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 Figuring a wonky die (Posted on 2006-04-24)
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.

 See The Solution Submitted by Jer Rating: 3.0000 (2 votes)

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 solution | Comment 4 of 5 |

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 5f = 2 e     |by stmt 6b + c = 1/2 |by stmt 3a+b+e=d+f   |by stmt 2a+b+e=10a+2e | by substitutionb = 9a+e    |algebraically`
`f^2 = be    |by stmt 44e^2 = be   |by substitutionb = 4e      |division`
`4e = 9a + e |substitution3e = 9a     |algebraicallye = 3a      |division`
`a = a          |identityb = 12 a       |substitutionc = 1/2 - 12 a |algebraicallyd = 10 a       |already given (stmt 5)e = 3 a        |found abovef = 6 a        |substitution`
`a+b+c+d+e+f = 1/2 + 20 a = 1`
`a = 1/40b = 12/40 = 3/10c = 8/40 = 1/5d = 1/4e = 3/40f = 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

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