VIDEO POKER

A Simple Problem at Triple Play

I was teaching at the Fiesta Rancho last Wednesday morning when two gentlemen approached me (either brothers or friends --- I'm not sure) and told me they had a disagreement about Triple Play and wanted me to arbitrate it for them. The first one said, "When you are dealt 3-of-a-kind, I know it is 23-and-a- half to one against ending up with 4-of-a-kind. He says the odds are triple that, namely 23-and-a-half to three, when you play Triple Play that you'll get a quad. I say it is more complicated than that. Whose right?"

The correct odds on each line are two chances in 47 (4.3%), or one chance in 23.5. The reason why you have two chances in 47 is that since you have been dealt five cards out of a 52-card deck, there are 47 left unaccounted for. There is only one card that will give you the quad, but you have two chances to get it. By way of terminology, that is 22-and-a-half to one or 23-and-a-half FOR one but not 23-and-a-half TO one.

If the chances of getting a quad are 2 out of 47, the chances to NOT get a quad are 45 out of 47. To figure the chances of getting a quad on the bottom line and non-quads on the upper two lines, we multiply 2/47 x 45/47 x 45/47 = 3.9%. There are equal odds to get a quad only on the middle line, or only on the top line. That is 3 x 3.9% = 11.7%. This is not the same as three times 4.255% (12.8%), which is what one of the questioners was voting for.

To get quads on 2 lines, the formula is 3 x 2/47 x 2/47 x 45/47 = 0.5%. The reason we multiply by the three at the beginning is that there are three equally likely ways we can get the two lines --- bottom and middle, bottom and top, and middle and top.

To get quads on all three lines, we multiply 2/47 x 2/47 x 2/47 = .008%, which is very slightly more than forget-about-it. To get three non-quads, you have 45/47 x 45/47 x 45/47 = 87.8% which is very close to seven times out of eight.

Since from this starting position there are only four possibilities relating to the number of quads, i.e., 0 quads, 1 quad, 2 quads or 3 quads, the sum of these probabilities must equal 100%. And they do. 87.8% +1.7% + 0.5% + .008% = 100%, allowing for rounding.

What if the question were how many full houses would you get starting from trips? There are 66 different ways (out of 1,081 possibilities), or 6.1%, to get a full house. You can count these out if you like, but it's easier to look on Bob Dancer Presents WinPoker or another computer program for the number. To not get a full house it is 100% - 6.1% = 93.9% = 0.939.

Using the same technique as before, to get zero full houses, it is .939 x .939 x .939 = 82.8%. To get one full house, it is 3 x .061 x .939 x .939 = 16.1%. To get two full houses it is 3 x .061 x .061 x .939 = 1%. To get three full houses, it is .061 x .061 x .061 = .02%. And as before 82.8% + 16.1% + 1% + .02% = 100%, allowing for rounding.

This isn't tough math, but I've had enough questions about it that I thought I'd spell it out.