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Topic: (difficult)Theoretical gambling puzzle
Replies: 29   Last Post: Jul 31, 2006 5:54 AM

 Messages: [ Previous | Next ]
 cuthbert Posts: 10 Registered: 6/26/06
Re: (difficult)Theoretical gambling puzzle
Posted: Jul 25, 2006 3:49 AM

Interesting puzzle and replies.

I agree that the answer depends upon whether you see the cards as they are presented or not. First, let's assume you see the cards as presented.

1) As soon as you're one dollar in credit, stop the game.
2) If you're never a dollar up, continue to the end of the pack.

In a long run of games, sometimes you will have a zero payoff (by rule 2). Most times, you will get a dollar (by rule 1). So the long run payoff is infinite, and there is no fixed number of cards after which you should stop the game.

>>>>>>>>>>>>>>>

Next, assume that you don't see the cards and you have to stop the game after some number of cards n. Is there an optimal n?

Let's assume there is an optimal n, such that the probability of there being more red cards than black in the n-sized sample is greater than 0.5. That is, in the long run, there will be a positive payoff.

Now consider a symmetrical game, Game 2, played in exactly the same way as Game 1, but with wins going to the black cards and fines imposed on the red ones.

Because the games are symmetrical, n is also the optimal sample size in Game 2. Since n is optimal in Game 2, the probability of there being more black cards than red cards in the n-sized sample is greater than 0.5.

Because the two games are played in the same way and the distribution of cards in the pack is random, then the probability p for an n-sized sample of there being more red cards than black cards is the same as the probability of there being more black cards than red cards. Let the probability of there being equal numbers of red and black cards be q. These three outcomes - more red than black, more black than red, equal number black and red - are exclusive and exhaustive. Therefore 2p + q =1. Now q > 0. So p < 0.5. But we stipulated that p > 0.5 in the optimal n-sized sample. That is a contradiction. Therefore there is no optimal n.

In other words, it doesn't matter how many cards you choose. Whether you're playing black-fines red-wins or the other way round, your expected payoff is zero, on the assumption that you don't see the cards as they are turned.

>>>>>>>>>>>>>>>>>>>>>>>>>>>

Summary:

If you see the cards as they are turned, expected long term payoff is infinite.
If you don't see the cards turned, expected long term payoff is zero.

Date Subject Author
7/21/06 nigel
7/21/06 Mary Krimmel
7/21/06 João Pedro Afonso
7/21/06 Earle Jones
7/23/06 João Pedro Afonso
7/24/06 Earle Jones
7/24/06 João Pedro Afonso
7/23/06 João Pedro Afonso
7/24/06 João Pedro Afonso
7/25/06 Eric Bainville
7/25/06 João Pedro Afonso
7/25/06 Eric Bainville
7/25/06 João Pedro Afonso
7/26/06 Eric Bainville
7/26/06 João Pedro Afonso
7/26/06 Eric Bainville
7/25/06 Eric Bainville
7/25/06 cuthbert
7/25/06 João Pedro Afonso
7/26/06 cuthbert
7/26/06 João Pedro Afonso
7/31/06 cuthbert1
7/25/06 Eamon
7/25/06 Eamon
7/28/06 João Pedro Afonso
7/28/06 mark
7/28/06 João Pedro Afonso
7/28/06 mark
7/28/06 João Pedro Afonso
7/28/06 mark