How do I calculate the probability of getting a 5 card straight in 7 card poker?

Question

So here's my progress on this problem. The first straight to examine is $\begin{pmatrix} A & 2 & 3 & 4 & 5 & n & m\end{pmatrix}$. If we start with all one suit, there are $\binom{47}{2}$ ways to obtain this hand. Next, we iterate through the suit of one card, say, the Ace, and we have $\binom{46}{2}$, then $\binom{45}{2}$, then $\binom{44}{2}$ ways to obtain this hand without repeating previous hands. To iterate through all suits of all cards, take the sum \begin{equation} \sum_{k,l,m,n,p=0}^{3}\binom{47-k-l-m-n-p}{2} \end{equation} and subtract $4\binom{47}{2}$ to discount straight flushes. Next, to count the hands of $\begin{pmatrix} 2 & 3 & 4 & 5 & 6 & p & q\end{pmatrix}$, $\begin{pmatrix} 3 & 4 & 5 & 6 & 7 & r & s\end{pmatrix}$, etc., we start with $\binom{43}{2}$ to obtain the first hand, so as not to double count hands of the form $\begin{pmatrix} A & 2 & 3 & 4 & 5 & 6 & m\end{pmatrix}$. Therefore we add 9 hands of the following form, subtracting the possible straight flushes. \begin{equation} 9 \left( \sum_{k,l,m,n,p=0}^{3}\binom{43-k-l-m-n-p}{2}\right) - 9\binom{46}{2} \end{equation} These sums may be reduced to one dimension by observing that the weight of each possible sum, r, for $k+l+m+n+p=r$ is equivalent to the coefficient of each term $x^r$ in the expansion $(x^0+x^1+x^2+x^3)^5$. The problem is that these equations describe 6412688 possible straight hands while there should be 6180020 possible hands (according to Wikipedia). I can not for the life of me figure out what hands I am double counting here. Any help is appreciated.

Answer 1

Well, in an ordinary 7-card poker, the standard deck card number is 52. I’ll do the calculations with that value. If you want to make it with another one, just replace it. This being said:

$\binom{52}{7}$ is the number of possible combinations of hand drawings in a 7-poker (please, excuse me if do not use poker’s terminology: i do not know it). If you want to get straights, you need 5 consecutive cards of any suit. There are four suits, so you have $4^5$ possible distinct suit combinations (think how the possible bit strings are calculated in binary). There are 10 disctinct straight number combinations, including [10,J,Q,K,A]. Summing all up, you have an amount of $$C_{(5)}=10\cdot4^5$$ possible 5-card straights combinations.

To get the probability of getting a 5-straight in a 7-poker game, we must consider that every one of the $\binom{52}{7}$ combinations of hands can contain one, and only one of the 5-straight combinations, but some of the 5-straight combinations will be repeated on the set of possible hands (i.e. [A,2,3,4,5,Q,Q] and [A,2,3,4,5,K, K], all hearts). When we have a 5-straigth in our hand, we are left with $52-5=47$ cards on the deck. Every 5-straight combination is repeated for each possible combination of the 2 additional cards, so we can consider only two-card combinations. This means that the number of possible 5-straight combinations per 7-hand is: $$C_{(7)}=C_{(5)}\binom{47}{2}$$ And so, the probability of getting one 5-straigth on a 7-hand game is: $$\frac{C_{(7)}}{\binom{52}{7}}=\frac{10\cdot4^5\binom{47}{2}}{\binom{52}{7}}≈0.083$$ It’s not a horrible bet to do, having in account that this is a 8.3% chance of having a straight in your first hand-drawing.