A Game of SNAP

**peterkelly01** · Apr 28th 2011, 06:21 AM

I came across this problem in the Mensa puzzle book cited below.

Book Title: The Mensa Puzzle Book
Author: Victor Serebriakoff (sometime Intl Chairman of Mensa)
Publisher: Treasure press (1991)
Page 195

At the risk of breaching copyright, I will quote the proble verbatim. I have added items in parentheses for clarity.

Two players shuffle a pack of (playing) cardseach as in snap. What are the chances that they will both lay down the same card at the same point in the deal, a SNAP deal in fact. What are the odds in the dealing of the whole (two) pack(s).

The question is not well posed but I interpret that the author is asking us to calculate the probability that at least one SNAP deal will occur when all 52 cards in each deck are dealt successively.

The author quotes a probility that no SNAP deal will occur. His answer is (51/52)**52
If my interpretation of the problem is correct, the quoted answer appears intuitively to be wrong. However I dont't know what the correct answer is, or how to obtain it!!

I have started my analysis as follows:

First Deal
Probability of no SNAP = 51/52

Second Deal
There are 51 cards in each deck prior to the second deal, but only 50 of the cards in a deck have a match in the other deck. Accordingly,

Probability of a SNAP deal = (50/51) times (1/51)
Probability of no SNAP = 1 - (50/51.51)

Third Deal

There are 50 cards in each deck prior to the third deal, but either 48, 49 or 50 of the cards in one deck may have matches in the other deck.

I'm stuck now. Is there an easier way to tackle this problem?

**Soroban** · Apr 28th 2011, 09:00 AM

Hello, peterkelly01!

Your suspicions are correct . . . They are wrong.

Their answer would be correct if the game were like this:

The two players select a card at the random from their respective decks.
Then the cards are returned to their decks, the decks are shuffled,
. . and they draw another random card.
And they repeat this procedure fifty-two times.

$\text{Then the probability of }no\text{ matches is: }\:\left(\frac{51}{52}\right)^{52}$

The game, as stated, involves the number of derangements,
. . arrangments in which no object is in its designated position.

This number has a complicated formula and a lengthy derivation.

$\text{However, for large }n,\;P(\text{no match}) \;\approx\;\frac{1}{e} \;=\;0.367879441$

$\text{Their answer: }\:\left(\frac{51}{52}\right)^{52} \:=\:0.36431352\:\text{ is quite close, but still wrong.}$

**peterkelly01** · Apr 29th 2011, 01:19 AM

Thank you for your answer. That's precisely the reasoning that I used to conclude that their answer might be wrong. I am not a mathematician, just a hobby puzzle solver. Deriving a complicated formula is not what I had in mind when I looked at this particular puzzle. I thought that there might be a ''trick'' that could be used to solve it, like some other problems in the same book.

Looks like it's a problem of a type suitable for a mathematics olympiad rather that a Mensa puzzle book.

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