Re: Another poker probability problem

Keith:

Thanks for the probabilities of the end state. I'm still looking for
the probabilities of the initial state - getting dealt three to a
straight flush (or even a straight for that matter; I think the
methodology is the same) in one's first four cards.

I understand that a straight flush is composed of exactly five cards so
getting dealt three towards any one straight flush is 5-choose-3 which
yields ten combinations. Where I start to lose my understanding is
where I know that there are ten distinct straight flushes in each suit,
so 5-choose-3 = 10 x 10 straight flushes = 100 combinations of the
first three cards, however I'm sure that calculation counts certain
combinations more than once (e.g. 4,5,6 of spades would be a valid
combination in three of the ten available straight flushes).

The two complicating factors are the fourth card *and* figuring out how
many distinct combinations of 5-choose-3 there are given that you're
choosing three cards out of five required to make the straight flush
out of 13 suited cards which can compose ten distinct straight flushes
- ten straight flushes which all overlap with each other.

If the three cards we end up choosing are the 6, 7 and 8 of spades and
I'm looking for the probability of choosing exactly three cards to a
straight flush (not four cards), the choices for the fourth card are
limited to 45 out of the remaining 49 cards as the 4, 5, 9 and 10 of
spades are undesirable (they would give us four cards to a straight
flush). So that yields 45 four card hands containing the 6,7,8 of
spades and a fourth card which would not further progress towards a
straight flush.

However if the three cards we end up choosing are the Ace, Deuce and 3
of spades, the number of acceptable selections for the fourth card is
now 47 cards as there are only two cards of the remaining 49 cards
which would further our progress towards a straight flush; the 4 and 5
of spades. So that yields 47 four card hands containing the ace, deuce
and three of spades and a fourth card which would not further progress
towards a straight flush.

I could continue such brute force counting, but that is quite labor
intensive. I could also write a program to enumerate all 270,725 four
card hands and examine them manually to get an answer - I'm actually in
the process of doing this but there would be situations I would like to
analyze which will yield much larger sample spaces which makes such
brute force inspection impractical.

So I'm trying to learn a quicker way than brute force; it will aid in
the development of strategy for other poker games that I might want to
play (like, say, Omaha - where you can play both high AND low, you get
four cards to start of which you must use exactly two in your high or
low hand).

I think I understand almost everything except the overlapping nature of
the straight and how to account for it in an algorithm.

I know I'm missing something... I just don't know what.

Thanks for bearing with me,
-Chuck

.

Relevant Pages

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