Fair Value of a Coin Toss GameDate: 01/21/2001 at 17:19:00 From: Raj Shah Subject: Pricing a coin toss game Hello. Here's my question: Suppose you play a game where you toss a coin untill Tails turn up. The payout of the game is 2^n, where n represents the number of 'Heads' that come up before a 'Tails'. For example, if you flip the sequence: H H H T, then the payout would be 2^3, or 8 dollars. If the first toss is a Tail, the payout is 2^0 or 1 dollar. I'm trying to find the fair value of the game. I started doing an expected value calculation, figuring the probability of each sequence multiplied by the payout. T : Probablity = 1/2 * 2^0 = 1/2 H,T : P = (1/2*1/2) * 2^1 = 1/2 H,H,T: P = (1/2*1/2*1/2) * 2^2 = 1/2 Now I'm stuck. I get an infinite sequence of 1/2+1/2+1/2,.... What does this imply? I'm trying to figure out the "fair value" of this game, or what you would pay to play the game over time. I would really appreciate your help, as I've been stuck for some time. Thank you, Raj Date: 01/22/2001 at 15:13:14 From: Doctor Twe Subject: Re: Pricing a coin toss game Hi Raj - thanks for writing to Dr. Math. This problem is called the St. Petersburg paradox. To determine a "fair value" (also called the expectation) for the game, we multiply the probability of each outcome by the payout, and sum the results. As you noted, the probability of getting the first tail on the kth toss is: P(k) = (1/2)^k and, with a payout of 2^(k-1), we get an expectation of: inf. inf. E(X) = SUM [(1/2)^k * 2^(k-1)] = SUM [1/2] = 1/2 + 1/2 + 1/2 + ... k=1 k=1 which adds to infinity. This means that over the long haul, no matter what price you set to play, the player will ultimately win money - in theory. In practice, even for small values of k, winning is highly unlikely, and having the patience (and money) to play the required number of games to win is unlikely. For example, to win $2^30 (just over $1 billion), you should get 30 heads in a row followed by a tail. The chances of this happening are less than 1 in 2 billion (1 in 2,147,483,648 to be exact.) I hope this helps. If you have any more questions, write back. - Doctor TWE, The Math Forum http://mathforum.org/dr.math/ |
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