A Bizarre Poker Hand and a Math Puzzle

Ed Brayton shows a YouTube video of a very unusual hand. For fun, I'll include the video as well.


My puzzle is this. Obviously -- do I even need to say that? -- we in Vegas have thousands of video poker machines. If you've never played them then this may be unfamiliar to you. But there are two methods of handing cards based on the cards held out of the five original cards. With each of those methods comes a prababilistic model one can use to establish the probabilities of a winning hand. But the really interesting mathematical question is how to determine which of the two methods is used by the poker machines. Or can it be determined?

First let me briefly describe the game (we'll limit it to straight poker, but joker poker is essentially the same problem but with one wild card added to the deck -- deuces wild is the same as straight poker). You are given five cards, anything including or better than a pair of jacks wins. You can keep any cards dealt or dump all five cards. Replacement cards are then dealt, and you win or lose based on those final five cards. Thus, at most ten cards in play from a deck of 52 (or 53 with the joker). The question is how are those ten cards handled by the machine? The two methods of dealing those cards are:

1) five cards are dealt and then five replacement cards are used based on the number of cards discarded. Thus at most ten cards are in play, and the replacement cards merely move forward in the stack to replace the disdcarded cards. For example the ten cards are numbered (unrelated to their actual value) 1,2,3,4,5,6,7,8,9,10. If the player keeps two cards, let's say in position 3 and 5, then cards 6,7,8, move into positions 1,2,4.

2) five cards are dealt and five replacement cards are underneath the five original cards, as if there were actually five stacks of two cards, with the second card hidden underneath the first card. Going back to our example: 1(6),2(7),3(8),4(9),5(10). If the player keeps the same cards, 3 and 5, then cards 6,7,9 are revealed.

Now in case one, it is obvous that it would be impossible for cards 9 and 10 to be revealed, but in case two it would be impossible for cards 8 and 10 to be revealed.

So herein lies the mathematical interest. Which situation would give a player (or the house) a stastical advantage? And more importantly, without knowing how the cards are dealt out, can one determine which method is used, even if no advantage is conferred by either method? Your knowns are what I sketched out above, plus the fact that no hand can be replayed to see if different cards are given out in the exact same situation.

The answer to the first question is faily obvious I would think, it is the second question that makes this puzzle interesting.