How much money are you willing to pay for the following game: You toss six coins
ID: 3359894 • Letter: H
Question
How much money are you willing to pay for the following game: You toss six coins. If you have more than 3 heads, we stop the game and give you h dollars, where h denotes the number of heads. If you have less than 3 heads, we let you toss all the coins again and then we stop the game. We will give you h2 dollars, where h2 is the number of heads you have observed in the second round.
Hint: You may say that I am not interested in gambling, so I will pay $0 for this game. That is not the right answer. The right approach is to make sure that if you play this game over and over, you can make money. Think about the weak law of large numbers. If on average you can make $10 in each game, then as long as you pay less than $10, you should be happy.
Explanation / Answer
Back-up Theory
Let X = number of heads obtained when 6 coins are tossed. Then, X ~ B(6, ½). So, probability distribution of X is as follows:
x
0
1
2
3
4
5
6
Total
p(x)
6C0(½)6 = 1/64
6C1(½)6 = 6/64
6C2(½)6 = 15/64
6C3(½)6 = 20/64
6C4(½)6 = 15/64
6C5(½)6 = 6/64
6C6(½)6 = 1/64
1
Now, to work out solution,
We will work out the expected return (pay-off) from the game and decision would be to pay an amount less than or equal to that expected value.
Let X = number of heads obtained when 6 coins are tossed initially and Y = number of heads obtained when 6 coins are tossed the second time when X < 3.
Case 1 X > 3
Pay-off (R) and corresponding probabilities are:
x
4
5
6
Total
R
4
5
6
-
p(x)
15/64
6/64
1/64
22/64
R x p(x)
60/64
30/64
6/64
96/64
Case 2 X < 3
P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = (1/64) + (6/64) + (15/64) = 22/64.
In this case, R will be equal to Y
y
0
1
2
3
4
5
6
Total
R
0
1
2
3
4
5
6
-
p(y)
1/64
6/64
15/64
20/64
15/64
6/64
1/64
64/64
R x p(x)
0
6/64
30/64
60/64
60/64
30/64
6/64
192/64
Case 3 X = 3
Since nothing is specified in the question, we will take the pay-off as 0.
Combining all the three cases, the expected value = (96/64) + (192/64)(22/64) + (0)
= 162/64 = 81/32.
Thus, the maximum money that can be paid to play the game is: $81/32 ANSWER
x
0
1
2
3
4
5
6
Total
p(x)
6C0(½)6 = 1/64
6C1(½)6 = 6/64
6C2(½)6 = 15/64
6C3(½)6 = 20/64
6C4(½)6 = 15/64
6C5(½)6 = 6/64
6C6(½)6 = 1/64
1
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.