I have an unfair six sided die with the following known characteristics:

1) The expected value is the same as for a fair die;
2) The probability of rolling either 1 or 2 or 5 is the same as the probability of rolling either 4 or 6;
3) The probability of rolling either a 2 or 3 is 1/2;
4) If the die is rolled twice in a row, the probability of getting 6 both times is the same as getting a 2 and then a 5;
5) The probability of rolling a 4 is ten times that of rolling a 1;
6) The probability of rolling a 6 is twice that of rolling a 5.

What is the probability of each side? This can and should be solved by hand.

If a, b, c, d, e and f are the probabilities of getting 1, 2, 3, 4, 5 and 6 respectively, then:

d = 10 a    |by stmt 5
f = 2 e     |by stmt 6
b + c = 1/2 |by stmt 3
a+b+e=d+f   |by stmt 2
a+b+e=10a+2e | by substitution
b = 9a+e    |algebraically

f^2 = be    |by stmt 4
4e^2 = be   |by substitution
b = 4e      |division

4e = 9a + e |substitution
3e = 9a     |algebraically
e = 3a      |division

a = a          |identity
b = 12 a       |substitution
c = 1/2 - 12 a |algebraically
d = 10 a       |already given (stmt 5)
e = 3 a        |found above
f = 6 a        |substitution

a+b+c+d+e+f = 1/2 + 20 a = 1

a = 1/40
b = 12/40 = 3/10
c = 8/40 = 1/5
d = 1/4
e = 3/40
f = 6/40 = 3/20

The expected value does turn out to be 3.5 as on a fair die, but that was not needed, as a+b+c+d+e+f, together with stmts 2 - 6 made up six equations in the six unknowns.

Posted by Charlie on 2006-04-24 13:46:37