Jack or Better Analysis

I was wondering if I could get your help on computing the probability distribution table for Jacks or Better.
I have been reading different books on the topic and none seem to resolve my issue.
I know that 52 choose 5 is 2598960, yet in every table that I have looked at for video poker, there are 19,933,230,517,200 outcomes. I was wondering why there are so many more outcomes and how to compute them.

Thanks.

I think one factor influencing the number of permutations is the draw. Your 2,598,960 number is correct for a 5 card stud game like Caribbean Stud, Let It Ride, or (most basically) 5 card stud. But JoB is a draw poker game, so there are additional cards to factor in. I'm pretty sure that's the "why." I have no idea how to compute the draw-dependent outcomes.

There are combin(52,5)=2,598,960 possible combinations on the deal. The reason my video poker return tables have almost 20 trillion combinations is you also have to consider what could happen on the draw. Here are the number of combinations according to how many cards the player discards.

0: 1
1: 47
2: 1,081
3: 16,215
4: 178,365
5: 1,533,939

The least common multiple of all those combinations is 5*combin(47,5)= 7,669,695. So the total number of combinations in video poker is 2,598,960 × 7669695 = 19,933,230,517,200 . For more on how to program video poker returns yourself, please see my page on Methodology for Video Poker analysis.

Thanks :)

one more question:
Why do the probability distributions vary for the different versions of Jacks or Better?
for example, why does a Royal Flush have 554,637,108 possibilities for 976-9-6 JoB but 500,790,444 for 90-9-6 JoB?

Wizard:

I looked at your linked page, and it made my head spin a little. I don't think all of that is necessary to clear up my confusion on this one point, so here's my question.

Suppose the rules limited the player to drawing a maximum of 1 card. I see combin(52,5) possible hands on the deal, which would still apply if 0 cards were drawn. For drawing 1 card from the remaining 47, it appears that you are saying there would be 47 times as many combinations possible. (Did I understand that correctly?)

Wouldn't there be 47 final combinations for each of the 5 dealt cards that you have the option of discarding to get that 1 out of 47 new card? Apparently I am missing something. Of course, my thinking goes on to find many more combinations under the real rules when 2 or more cards are drawn. Am I double (quintuple) counting here?

You have to weight the draw so that the total combinations works out to 7,669,695. For example, if the player drew 1 card, you would multiply the combinations by 7,669,695/47 = 163,185.

I have another question.
I read your analysis and I am confused as to why the total number of outcomes is multiplied by the LCM of combin(47,0) through combin(47,5) and not by the sum of combin(47,0) through combin(47,5).