I'll offer you a bet. Each of us shuffles a deck of cards. Then we each simultaneously turn over our cards one at a time. Whenever the two cards match, you give me a dollar. If we get through the entire deck without any matches, I give you three dollars. We'll do this several times (consider an endless series of games.)
Would you take the bet? What's the break-even value of a non-matching deck to make it worthwhile?
You can test this with a single deck of cards by shuffling and dealing against a predefined order, such as Ace-King in each suit.
Source: Original.
# Amt
36583 -3
36959 1
18397 2
6178 3
1509 4
324 5
45 6
5 7
total
-9501 from me to you
So in this case, I'd be making 9.5 cents per game.
The expected number of matching cards is 1; that is, for N games, you would expect to see N matching cards. Even in the simulation of 100000 games above, the number of matching cards is 100248. Do this with an actual deck a few times; it's an interesting phenomenon. I had expected this meant you'd be paying out $N, and that would offset any non-matching payouts. But I didn't account for the frequency of a non-matching deck showing up.
Update 12/21/2006. Alan O'Donnell found a simple analysis for this:
Probability of a match on each card turn is 1/52, so an average return of $1/52. Therefore, with 52 card turns in a game, the average return on the whole pack is $1! The probability of no matches is (51/52)^52 = 0.36431352 So the average payout over infinite games is $3 x above = $1.09294056 For the average payout to equal the average return (ie $1), the price of the game should be $1/0.36431352 = $2.74488852.
Update 2/5/2007. Dan Chirica thought Alan's analysis might be a little off and found a different relationship by letting a computer program find the total number of matches for 52 cards and adding them up, and dividing that into the number of non-matching games. He ended up with the price being closer to $2.71828