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I am trying to calculate the probability of getting a particular straight after drawing $x$ cards from a deck. Say, for this example, that the straight I am attempting to obtain is "2-3-4-5-6". I know that the deck only has 40 cards remaining, and there are two $2$s, three $3$s, one $4$, four $5$s, and three $6$s. Additionally, I already have a $2$ in the set of cards I am trying to form the straight from, so there is no need to calculate the odds of getting another one. After drawing 20 more cards from the deck, what is the probability that I will have the cards required for a straight.

With this question, here is how far I've been able to get:

$P(2-3-4-5-6) = 1 - (C(38, 20) + C(37, 20) + C(39, 20) + C(36, 20) + C(37, 20) - ?)/C(40, 20)$ (explained below)

I am just having trouble figuring out how to calculate the "overlapping" probabilities that have been calculated twice (denoted by "?" above). For example, the chance of "not getting a 2" and "not getting a 3" both account for the chance of "not getting a 2 and not getting a 3", so this overlap would have to be subtracted. I am looking for a simple (or as simple as possible) "rule of thumb" for doing this, as per the other "pieces" of the equation explained below.

To help with the understanding of the above equation, here is where I draw my numbers from:

A = # of cards remaining in deck B = # of cards to draw from deck C1 = # of c1s remaining in deck C2 = # of c2s remaining in deck C3 = # of c3s remaining in deck C4 = # of c4s remaining in deck C5 = # of c5s remaining in deck D = Probability of all mutually exclusive parts of "probabilities" calculated P(c1-c2-c3-c4-c5) = 1 - (C(A - C1, B) + C(A - C2, B) + C(A - C3, B) + C(A - C4, B) + C(A - C5, B) - D)/C(A, B)
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  • $\begingroup$ You are right, this is quite a bit harder than regular probability due to the overlaps. Problems like this in mathematics are often referred to as the Coupon Collector's Problem - en.wikipedia.org/wiki/Coupon_collector%27s_problem $\endgroup$ Commented Mar 22, 2016 at 4:29

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You already have a $2$, so let's forget about those. There's only one $4$ left; the probability that you'll get it is $\frac12$. If you do, you draw $19$ more cards out of $39$, of which $3$ are threes, $4$ are fives and $3$ are sixes. The probability to get $k$ particular threes, $l$ particular fives and $m$ particular sixes is

$$ \frac{\binom{39-k-l-m}{19-k-l-m}}{\binom{39}{19}}\;, $$

so by inclusion-exclusion the probability to get at least one three, five and six is

$$ \binom{39}{19}^{-1}\sum_{k=1}^3(-1)^{k+1}\binom3k\sum_{l=1}^4(-1)^{l+1}\binom4l\sum_{m=1}^3(-1)^{m+1}\binom3m\binom{39-k-l-m}{19-k-l-m}=\frac{818955}{1148147}\approx71\%\;. $$

As a simple check, this should be roughly the product of the three probabilities of getting a three, a five or a six, respectively, which is

$$ \left(1-\frac{\binom{20}3}{\binom{39}3}\right)^2\left(1-\frac{\binom{20}4}{\binom{39}4}\right)=\frac{240693278}{333853923}\approx72\%\;, $$

slightly higher since the cards anticorrelate. Now multiply by $\frac12$ for the four, so the probability for the straight is

$$ \frac12\cdot\frac{818955}{1148147}=\frac{818955}{2296294}\approx36\%\;. $$

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