Consider the following game - a fair coin is tossed repeatedly in independent trials until a head comes up. If a head comes up on the n-th toss, then the payoff from this gamble is (e^2)^n. Consider an individual with utility function ln(m), where m is the amount of money won. If this individual is an expected utility maximiser, what will be the maximum amount that he will be willing to pay as an entry fee to play this game? Briefly explain the result. note - do not leave any incomplete calculations, complete all the calculations to the final values and answer the question in detail
Question
Consider the following game - a fair coin is tossed repeatedly in independent trials until a head comes up. If a head comes up on the n-th toss, then the payoff from this gamble is (e^2)^n. Consider an individual with utility function ln(m), where m is the amount of money won. If this individual is an expected utility maximiser, what will be the maximum amount that he will be willing to pay as an entry fee to play this game? Briefly explain the result. note - do not leave any incomplete calculations, complete all the calculations to the final values and answer the question in detail
Solution
The game described is a geometric distribution with parameter 0.5 (since the coin is fair). The expected payoff from the game can be calculated as the sum of the payoff times the probability of that payoff occurring.
The payoff from the game if a head comes up on the n-th toss is (e^2)^n. The probability of this happening is (1/2)^n (since each toss is independent and the probability of getting a head on any toss is 0.5).
So, the expected payoff from the game is the sum from n=1 to infinity of [(e^2)^n * (1/2)^n].
This is a geometric series with first term a = e^2/2 and ratio r = e^2/2. The sum of an infinite geometric series is a / (1 - r), so the expected payoff from the game is (e^2/2) / (1 - e^2/2).
The individual's utility from a payoff of m is ln(m). So, the expected utility from the game is the sum from n=1 to infinity of [ln((e^2)^n * (1/2)^n) * (1/2)^n].
This simplifies to the sum from n=1 to infinity of [n * ln(e^2/2) * (1/2)^n].
This is a geometric series with first term a = ln(e^2/2) and ratio r = 1/2. The sum of an infinite geometric series is a / (1 - r), so the expected utility from the game is ln(e^2/2) / (1 - 1/2) = 2 * ln(e^2/2).
The individual is an expected utility maximiser, so he will be willing to pay an entry fee up to the amount that makes his expected utility from the game equal to his utility from not playing the game (which is ln(m), where m is the amount of money he has before paying the entry fee).
So, the maximum entry fee he will be willing to pay is the amount m such that ln(m) = 2 * ln(e^2/2).
Solving this equation for m gives m = e^(2 * ln(e^2/2)) = (e^2/2)^2 = e^2/4.
So, the maximum amount that the individual will be willing to pay as an entry fee to play this game is e^2/4.
Similar Questions
Suppose a box contains 20 coins- 14 of them are fair i.e. both Head (H) and Tail (T) are equally likely, 3 of them have both sides H, and 3 of them have both sides T. A coin is picked randomly and tossed- if its H, you get Rs. 1700; else you get 1200. You are not allowed to inspect the coin, you can just see the outcome of tosses. Further suppose first toss happens to be T. Now what is the maximum amount that you will be ready to pay to play this game for the second toss (same coin)? Assume that you are a expected utility maximiser with a risk averse attitude.
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