the hand has a three of a kind, there would actually be an argument behind the hand quality comparison.

[@gilliganallmighty3]

In the 3 of a kind v.s. 2 pair paradox, ranking a natural 3 of a kind v.s. one using a wild can be used.

[@TheAgentAPM]

Joker brings to Poker what green branches bring to Hackenbush.

[@jakebimbi]

the best way to resolve this is to play the best 6 card hand vs the best 5 card hand.

[@pbanthonyv]

You simply rank the hand based on the number of possible hands when the player attempts to achieve the hand. So three of a kind would be worth more in this situation.

[@davidkrcil9326]

I mean, this can be solved quite easily by not mixing in the choice element. Simply put - if desired the two pairs is more likely than three of a kind, thus three of a kind is more rare and is ranked higher.

[@glowstonelovepad9294]

Make it so that if you have a joker and 1 pair, you have to use the joker as a card you don't already havw.