Imaginary poker game probability question
 

Hold'Em and 7 card stud are games where you make your best 5 card hand from 7 cards. In this game, you are more likely to get a straight than a flush. What if we created a game of 9 card stud, where you made your best 5 card hand from 9 cards? Are flushes still less likely than straights? What if it was 11 card stud?

I am convinced that eventually flushes become more likely because in 17 card stud, you would always have a flush, but not always have a straight. In fact, you could be dealt 44 cards and still not have a straight.

I lack the required probability expertise to calculate the odds. Can someone teach me?

Re: Imaginary poker game probability question
 

Yes you're right that if there are 17 cards , then you'll always be able to make a flush from the pigeonhole principle . 17=4*4+1 .

Lets take a look at a 9 card stud game where you can select 5 cards to make the best hand .The number of hands in which you have a flush in clubs is :
13C5*39C4/52C9 = 0.028
13C6*39C3/52C9=0.00426
13C7*39C2/52C9=0.000345
13C8*39C1/52C9 = 0.0000136
13C9*39C0/52C9= 1.94*10^-7

The same is true for all the other suits so we multiply the total by 4 . Since you're dealt 9 cards , there is no possible way for you to be dealt 2 different flushes , so we can simply take 0.0326*4= 0.13 to be the probability that you'll get a flush .

The number of ways of getting a straight in 9 stud is more involving and has to be broken into cases . I can only think of a brute force way of figuring this out and it's very time consuming . You would have to figure out the total number of ways of having exactly 5 card straights , 6 card straights ... 9 card straights and divide this total by 52C9 .

For instance , AAAAA2345 would be exactly 5 cards to a straight . Notice that you can have multiplicities in rank . If anyone else can think of a better way , let me know .

Re: Imaginary poker game probability question
 

This is off the top of my head, so if I am wrong - sorry.

How about figuring out the total number of possible hands that can be dealt w/ 9 cards (52C9) to start. Now we need to figure out how manu of those hands contain straights.

if we know how many possible 5 card straights there are (this should be easy to calculate or look up on the internet). For each straight, we have (52-5)C4 possible ways of making the straight. We then multiply this by the number of possible five card straights. Finally divide this by the number of 9 card hands that we first calculated.