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Interesting Probability Problem

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Here is one for all of the probability geeks out there.  A friend of mine posed this to me over the weekend:

Given a standard shuffled deck of 52 cards, you flip through the deck starting at the top until you reach the first Ace (A).  What position does that A need to be in to give an equal probability of running into another A or a 2 after that?  (I haven't specified whether you have already seen a 2, or not, prior to flipping the first A).

For example, if the first A is the 20th card flipped and I keep flipping the cards, am I more likely to get to another A or a 2 first?  The answer for this is A. 

So what is the magic position for the first A so that the probability of flipping an A or a 2 after that is equal.

I originally thought 13.  Well, it turns out that I am wrong (not a surprise).  After multiple computer simulations, it seems that the number is closer 12.  (Actually, it seems to be between 11.5 and 12).

My problem is that I do not know how to write a probability equation for this problem, so I am hoping someone out there could provide some insight.

Comments

  • Anonymous
    March 06, 2008
    Let x be the postion of the first ace.  I believe you are looking for x to satisfy: 3/(52-x) = C(48,x)/C(52,x) * 4/(52-x) + C(4,1)*C(48,x-1)/C(52,x) * 3/(52-x) + C(4,2) * C(48,x-2)/C(52,x) * 2/(52-x) + C(4,3) * C(48,x-3)/C(52,x) * 1/(52-x) If this is the case, the x = 13 is the solution.

  • Anonymous
    March 06, 2008
    I can also only get 13 with my limited probability knowledge.  But when this scenario is run through over 10 million iterations with a computer program, the answer actually seems to be somewhere slightly lower than 12. I just don't get it. The prob of getting a 2 at any point in the deck is 1/13. The prob of getting another ace is 3/(52-x) where x is the position of the first ace. So to find where these are equal, solve for x where 1/13 = 3/(52-x).  This gives 13 also, but the program says something different.

  • Anonymous
    March 12, 2008
    A few weeks ago, I wrote a post about a probability problem that I found interesting.  The basic

  • Anonymous
    March 12, 2008
    A few weeks ago, I wrote a post about a probability problem that I found interesting.  The basic