# Straight Flush probability with a huge hand.

It's easy to calculate the probability of a straight flush when you're dealt $5$ cards. I'd like to ask for the probability of the same when you're dealt half the deck.

I seek $P(\text{straight flush})$ of any suit. We have $52$ cards, and $26$ are dealt. A straight flush is $5$ consecutive cards of the same suit. An Ace may come either before a $2$ or after a King but not both.

• This actually seems like a job for a recursive process. I'm unsure whether it would be easier to work out the probability of a straight flush for $n$ cards drawn and increase it towards 26, or to always draw half a deck and grow the deck from 20 cards to 52. – Glen O Jun 7 '15 at 15:36

It may be possible to get a computer to count it.
Each suit is disjoint from the other suits.
We only want the probability of zero straight flushes.
The 'distribution' is how many spades, how many hearts, and so on.
$$P(a,b,c,d)=\frac{{13\choose a}{13\choose b}{13\choose c}{13\choose d}}{52\choose 26}$$ Then, given $x$ cards in a suit, what is the chance of no straight flush in that suit? Some easy values are $Q(0)=Q(1)=...Q(4)=1$, and $Q(13)=0$. There are $2^{13}=8192$ subsets of thirteen cards. It is simple to get a computer to run through all 8192, counting how many cards and whether there is a straight flush. Lastly, get the machine to calculate $$\sum_{a+b+c+d=26}P(a,b,c,d)Q(a)Q(b)Q(c)Q(d)$$ There are $29\choose3$ ways that four non-negative numbers can sum to 26; the cases with $a>13$ can be covered by setting $Q(14)=...=Q(26)=0$.
The reason for $29\choose3$ is a stars-and-bars argument. You have 26 cards, and three suit-dividers. They fill 26+3=29 places. Choose the three spots for the suit-dividers, and put unmarked cards in the other 26 spots. Now you know how many of each suit there are. All those to the left of the first divider are spades; all those between divider 1 and divider 2 are hearts, then diamonds; and all those to the right of the third divider are clubs.

You've posed a difficult problem my friend. These types of problems are normally solved with inclusion-exclusion. Consider the event of the A2345 of hearts being included in this 26 deck hand. It is not too difficult to compute the probability of this specific event arising, by counting the number of hands including these five cards, and dividing by the number of all hands, getting $$\frac{{47\choose 21}}{{52\choose 26}}\approx 2.5\%$$

There are 35 similar events, corresponding to the 36 possible straight flush hands. However, what you asked was the probability of at least one of these events happening, i.e. the probability of the union of these 36 events. Enter the inclusion-exclusion principle. Now we need to compute the probability of the intersection of every subset of the set of these 36 events. For example, the intersection of the event above, and the event "56789 of hearts", is the event "A23456789 of hearts", which has probability $\frac{{43\choose 17}}{{52\choose 26}}$. Or, the intersection of the event "A2345 of hearts" and "56789 of clubs" is the event of both of those, which has probability $\frac{{42 \choose 16}}{{52\choose 26}}$.

In easier problems, these subsets have a nice pattern to find, and/or they quickly max out in size. For example, if you were drawing a deck of six cards, TWO of the events might happen at the same time, but no more. Hence the probability of all subsets of size three, four, ..., 36 is all zero.

In the OP however, the calculation of these probabilities has many tedious cases, which quickly devolves into tedious subcases. The intersection of a pair of events is not too bad -- some pairs of five-card hands intersect in zero cards, some in one, some in two, some in three, and some in four. Each of these probabilities is not too bad to compute, and I've done two examples above. In a few minutes we can handle all ${36\choose 2}$ pairs of events.

But we now need to consider the intersection of three of these events. Now it's quite complicated indeed. Perhaps the three hands are all nonintersecting. Perhaps the first two hands intersect in two cards, but no other intersections occur. Perhaps the first two hands intersect in one card, and the last two hands intersect in one card as well. And all of these can happen, because with 26 cards there is a lot of room for cards! Hence this method suffers from combinatorial explosion and is more or less hopeless to compute by hand, without careful calculation of many possible cases that could take hours. Sadly a computer can only help with the arithmetic, not with the modeling.

A related question that is easily answered is the expected number of straight flushes in your 26 card hand. This can be found using linearity of expectation, hence we multiply the expected number of the (A2345 of hearts) event, by 36. The result is $$\frac{759}{833}\approx 0.9111$$ i.e. on average you will find approximately $0.9$ straight flushes in each hand of 26 cards. However this averages hands with zero, and hands with seven, in a manner that can only be fully understood by using the inclusion-exclusion principle as above.