VIDEO POKER

Still Another Interesting Question in Deuces Wild

In my March 19 column (available a few mouse-clicks away), I asked the following question: Is it better in Deuces Wild to start from a natural 4-of-a-kind or a 4-of-a-kind including one wild card? The answer, it turned out, was that both starting combinations were equal.

A reader, Alan Peterson, emailed me and brought up a similar kind of hand. In all Deuces Wild games where the one coin return for full houses is 3, you play a hand such as 55773 by holding either the pair of fives OR the pair of sevens, but not both. In a hand such as 55678 you would also hold the pair. Peterson's question was as follows: WinPoker shows that holding a pair while starting from 55773 is slightly more valuable than holding a pair from 55678. Why?

I found the question interesting. I didn't know the answer offhand, so I needed to work it out. Perhaps a discussion of how I worked it out would be useful to some players.

It was obvious to me at the start that drawing 3-of-a-kind, 4-of-a-kind or 5-of-a-kind would be equally likely whether we held the fives from 55773 or 55678. After all, the only cards we are concerned with in these cases are the fives and the deuces, which I will refer to as "hits", and the other 41 cards in the deck are equally "misses". So to get a 3-of-a-kind, we need to figure out the odds of getting 1 out of 6 hits and 2 out of 41 misses --- and then subtract off the full houses that arise when one miss is the same rank as the other miss. Turns out you have 4,578 out of 16,215 (about a 28% chance) ways to do this. How do I know this? I looked it up on WinPoker. Some other time I'll explain how WinPoker came up with these numbers. The number of full houses, however, can vary between the starting positions. To start with, let's look at 55773. It should be obvious that if we just hold the fives, it is equally easy to get a full house with fives and nines as it is with fives and tens, or fives and jacks, or fives and queens, etc. It is harder to get a full house with fives and threes because there are only three 3s still in the deck. And it will be harder still to get a full house with fives and sevens because we are throwing away two sevens. So let's count these types of hands separately.

a. 55577 --- There are two ways to get this hand. If we started with the two black fives, for example, we could draw the 5h and the remaining two sevens and we could draw the 5d and the remaining two sevens.
b. 55W77 --- Remembering that I use the W to indicate a wild card (in this case a deuce), we have 4 ways to draw this hand. Why four? Because there are four deuces, and each of them can be paired with the remaining two sevens.
c. 55777 --- This hand is impossible, simply because there are only two sevens remaining in the deck.

So of the 16,215 hands we can draw by holding the fives from 55773, 6 of them will be full houses with fives and sevens.

Now let's look at full houses with fives and threes --- remembering that we discarded a three before the draw.

a. 55533 --- There are six ways to get this hand. There are the two ways to draw a five and three ways to draw a 3, so there are 2x3=6 ways they can be combined. Why three ways to draw a 3? Let's assume we started with the 3d and threw it away. The three remaining threes would be the 3h, the 3s and the 3c.
b. 55W33 --- There are twelve ways to get this hand. Two different fives times four different deuces times three ways to get the 3s.
c. 55333 --- There is exactly one way to get this hand. There are only three 3s in the deck and we are drawing three cards.

Adding these together, we come up with 19 ways out of 16,215 to get a full house involving fives and threes when we hold 55 from 55773.

Now let's look at, for example, a full house involving fives and kings. Since there were no kings in the initial deal, all four of them are still in the deck.

a. 555KK --- There are twelve ways to get this full house. The two ways to draw a 5 times the six ways of drawing KK. Why six? Count them. There are KcKd, KcKh, KcKs, KdKh, KdKs, and KhKs.
b. 55WKK --- There are 24 ways to get this full house. The four deuces times the six ways of drawing a pair of kings.
c. 55KKK --- There are four ways of drawing three kings when you start with four of them, i.e. KcKdKh, KcKdKs, KcKhKs, and KdKhKs. An equivalent way to count them is all except the Ks, all except the Kh, all except the Kd and all except the Kc.

So there are 40 (i.e. 12 + 24 + 4) ways of getting a full house with fives and kings. Note that from 55773, there are 9 ranks equivalent to the kings --- i.e. A, 4, 6, 8, 9, T, J, Q, and K. None of these cards were in the original five.

To add this all up, we have 6 ways with 5s and 7s, 19 ways with 5s and 3, and 9x40 = 360 ways to get a full house with 5s and a rank not originally dealt. This adds up to 385 ways out of 16,215 to get a full house.

From 55678, we don't have to start all over again. We have already worked out that you'll get a full house with the 5s and either the 6, 7, or 8 nineteen times each. The logic for this is exactly the same as the earlier discussion of fives and threes --- because in all of these cases there are three cards of the designated rank still in the deck because we threw away one. And to get a full house with fives and a rank not originally dealt, there are 40 of these --- for exactly the same reason that there were 40 full houses with fives and kings earlier. There are eight ranks not originally dealt (i.e. A, 3, 4, 9, T, J, Q, and K), so we have 3x19 + 8x40 = 377.

Since there are 385 full houses holding a pair from 55773 and there are only 377 full houses holding a pair from 55678, then the first is slightly more valuable. Not earth shattering news, to be sure. This is the kind of situation where understanding the process by which the answer is obtained is more valuable than the answer itself.