Calculating probability of an 11-card straight

Question

My friends and I were playing a card game where you're dealt 11 cards (it was one of those "family games", so I'm not sure what we were playing. It was kind of like Canasta).

Anyway, on one of my deals my friend was dealt an 11-card straight, 3-K. Before dealing, I shuffled 7 times.

What are the odds of this happening? I never took probability courses, but it seems exceedingly rare from a layman's perspective.

I'm interested in the result, but I'm also very interested to see how the result is produced! I'm just starting to get back into math after a series of awful high school teachers.

Thank you!

Answer 1

For just a 3-K straight, the probability is $$\frac{4^{11}}{\binom{52}{11}}=6.944×10^{-5}$$ The numerator arises because there are 4 possible suits for each card in the straight; the denominator is just the number of ways to deal an 11-card hand from the full deck.

If any 11-straight was allowed, the above probability would be multiplied by four because there are four such straights (A-J, 2-Q, 3-K, 4-A), leading to a probability of $2.778×10^{-4}$.

Answer 2

Number of ways you can choose $11$ cards from a deck of $52$ cards is $\binom{52}{11}=\frac{52!}{11!41!}$. Number of ways those a set of $11$ cards contains $3$ K's, is (the number of ways of choosing $3$ K's from the $4$ K's)$\times$(the number of ways of choosing $8$ cards from the remaining $(52-4)$ cards)$=\binom{4}{3}\times\binom{48}{8}$. Thus the required probability is $$\frac{\binom{4}{3}\binom{48}{8}}{\binom{52}{11}}$$