Joined: Jan 11, 2013
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January 11th, 2013 at 11:37:21 AM permalink
This ridiculous hypothetical question came up in an argument about a totally different card game, but I am completely stumped as to how to even start figuring out the answer, much less what that answer is.

Alex and Burt are sitting down to play a game of poker. For some reason -- maybe both of them are notorious for spilling drinks and don't want their deck ruined by the other guy, I don't know -- each of them will be drawing from their own deck of cards instead of one community deck. Alex brought a standard deck of 52 playing cards, with an Ace through King of each of the four suits. Burt, because he is a jackass, opened up several identical decks at home and combined them so that his deck has 52 cards, but some of them are duplicates of the same card, and some are left out.

Let's say, he has 2 each of the ten, jack, queen, king, and ace of each suit, and he does not have the two, three, four, five, or six of each suit. Burt does not want to reveal what a cheating jerk he is, so when he reveals his hand, he will never reveal two copies of the same card; if he has two copies of the queen of hearts and one queen of diamonds, he will just flip over the QD and one QH and claim one pair of queens, and discard the other one unseen by his opponent. Also, assume Burt is quick enough to hide any card he discards so that Alex can't see it's a duplicate of something in his hand, because that makes the hypothetical scenario work.

My friend says Burt has no advantage in this situation, as overall the odds of him drawing the second copy of a card he needs to complete a hand when he doesn't have the first are always equal to the odds he draws a duplicate copy of a card he already has. I think that if they are playing five-card draw, this is true, but that if they are playing stud poker, where Burt draws more than five cards and then discards to the best five-card hand, Burt has an advantage.

I have no idea whatsoever how to compute what that advantage would be.

Short version: How do I calculate the odds of drawing each hand from a deck of 52 playing cards that has been modified to contain multiple copies of certain cards and no copies of others, with the stipulation that no "claimed" hand can have duplicates of the same card? Would those odds be more advantageous than the odds for a normal deck of cards in a game of six, seven, eight, etc. -card stud?
Joined: Mar 11, 2010
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January 11th, 2013 at 11:55:48 AM permalink
I think Burt always has an advantage - he's much more likely to get paired cards. Here's an example:

Let's say both players start with 3 queens of different suits. Both players discard the other two cards. Alex has a 1 in 47 chance of getting the 4th queen on the first draw card, and if he doesn't get it there, he has a 1 in 46 chance of getting the final queen on the second draw card. Burt, that cheating jerk, has a 2 in 47 chance of getting a queen in the 4th suit on the first draw card and a 2 in 46 chance of getting it on the second draw card. Burt is twice as likely of winding up with 4 of a kind, even if it is still a longshot for him to do so.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
Joined: Aug 31, 2010
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January 11th, 2013 at 12:05:59 PM permalink
Short answer: the typical combinatorial analyses used for poker hands assume equal card frequencies, but the basic idea can still be used with unequal card frequencies. You just need to factor those in. E.g. the probability of a Royal Flush is normally
C(4,1) / C(52,5)
or in English
Pick one of 4 suits * only 1 way to make a royal straight / total possible combos
Let's expand that:
C(4,1) / C(52,5) = C(4,1) * C(5,5) / C(52,5)
Because C(5,5) = 1, and that basically says that you need to pick all 5 of the 5 royal ranks (A..10)

In your deck the chances of a Royal would be
C(4,1) * C(5,5) * C(2,1)^5 /C(52,5)
which means
Pick one of 4 suits * pick 5 of 5 royal ranks * for each royal rank, pick one of the 2 available cards / total possible combos

The general rule is that wherever more than one option is available to fulfill a criterion, that option needs to be accounted for in the numerator.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice." -- Girolamo Cardano, 1563
Joined: Jan 11, 2013
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January 11th, 2013 at 6:36:11 PM permalink
Thanks for the help! Edifying.

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