Three from Four Poker Rankings

I recently received the following letter from one of my readers. Here it is:

Dear Mr. Catlin,

I'm trying to find out the correct Poker ranking (frequency?) for a 3-card hand selected from 4-cards. We played this on a recent golf outing to South Carolina where it was called "Git You One" - each player being dealt 3-cards, then having the option to commit to the pot by accepting a fourth card. The player either accepted the card or pushed it on to the player on his left.

They played the regular 3-card hand ranking but I'm not sure that's mathematically correct. Can you help?

Thanks,
Don

Well, Don, I can try. The fact that you can essentially fold your hand might affect the playing strategy and that could affect the frequencies. Since I am not sure about the details I am going to ignore that possibility and produce the frequencies for a three-card hand chosen from four cards.

Poker hand rankings are determined by frequencies; the less frequent the hand the higher the ranking. These are tricky calculations, so tricky in fact that I prefer to do this using a computer program. So let me first describe the structure of the program; this is important in order to understand some of my later remarks.

My program begins by cycling through all four-card hands dealt from a fifty-two card deck. There are 270,725 of them. For each such hand there are four possible three-card hands. My program checks each of these and selects the highest-ranking hand. Do you see a potential problem here? We must assume a ranking in order to determine one and our assumption certainly affects the final outcome. Nevertheless, let's forge ahead and see what happens. I am going to assume the usual three card Poker hierarchy and run my program.

In order to check to see that the program is working properly I like to check a few of the easier hands to make sure the program produces the same results. For the Royal Flush there are four ways to choose the suit, only one way to chose the hand, and then we can fill the hand in with any of the 49 remaining cards. Hence we would expect to get 4 x 49 or 196 Royal Flushes. For the Straight Flushes we have 4 ways of choosing the suit and then 11 ways of choosing the high card in the hand. This done, there is only one way to choose the hand and then 48 ways to choose the remaining card. Why not 49? Suppose the hand was JT9 of hearts. The Queen of Hearts is one of the 49. If we choose that card then the Straight Flush would be the QJT and that has already been counted. Hence we would expect our answer to be 4 x 11 x 48 or 2112.

Finally, let's look at triples. There are 13 ways of choosing a packet of four equally ranked cards. This done there are 4 ways to choose three of these cards. If we make the fourth card one of the 48 cards unequal to the 4 we chose first then we have a hand with 3 equally ranked cards and one that is unequal to these. However, there is one more hand to consider, namely, all four cards equally ranked. Hence our calculation looks like 13[4 x 48 +1] or 2509.

Well, I think you can see how tricky this gets. Let's run the program and look at the results. Here they are:

The hierarchy in the table is consistent with our assumption. Suppose, however, we had assumed that flushes were better than straights. Let's try it. The program produces the same frequencies as before with the exception that straights occur 25,260 times and flushes occur 45,168 times. Should this surprise us? No! If we choose flushes over straights then any four-card hand that contains three-card hands of both types will choose the flush rather than the straight. We could have guessed the result without running the program.

The conclusion here is that if we assume a hierarchy and the program produces results consistent with our assumption then the frequencies produced by the program are the correct ones. Thus the above table is the correct table. Don, I hope this addresses your concerns.

There is an amusing footnote to all of this. I decided to see if I could hand calculate the number of Flushes. My calculation went like this. There are 4 suits each containing 13 cards. We can either choose 3 suited, C(3,13) ways [C(k, n) is the number of combinations of n things taken k at a time] which is 286 and then choose one of the 39 off suit cards or choose 4 of the suited cards which is C(4, 13) or 715. So I calculated 4 x [286 x 39 + 715] which is 47,476. Of course all of the Straight Flushes have already been counted; there are 2,308 of them [including the Royals]. So 47,476 - 2,308 = 45,168. This is not what my program calculated.

Is my program wrong? No, my calculation is wrong. Why? Where have we seen the number 45,168 before? When I ranked Flushes above Straights that is the number my program calculated. Aha. In a hand like 8C, 7H, 6H, 3H one should choose the three card Straight rather than the three card Flush. My calculation ignored this. I told you this stuff is tricky. See you all next month.

Don Catlin can be reached at 711cat@ccomcast.net