I did not see this one here..so here it goes :Cards from an ordinary deck of 52 playing cards are turned face up one at a time. If the 1st card is an ace, or the second a deuce, or the third a three, or ...., the thirteenth a king, or the fourteenth an ace, and so on, we say that a match occurs. Note that we don not require that the (13n+1)th card be any particular ace for a match to occur but only that it be an ace. Compute the expected number of matches that occur.

Hey noexpert, where you get these problems from? Not interviews you've had, is it.

Let A_{k}=card k is a match, k=1,...,52 and T=total number of matches. Then P(A_{k})=1/13 and so that

Simulation with 100E6 tests gives 4.00038028.

Yes. the answer is 4.

Prob the first card is an ace is 1/13Prob that second card is 2 is the sum of 2 events:1. (prob first card was not 2) and (prob second card is 2) and2. (prob first card is 2) and (prob second card is 2) i.e., 12/13 * 13/51 + 1/13 * 3/51 = 1/13 !!Prob that third card is 3 is the sum of 4 events: 1. (prob first card is not 3) *(prob second card is not 3) *(prob third card is 3) 2. (prob first card is not 3) *(prob second card is 3) *(prob third card is 3) 3. (prob first card is 3) *(prob second card is not 3) *(prob third card is 3) and 4. (prob first card is 3) *(prob second card is 3) *(prob third card is 3)i.e., 12/13 * 12/13 * 4/50 + 12/13*1/13*3/50 + 1/13*12/13*3/50 + 1/13 *1/13 * 2/50 = 1/13 !!!This just goes on with the prob of each match being 1/13Therefore, the expected no. of matches with 52 cards = 52 * 1/13 = 4

I like your approach Moose !

Prob that any particular card lands in a matching space * number of cards: 4/52 * 52 = 4

QuoteOriginally posted by: phvanProb that any particular card lands in a matching space * number of cards: 4/52 * 52 = 4I am doubtful it is that simple. As Moose has shown, you don't always have 4 cards to draw from. Sometimes a few of them might have been drawn already. On the other hand if we have S suits of cards and there are C cards in each suit, where both S and C are integers, we can consider the case that C is infinite so that the cards drawn do not deplete the deck in any significant way. In this case, prob(a draw resulting in the right card) = 1/C since there are a total of S*C numbers, the expected no. of matching draws is S*C/C = S.So the expectation is just the no. of suits in the deck. In our example this is 4. If there is only one suit, this is 1, 10 if there are 10 suits etc. Is there any reason why the answer is so simple?

It is that simple.What if you didn't go through the cards in order. Assume you look at only the 27th card. What is prob it is an Ace? (it's match).It's 1/13. What is the prob that the 42 card is a match:1/131/13 x 52 = 4

QuoteOriginally posted by: NE1QuoteOriginally posted by: phvanProb that any particular card lands in a matching space * number of cards: 4/52 * 52 = 4I am doubtful it is that simple.It really is. Here's a way to look at it. How many permutations of the deck are there? There are . How many of these permutations have the kth card as a match? Well, there are four possible cards to match (one of each suit) and ways to order the remaining cards. There is no double-counting here, so the probability of a match on the kth card is . That's it.QuoteIs there any reason why the answer is so simple?Everything is when looked at in the right way. Relevant here is the following:"Everything should be made as simple as possible, but not simpler."---Albert Einstein

QuoteIt really is. Here's a way to look at it. How many permutations of the deck are there? There are . How many of these permutations have the kth card as a match? Well, there are four possible cards to match (one of each suit) and ways to order the remaining cards. There is no double-counting here, so the probability of a match on the kth card is . That's it.QuoteIs there any reason why the answer is so simple?Everything is when looked at in the right way. Relevant here is the following:"Everything should be made as simple as possible, but not simpler."---Albert EinsteinOk fine if you put it like that. I did find the last two lines incredibly patronizing though! I guess I deserved it.

QuoteOriginally posted by: NE1Ok fine if you put it like that. I did find the last two lines incredibly patronizing though! I guess I deserved it.I truly did not mean for it to come across as patronizing or snarky, or negative in any way whatsoever. I'm really sorry that it came across that way. I can see how you might have interpreted it as "damn you baboon, just look at it in the right way and you'll see how simple it is." I did not mean it in that way. I meant that for every problem there is a way of looking at it in which it becomes simple; our goal should always be to find that way.