- 25 Oct '11 09:31Start with two standard decks of 52 playing cards. Shuffle each of them. Then turn open the cards one by one from each deck, going through the whole deck. Find the probability that at some point you'll turn the same card open from both decks, getting for example a pair of jack of clubs.

I wonder if this is equivalent to the (imo far easier) problem of taking a random card from each deck, comparing them, returning them to their decks and doing the same again, 52 times. - 25 Oct '11 10:52 / 2 editsYes, standard 52-card decks each.

Draw 1st of each; compare. 2nd of each, compare. etc.

In exact terms if the cards in deck A are a(i) for i = 1,..,52 in some order and the cards in deck B are b(j) for j = 1,...,52 in some order, find the probability that a(k) = b(k) for some k in {1, 2, ... , 52}.

It is easy enough to check what happens by using only a couple of cards.. but doing the whole thing by a brute force method doesn't really work, and monte carlo gives only an estimate. Given that the first numbers oscillate, it would seem it is 0.6'ish.. and it would seem likely that as the deck size goes up, there is a value that the probability approaches.

Deck of 2 cards: 1/2 = 0.500

Deck of 3 cards: 4/6 = 0.667

Deck of 4 cards: 15/24 = 0.625

Deck of 5 cards: 76/120 = 0.633

Deck of 6 cards: 411/720 = 0.571

..

Deck of 52 cards: ? - 26 Oct '11 10:18I agree with both observations, and actually did the calculations on tiny decks that way, looking for a pattern. It makes a very tricky equation. My attempts at making it pretty have so far hit into circular proofs. I've tried to find it by means of "exactly n card in the right place" and "at least n cards in the right place", as

p(exactly 1 card in the right place) = p(at least 1 card in the right place) - p(at least 2)

and it would seem that the probabilities of at least n cards in the right place are easier to calculate than the exactly n cards in the right place ones.

So far, no luck though. - 26 Oct '11 12:52

Depends on the quality of your shuffling*Originally posted by talzamir***Start with two standard decks of 52 playing cards. Shuffle each of them. Then turn open the cards one by one from each deck, going through the whole deck. Find the probability that at some point you'll turn the same card open from both decks, getting for example a pair of jack of clubs.**

I wonder if this is equivalent to the (imo far easier) problem of t ...[text shortened]... om each deck, comparing them, returning them to their decks and doing the same again, 52 times.

(No, that is actually an interesting point. Most people can't shuffle worth a dime, and most people underestimate how badly they shuffle. That's why casinos are meticulate about specifying how their decks are shuffled - often with a machine instead of a human.)

Assuming a perfect shuffle, the odds are equal to that of shuffling only one of the decks and then doing the same test (this is left as an exercise for the reader ). I don't think it's equivalent to taking a card from each, comparing them, and reinserting them; but it is trivially equivalent to taking a random card from each deck, comparing them, and*discarding*them until you run out.

And no, I don't know what the numerical answer is. It should be quite high, though.1/52, plus 51/52*1/51, plus 1-that times 1/50... ad nauseam. Not doing it. Too many numbers. That\'s what we have computers for.

Richard - 26 Oct '11 14:02Mathematicians shuffle perfectly, and a true mathematician does not go into a casino to win. ^_^

So yes, a perfect shuffle was intended. Could you elaborate a bit on the answer you had, as the way I read it, it gives a 100% chance? I agree that the figure is fairly high. About 0.6 or so. - 28 Oct '11 14:16

True - the only reliable(!) way to win in a casino is to run one.*Originally posted by talzamir***Mathematicians shuffle perfectly, and a true mathematician does not go into a casino to win.**

**So yes, a perfect shuffle was intended. Could you elaborate a bit on the answer you had, as the way I read it, it gives a 100% chance?**

Yeah, I think I had the probability for the second card wrong. It should be 1/52 times the inverse of the previous, not 1/card number times ditto. And I'm still not crunching the numbers

Richard