Pair of Jacks

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 taking a random card from each deck, comparing them, returning them to their decks and doing the same again, 52 times.

I presume the two decks are independent of each other but retain their original 52 cards?

And do you turn over 2 new cards at the same time, one from each deck?

Or do you alternate flipping one card from one deck and then one from the other, and so on?

Yes, 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: ?

Hmm
It's the same as the number of ways of arranging the numbers 1..N such that no number is in the right place (divided by the total number of ways of arranging them), but this is a tricky thing to calculate.

I 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.

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.

Depends on the quality of your shuffling 😛
(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. Reveal Hidden Content

Richard

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? I agree that the figure is fairly high. About 0.6 or so.

Originally posted by talzamir
Mathematicians shuffle perfectly, and a true mathematician does not go into a casino to win.

True - the only reliable(!) way to win in a casino is to run one.

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