A poker problem
In Straight 5 card poker every possible hand is ranked inversely to the probability of being dealt. For example, the royal straight flush is the least likely hand ( 2 5 9 8 9 6 0 4 ) and beats all others.
In calculating the probability for each of the various hands, it is clear why a four of a kind beats a full house . The specific probability of getting a four of a kind ( 2 5 9 8 9 6 0 5 2 4 ) is lower than that of getting a full house ( 2 5 9 8 9 6 0 3 7 4 4 ) .
If the ranking of poker hands must ensure that a hand with lower probability ranks above one that is easier to get, consider what happens when two wild-card jokers are introduced to the deck.
Among four of a kind and full house , which should now be considered the higher ranking hand?
Note: Wild-card jokers can represent any other playing cards.
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1 solution
If 2 jokers are added to a standard deck of 52 cards, the odds of getting a four of a kind and a full house will be exactly the same. Without the jokers, the full house is more likely, which is why four of a kind beats a full house.
This is based on drawing 5 cards from a shuffled deck that hasn't been drawn from yet, and it's assumed that the player wants the highest value hand that can be got from the jokers, if he gets any.
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The point is whichever hand is ranked higher becomes easier to get, therefore it's not possible to rank them consistently.
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Okay, this is an interesting subject, because it's not about odds of drawing a full house or a four of a kind from the deck, supplemented with wild cards. It's about "okay, who should get the pot?" Let's start with the simpler scenario where only 1 joker has been added to the standard 52 card deck. Then we can still keep the ranking of the hands, where four of a kind beats a full house. Since there is only 1 joker, whoever has it can use it to his advantage and legitimately use to create a four of a kind if he can, and beat a full house.
The fun starts when there's 2 jokers, because then it's possible that we have 2 players that each have a joker, a three of a kind, and a 5th unmmatched card. Who wins? How is that decided? If the players still abide by the traditional ranking of hands, then both players will strive to form a four of a kind, and the one with the higher card takes the pot, and there is no argument. In spite of the fact that getting a full house is equally likely.
The key is the way the question is worded, "what hand should now be considered to be the higher ranking hand?" A game is a game, and people can agree to rules that aren't necessarily logical, as for example saying that a three of a kind still beats a pair even when there's a joker in the deck (it actually becomes easier to get than a pair!). Rules nevertheless still need to be agreed to before play. To be more fair and consistent, the ranking of hands should be altered to fit the number of jokers in the deck, based on mathematical odds, but we do have this oddity that for 2 jokers, the odds of getting a four of a kind and a full house is exactly the same. In that case, some tiebreaker, like having the highest card, will have to decide.
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@Michael Mendrin – I get what you are saying but I'm not sure this is where I want to go with the problem.
I mean for there to be a key condition for ordering hands: That a lower probability hand is ranked higher. Then the question of what should be the lower probability hand becomes impossible to answer. (Partly because they have the same probability if considered separately and partly because the one ranked higher becomes more likely.)
Is the current wording not good? I see it has 0 solvers in 12 attempts, so maybe not.
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@Jeremy Galvagni – To me this is an interesting problem, but it's not defined well enough. For example, if we simply ask "given 2 jokers in the 52 card deck, what's the probability of drawing a full house, and drawing a four of a kind?" Then that's readily analyzable with an unambiguous answer. But you are suggesting that the game could end up resembling the sissors, rock, paper game, and I'm trying to figure out how to do that. Let me think of a simpler game with much fewer cards that could be like that. Here's an extremely simple game: There's only 2 jokers in the deck.. that's it, just 2 cards in the deck. Both players get a joker. One puts it down and declares what it will be... say, a rock. The other will put down his joker and say it's a paper! I think this is where you are trying to go with this. One cannot assign a ranking for sissors, rock, and paper.
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@Michael Mendrin – For your game, I'd ask "which is best, rock, paper, or scissors?" Hence the answer "It is impossible to determine."
Here was my reasoning that a four of a kind is still the better hand even with two jokers added:
Ways to get 4 of a kind that can't be a full house:
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natural 4 of a kind with 1 unmatched card: ( 1 1 3 ) ⋅ ( 4 4 ) ⋅ ( 1 5 0 ) = 6 5 0 ways
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natural 3 of a kind with 1 joker and 1 unmatched card: could be a full house, so not counted
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natural 2 of a kind with 2 jokers and 1 unmatched card: could be a full house, so not counted
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total: 6 5 0 ways
Ways to get a full house that can't be 4 of a kind:
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natural full house with no jokers: ( 1 1 3 ) ⋅ ( 3 4 ) ⋅ ( 1 1 2 ) ⋅ ( 2 4 ) = 3 7 4 4 ways
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natural 3 of a kind with 1 joker and 1 unmatched card: could be 4 of a kind, so not counted
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natural 3 of a kind with 2 jokers: could be 4 of a kind, so not counted
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natural 2 pairs and 1 joker: ( 1 1 3 ) ⋅ ( 2 4 ) ⋅ ( 1 1 2 ) ⋅ ( 2 4 ) ⋅ ( 1 2 ) = 1 1 2 3 2
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natural pair with 2 jokers and 1 unmatched card - could be 4 of a kind, so not counted
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total: 1 4 9 7 6 ways
There are less ways to have 4 of a kind (that can't also be a full house) than a full house (that can't also be 4 of a kind), so 4 of a kind should still be ranked the higher hand.
Where did I go wrong?
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What do you do with the 'not counted' ways? There are a lot of them and they have to count for something don't they? My argument is if you get to choose what kind of hand they become, you would want to count them as the higher value hand. Then this hand is not the harder of the two to get. It's sort of a paradox.
One way out of it would be to value the hands the same. You seem to be saying that being dealt say a natural 3 of a kind with 1 joker and 1 unmatched card should count as neither.
(Incidentally, 3 of a kind with 2 jokers is 5 of a kind so really should not be counted.)
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Why can't we count a 3 of a kind using 2 jokers? The whole point of having jokers in the deck is to make it easier to get the higher ranking hands, making the game more lively, if less professional.
I've gotten Royal Flushes in my lifetime, but I'm sure it was helped with wild cards.
Furthermore, remember, players are allowed to discard cards and ask for replacements, which alters the odds of getting the higher ranking hands.
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@Michael Mendrin – I think you're getting off topic. Yes, high ranking hands are easier, but so are low ranking hands. I'm asking about two specific hands. The idea for this problem comes from https://www.tandfonline.com/doi/abs/10.1080/0025570X.1996.11996455
In draw poker, you can get replacement cards. This is straight poker.
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@Jeremy Galvagni – Jeremy, as I said, a game is a game and it has agreed "rules"---however nonsensical the rules may be. But I do understand your point...if we're going to have straight poker with 2 jokers added, and if we insist that the the ranking of the hands are to be determined by probability of gettting them, then we have a paradox. Let me think of a simpler version of this to make this more apparent. Very interesting, for a lot of reasons.
Somehow I think the wording of your problem needs to be reworked, so that readers can better understand this? That itself is already a problem, how to word it just right?
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@Michael Mendrin – I'm glad you agree there's a good problem here. Problems can be so hard to word correctly, as we all know. I appreciate your help so far. If one of you would please report this problem as unclear I can work on it there. That seems like a better place. Apparently, I can't report my own problem.
David's link goes into detail with a very closely related paradox.
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@Jeremy Galvagni – the strangest thing about this problem, I'n not even sure if it's even improperly worded! Very strange problem, indeed!
@Jeremy Galvagni – Here is a website that supports Jeremy's conclusion: http://datagenetics.com/blog/september32016/index.html but for a similar scenario comparing the rankings of 2 pairs versus 3 of a kind where there is one joker.
It's actually counted for both, but we can set the "not counted" aside since it would be in the middle ground of the intersection in the Venn diagram.
Strictly 4+1 = 13 × 50 = 650
Strictly 3+2 = 14976 (from David's)
3+1+J = 52 × 48 × 2 = 4992
2+1+JJ = 78 × 48 × 1 = 3744
3+JJ = 52 × 1 = 52
Total ambiguous hands = 8788
Let's say you want to lump all the ambiguous hands into 4+1, even with your own assumptions of 5 of a kind trumps 4+1 above, the total probability is still lower for strictly 4+1 and ambiguous hands since the count is 650 + 8788 = 9438 < 14976.
It is impossible to rank these two hands according to which is less likely.
Consider being dealt a hand with a three of a kind as well as a joker and some unmatched fifth card. The joker can be used to make either of the two hands. There are so many such hands that whichever is given the higher value, it becomes easier to get!