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.
Side Probability
-----------------
1 0.025
2 0.300
3 0.200
4 0.250
5 0.075
6 0.150
------
1.000
The expected value is 3.5 (same as a normal die).
I labeled the sides a-f (a is side 1, b is side 2, etc). From the information given, you can create the following equations:
(1) a + b + c + d + e + f = 1
(2) a + b + e = d + f
(3) b + c = 0.5
(4) f^2 = b * e
(5) d = 10a
(6) f = 2e
You can then combine these equations to find the proper values. From (6) we know that f = 2e. From (4) and (6) we know that b = 4e. Substituting into (2) we find that a + 5e = d + 2e, or a + 3e = d.
Combining this with (3) and (6) and substituting in (1) we get a + 0.5 + d + 3e = 1, or 2d + 0.5 = 1, so d = 0.25.
From (5) we then know that a = 0.025. Knowing that a + 3e = d, we can figure out that e = 0.075. Then from (6) we know that f = 0.15. Finally from (4) we figure out that b = 0.3, and from (3) that c = 0.2. Voila!
The fact that the die has the same expected value as a fair die seemed to be of no use other than to confirm that the solution actually works.
Note: Sorry about the multiple posts, I'm not sure what happened. :)
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Posted by tomarken
on 2006-04-24 13:36:49 |
Solution (Full)
