1. ## Unfair dice

Let S1 and S2 be the results from throwing two dice. Let X = S1 + S2. Now suppose that the dice are unfair, and that they give the values 1,...,6 with probabilities p1, . . . , p6 and q1, . . . , q6, respectively. Write down the values of P(X = 2), P(X =7)and P(X =12). By comparing P(X =7) with root(P(X =2)P(X =12)) and applying the arithmetic-mean–geometric-mean inequality, or otherwise, show that the probabilities P(X = 2), P(X = 3), . . ., P(X = 12) cannot all be equal.

It's the last bit I can't do... I just can't get anything to work.

Any help would be greatly appreciated.

Thanks

2. Originally Posted by Mathmo55
Let S1 and S2 be the results from throwing two dice. Let X = S1 + S2. Now suppose that the dice are unfair, and that they give the values 1,...,6 with probabilities p1, . . . , p6 and q1, . . . , q6, respectively. Write down the values of P(X = 2), P(X =7)and P(X =12). By comparing P(X =7) with root(P(X =2)P(X =12)) and applying the arithmetic-mean–geometric-mean inequality, or otherwise, show that the probabilities P(X = 2), P(X = 3), . . ., P(X = 12) cannot all be equal.
Justify the following: $P(X=7)\geq p_1q_6+p_6q_1\geq 2\sqrt{p_1q_6p_6q_1}=2\sqrt{P(X=2)P(X=12)}$ hence a contradiction if $P(X=7)=P(X=2)=P(X=12)=\frac{1}{11}>0$.