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My professor threw out this question for the class as an extra credit question. I'm not asking for an answer, but rather, a nudge in the right direction to solve the problem. I'm currently stuck because this question is not discrete because it doesn't deal with a finite amount of stuff.

I mean, sure, I could code a little program and let it run though a million simulations to figure it out, but there has to be some math behind it, right?

The question: Suppose you have a truly shuffled deck of cards that is infinitely large and holds all 52 standard cards in equal probability (that is, each card has an equal chance of being drawn and the chance does not change at all if you draw any cards beforehand). Figure out the average amount of cards you have to draw (one at a time) before you get a 5-of-a-kind of anything (eg 5 As, 10s, Ks, etc).

user2837858
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4 Answers4

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Note that choosing a card from a well-shuffuled stack of infinitely many cards is mathematically the same as drawing a card from a stack of $52$, recording the card you get, and replacing your card into the stack.

That is, we note that the probability of drawing any given card is $1/52$, regardless of how many or which cards have already been chosen.

Ben Grossmann
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Since you are asking for a hint, this is the Coupon collector's problem.

Guy
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I'd probably not solve it how your professor wants you to....

First, it seems you effectively have $13$ cards. You have $13^5$ states of having not succeeded (having drawn zero through four of each card type), and $1$ state of having succeeded. That's a 371,294 state transition matrix.

http://en.wikipedia.org/wiki/Absorbing_Markov_chain

See the section on expectation. I'd solve it exactly with a computer using these formulas; however, if you are trying to solve it without a computer then it is a matter of figuring out how to simplify the matrix inverse expression, which doesn't look fun at all.

DanielV
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  • Actually, you don't need to distinguish between ranks that you've drawn the same number of times, so there are only $\binom{13+5-1}{5-1}=2380$ states, the number of ways to distribute $13$ balls into $5$ bins. – joriki Jun 18 '16 at 00:25
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    @joriki Took me a minute to figure out what you were saying, but that is a good point. I'll keep that in mind for future problems. – DanielV Jun 18 '16 at 07:03
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Your proposal to simulate is a good one. You can get a quick answer that way. There may be a clever way to get an analytic solution, and if so the users of this site will no doubt provide it.

I suggest that in order to get some insight, you do two things. Do the simulation and study the distribution of the number of cards it takes to get the 5 of a kind. What does it look like?

The second thing is to consider simpler questions. So, what is the expected number of cards to draw to get a pair? To get 3 of a kind? Perhaps in solving these you can gain intuition as well as work your way up to the original problem.

soakley
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