Matching cards

[bonanova]

A deck of cards is shuffled and dealt face up in a single row. A second deck is shuffled and dealt face up in a single row beneath the first row, making 52 coloumns each containing two cards. On average, how many pairs of cards will match?

[Thalia]

I've always struggled with this kind of probability but...

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If you pull a card from each deck, the odds of a match are 1/52. Doesn't that mean that out of 52 attempts, there should be an average of 1 match?

[MikeD]

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If both decks had one card then the average would be 1 pair (100%)
if there were two cards then the average would be 1 pair (50% = 2 and 50% = 0)
if there were three cards then the average would be 1 pair (16.7% = 3 pairs, 50% = 1, and 33.3% = 0)
and so on, so in a deck of 52 cards the average would still only be 1 pair.

 

[bonanova]

@Thalia and @MikeD, two good approaches, same right answer. Kudos. 

[flamebirde]

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So the chances that the first two will match is 1/52. That much should be fairly obvious I think.

If they do, then the chances of the next two matching is 1/51, and if they do then the chances of the next two are 1/50, and so on.

If a pair doesn't match, then we're looking at an issue -- at least one more pair is guaranteed not to match... but there's some finicky stuff going on there, since you could have a loop of 3 or more unmatched pairs if it chains down a bit. So at minimum if you draw a non-matching card then you have at least 2 unmatched pairs right there and potentially more. Not to mention that you could spawn different "loops" in one run through.

if a pair doesn't match, then the chances of the next pair matching are... what? That's where the riddle lies next I guess. If we number the first deck 1-52 and the second 1'-52' and the very first pair is 1-2', then we know that the next pair (the one starting with 2-) must also not match. Then there's a 1/51 chance that the loop ends there -- a 1/51 chance that the next pair is 2-1', in which case the probabilities continue as before. But there's also a 50/51 chance that the chain extends, and we draw something like 2-5', which means that now the fifth pair (the one starting with 5-) is guaranteed to be a non-match with a 1/45 chance of terminating.

I'm sure there's an easier way to think about it, but I'm gonna let it percolate through the back of my head for now.

Seems similar to that airplane seating problem a while back, I'll see if I can't find and link it here.

Edit: here it is:

 

 

Edited June 15, 2019 by flamebirde
Found the thread I was talking about.

[andaviking@gmail.com]

2

[bonanova]

Clue:

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Does it matter that the first deck was shuffled?

Big clue:

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Are the chances different for different pairs?