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You have two shuffled decks of cards (52 cards each).

You put them next to each other and deal two at a time, one from each deck.

What is the probability that there will be at least one match? That is, the probability that the two cards turned up will be the same at some point in the game?

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  • $\begingroup$ Many answers to this question on the web. Not to degrade these answers below, congrats to them. But I would propose to transform this to the next level: that being, 2-2 cards turned up at a time, doing until the end, and looking for probability of at least one exact match, that meaning first to first and second to second at same time. $\endgroup$ – FIreCase Jan 5 at 21:54
  • $\begingroup$ @FIreCase I agree. I wasn't looking for karma: simply pointing out the triviality of the solution. $\endgroup$ – Don Thousand Jan 5 at 22:44
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The answer is quite simple when you know the concept of a derangement, since the only time this doesn't happen is if the first deck is a derangement of the second.

In other words, we are looking for

$1-\frac{!52}{52!}\sim1-\frac1e$

There are online calculators if you want the exact value, which is approximately

$63.2\%$

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63.57%

The easiest approach is if we first calculate how often there are no match at all.

If we match a card from deck 1 with one from deck 2, it doesn't matter what card 1 shows, there are always 51 chance in 52 for it to not match with the card 2. Since this happen 52 times in a row, we have 52 attempts with a probability of 51/52 that all must go through.

The chance that there are no match is therefor (51/52)^52 =0.3643 or 36.43%
The remaining percent must be all the times it does match. 100% - 36.43% = 63.57%

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    $\begingroup$ This is not really correct. You somehow ended up very close to the correct answer, but the approach is not right at all. There isn't a $\frac{51}{52}$ chance everytime unless the cards are being drawn with replacement. See the derangement article I linked to see the actual calculation. $\endgroup$ – Don Thousand Jan 5 at 22:46

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