Solving a Gambling Game: Average & Median Winnings

[Sam_]

In a gambling game, a gambler wins 0.5 for each 1.0 of the bet if a coin toss is "heads," and loses 0.4 if the coin toss is "tails." So if he bets five Australian dollars, then he wins 2.5 if the coin is heads, and loses 2.0 if it is tails. Now, suppose someone plays this game repeatedly with the following strategy. He begins with a gamble of one Australian dollar. Whatever he has after this gamble (1.5 or 0.6), he bets the entire amount on a second coin toss. After this, he can have 2.25, 0.9, or 0.36, but he then bets the full amount on a third coin toss. And he does this repeatedly until N coin tosses are complete.

What is his average winning (including the one Australian dollar he began with) after N coin tosses? As N goes to infinity, what is his average winning?

What is his median winning (including the one Australian dollar he began with) after N coin tosses? As N goes to infinity, what is his median winning?

 

[davee123]

Interesting! The mean value appears to approach infinity, while the median value approaches 0!
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DaveE

 

[Sam_]

That is correct.

can you please explain how you got that?

 

[Werg22]

Sam_, notice that if the gambler wins with a bet x, his total amount of money becomes 1.5x. If he loses, then it becomes 0.6x. So if there are N tosses, the money he gets, y(x), is so that y(x) = x*1.5^a * 0.6^b where a + b = N. So a = 0,1,2, 3 ... N and b = N, N-1, N-2...0. So there are N + 1 ways of having a + b = N. That said, we look at 1.5^0*0.6^N + 1.5^1*0.6^N-1 + ... + 1.5^N*0.6^0. This is a geometric sum of N +1 terms with the first term equal to 0.6^N and the argument being 1.5/0.6. So that sum is equal to 0.6^N * ((1.5/0.6)^N+1 - 1)/(1.5/0.6 -1). This really simplifies to 5/3 * 1.5^N - 2/3 *0.6^N. Taking the average, we get 5/3 * 1.5^N - 2/3 *0.6^N / N + 1. From this it's clear that the average as N goes to infinity is infinity: 0.6^N / N + 1 obviously goes to 0, and 1.5^N / (N + 1) goes to infinity since 1.5 > 1 and the order of magnitude of e^x is greater than x + 1.

Now considering the median: from the geometric sum above, we have an increasing sequence of N + 1 terms. So if N is even, the median is 0.6^N * 1.5^(N/2 + 1) = . If N is odd, the median is the average of 0.6^N * 1.5^(N+1 /2) and 0.6^N * 1.5^(N+1 / 2 + 1). But in all cases the we get a development that yields numbers lesser than 1 to unbounded powers. So the median goes to 0.

 

[Sam_]

I think you have the right idea, but there are some details. There are N+1 ways to have a+b=N, but they do not all have equal probability. To have a=b=N/2 is much more probable than to have a=0 and b=N. The probability of a=M and b=N-M is {(N)!/[(M)!(N-M)!]}/(0.5^N), so to find the mean you need to weight by these probabilities, not give equal weight to all values of M as you did.

But I think an easier way to find the mean is to call X(i) the multiplier of gamble i, so X(i) is 1.5 if he wins, and X(i) is 0.6 if he loses. Then he has X(1)*X(2)*...*X(N-1)*X(N). The problem is to find the mean of this quantity. But if random quantities are independent, then the mean of their product is the product of their means. And the mean of X(i) is the same for all values of i, so the mean of his winnings after N gambles is some number raised to the power of N.

 

[Sam_]

So? any takers?

Come on!

 

[doodle]

Answer is hidden.

The probability of getting n heads and (N-n) tails in N toss is
P(n) = \binom{N}{n} (0.5)^N
and the corresponding multiplier of gamble is
X(n) = (1.5)^n (0.6)^{N-n} = (0.6)^N (2.5)^n

The mean of this measure is
E{X} = \sum_{n=0}^N P(n) X(n) = (0.3)^N \sum_{n=0}^N \binom{N}{n} (2.5)^n = (0.3)^N (1+2.5)^N = (1.05)^N[/color]