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An Interesting Question about Toking3 August 2007
Recently one of the readers of Casino City Times, Andrew S., sent Frank a question involving the toking of Texas Hold'Em dealers. Frank sent the question to Bill Burton, who in turn sent it to me. So, Andrew, I guess you're stuck with me. Here is Andrew's letter to Frank:
The rules were not quite clear to me so I wrote to Andrew for some clarification. Having done this let me state the rules precisely. If on the flop at least two cards of the three ranks chosen show up, the dealer is toked $1. This includes two or three cards of the same rank. If three cards of three different ranks show up the dealer is toked $5.
To analyze this proposition I will use the notion of a combination and use the notation C(k, n) to represent the number of combinations of n things taken k at a time. I have discussed this idea before on this web site. FYI, the formula is
C(k, n) = n! / [(n - k)!k!] where n! = n(n - 1)(n -2) ... 3 x 2 x 1
To begin with, this proposition must be calculated using a full 52-card deck. You may think this strange since there are less than 52 cards in the deck when the dealer deals the flop. However since the dealer has seen none of the table cards at this point, the fact that the cards are on the table rather than in his deck is irrelevant. Here is another way to look at it.
Suppose we play the game a bit differently. The dealer deals out the three-card flop at the start of the game but doesn't turn the cards over. Then play continues as usual. The flop is exposed only after all of the initial two-card play has been completed. This game is equivalent to the usual Hold'Em game but it is clear that the flop has been dealt from a 52-card deck.
Next, note that the order of the cards in the flop is unimportant. That is why the analysis uses combinations rather than permutations. This remark and the remarks above apply to the second player's proposition as well.
How many ways can we get three of the dealer's choices on the flop? Well, there are twelve cards to choose from so there are C(3, 12) or 220 possibilities. However, 4 x 4 x 4 or 64 of these are flops with no matches. Thus there are 220 - 64 or 156 $1 tokes when three appear.
For exactly two there are C(2, 12) or 66 ways of choosing two of the twelve and C(1, 40) or 40 ways of choosing the other card; the product is 2640. For exactly one we have C(1, 12) x C(2, 40) or 9360 and for exactly none we have C(0, 12) x C(3, 40) or 9880. If you're wondering why I calculated these last two it is to check my calculations. All of my numbers, if correct, must add up to C(3, 52) or 22,100, which is the number of possible three-card flops. The results are tabulated below.
The expected return to the dealer is 3,116 divided by 22,100 and is approximately 0.141.
The calculations for the other player are similar to those above except that 5 cards are used. Here is my tabulation of the results:
The expected return to the dealer is 844,272 divided by 2,598,960 and is approximately 0.325.
Altogether the two propositions have an expected return of 0.325 + 0.141 or 0.466 per game. Assuming 20 games per hour this results in an expected return to the dealer of approximately $9.32 per hour.
So, Andrew, I hope this clears up any questions you have. See you next month.
Don Catlin can be reached at firstname.lastname@example.org
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