Originally Posted by

**Grandad** Hello everyoneHere's the reason the answer is $\displaystyle \frac{12}{51}=\frac{4}{17}$, and it's very simple.

Call the 'twin' cards 'a' and 'b', as Aquafina has suggested, and imagine that all 52 cards have been placed face down in a line, in the order in which they are about to be dealt. Number the cards 0 through 51 according to their position in the line.

Then the first player will get cards 0, 4, 8, ...; the second 1, 5, 9, ...; and so on. Using modular arithmetic notation, the first player has cards numbered 0 (mod 4), the second has cards numbered 1 (mod 4), the third 2 (mod 4) and the fourth 3 (mod 4). There are, of course, 13 cards of each number (mod 4).

Well, card 'a' will be in one of these sets - in other words it will have a certain number mod 4 - and we simply want the probability that card 'b' has the same number mod 4. Since there are 12 such cards out of the remaining 51, the answer is obviously $\displaystyle \frac{12}{51}$.

Grandad