Find the probability of flopping a straight, given you have 2 hole cards that can flop a straight

Question

Find P(flop straight | dealt 2 hole cards that can flop a straight)

At first I simply did: $\frac{12}{50} \times \frac{8}{49} \times \frac{4}{48} = 0.003$ but that seems awfully low.

Then I tried combinatorics with ${}^{11}C_3 \times {}^{4}C_1 \div {}^{50}C_3 = 0.037$

I then realize that some hole cards have a bigger range. E.g: 2-6 suited must flop 3-4-5 to get a straight, whereas J-10suited can flop Q-K-A, or 9-Q-K or 8-9-Q or 7-8-9

How do I solve this problem? Thank you.

Answer 1

It isn't clear to me what game is being played, so I assumed that the game is five-card stud. It will be easy to adapt the calculations to a different game, since the only difficulty is counting the number of two-card combinations that could result in a straight.

Let $T$ be the event that the first two cards could be part of a straight and let $S$ be the event that a straight is flopped. We are asked to compute $$\Pr(S|T)=\frac{\Pr(S\cap T)}{\Pr(T)}=\frac{\Pr(S)}{\Pr(T)}$$

$\Pr(S)$ is easy to compute; there are $10$ possible straights, from $5$-high through Ace-high, and each can occur in $4^5-4=1020$ ways, so $$\Pr(S)=\frac{10200}{\binom{52}{5}}=\frac5{1274}$$ To compute $\Pr(T)$, we need to know the number of two-card patterns that could be part of a straight. Say there are $t$ such patterns, ignoring suits. Then $$\Pr(T)=\frac{16t}{\binom{52}{2}}=\frac{8t}{663}$$

The problem is to compute $t$ systematically, so that we can be sure we have counted each pattern exactly once. I think we ought to organize the two-card hand according to the number of possibilities (that is, 3-card combinations, ignoring suits) to complete the straight.

• One-chance hands: A$2$ through A$5$, $10$A through KA, and $x,x+4$ where $x$ runs from $2$ through $9$, giving $16$ patterns.
• Two-chance hands: $23$ through $25$, $10$K through QK, and $x,x+3$ where $x$ runs from $3$ through $9$, giving $13$ patterns.
• Three-chance hands: $34$, JQ, and $x,x+2$ where $x$ runs from $3$ through $10$, giving $10$ patterns.
• Four-chance hands: $x,x+1$, where $x$ runs from $4$, through $10$, giving $7$ patterns.

To check, we count how many straights are produced: $$16\cdot1+13\cdot2+10\cdot3+7\cdot4=100$$ On the other hand, there are $10$ possible straights, and $\binom52=10$ ways of choosing two cards from one of them, so again we get $10\cdot10=100$, and the count appears to be correct.

Now, $t=16+13+10+7=46$ and $$\Pr(S|T)=\frac{5\cdot663}{1274\cdot8\cdot46}=\frac{255}{36064}\approx0.00707$$