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 20060424 13:46:37 