perplexus.info :: Games : Matching pairs

Matching pairs (Posted on 2002-04-23)

A standard deck of 52 cards is shuffled, and we take cards from its top in pairs.

If both are of a red suit, I take both cards. If both are black - you take them. If they are different colors - we put them in the discard pile.

If at the end of the deck you have more cards than I, I will pay you a dollar.

If I have more cards, or if we have the same amount of cards, I pay you nothing.

What is a fair "admission price" to play this game?

(from techInterview.org)

Submitted by levik

Rating: 3.5714 (7 votes)

Solution: (Hide)

When the game is finished, let's say you took X pairs, I got Y pairs, and Z pairs were discarded.
That means there were 2*X + Z black cards, and 2*Y + Z red cards played. Since in a standard deck, there is an equal amount of red and black cards in it, therefore, 2*X + Z = 2*Y + Z and X = Y.
We would always get an equal number of cards at the end of the game, and therefore I would always win. This game is not fair at all, and the only fair admission price for it is nothing.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Answer	K Sengupta	2008-03-20 09:50:17
re: .....	nick lygnos	2006-03-08 14:12:10
i got it!	nick lygnos	2006-03-08 14:08:54
i beg to differ.	Amon	2005-04-30 02:35:29
.....	Mike	2002-05-10 14:56:42

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