Probability and Snap

Question

I have a question about a modified (/simplified) version of the card game Snap:

There are 20 playing cards, such that each card has a number from 1...10, so there are 2 cards with each number. The players divide the cards equally between each other, and place a randomly selected card on the table. What is the probability that they place the same card (i.e. same number) on the table on the first try if:

1. They divide the cards such that both players have a set of cards with number 1 to 10

2. They shuffle all 20 cards together and randomly divide them

I worked out that the answer for 1. is 1/10 (I am not sure it is correct), because whatever card is placed on the table by player 1, there is a 1/10 chance of picking the same numbered card. On the second one however, I am slightly confused about how to approach the problem. Any help would be appreciated! :)

In advance, thank you!

Answer 1

Suppose player A is putting down a '1', in the first case the other '1' is any one of the ten cards belonging to player B. So there's a 1 in 10 chance of it being B's first card.

In the second case the other '1' is any one of the ten cards belonging to player B or any one of the remaining 9 cards belonging to player A. So there's a 1 in 19 chance of it being B's first card.

Answer 2

In the second case, each pair of cards is equally likely. There are ${20\choose{2}}=190$ ways to choose two cards out of twenty, of which $10$ are same-numbered pairs; so the probability is $10/190=1/19$.

This is less than the probability of a same-numbered pair in the first case, because some same-numbered pairs are likely to be held by a single player and therefore "out of the running". An interesting secondary exercise is to determine how many same-numbered pairs are expected to be held by a single player.