This is a very nice question on Math StackExchange. The solution below is based on the answer by Tanner Swett.

Two dice A and B are each numbered 1 to 6. But each dice is unfair: the numbers 1 to 6 do not all show with equal chance. If you roll the two dice, their sum will be a whole number from 2 to 12. Prove it is impossible that all of the sums can occur with equal chance.

As usual, watch the video for a solution.

Or keep reading.

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.**Answer To Impossible Dice Sums**

(Pretty much all posts are transcribed quickly after I make the videos for them–please let me know if there are any typos/errors and I will correct them, thanks).

We will utilize a proof by contradiction. Suppose it is possible to get all sums with equal chance.

Let dice A have probabilities for its rolls *p*_{1}, …, *p*_{6} and dice B have *q*_{1}, …, *q*_{6}. Each probability is a non-negative number not exceeding 1.

Rolling the pair gives a sum between 2 and 12, so each sum must have probability 1/11. Consequently the sums of 2 and 12 have a probability of 1/11, and each can occur only if both dice show 1 or 6, respectively. Thus we have:

*p*_{1}*q*_{1} = 1/11*p*_{6}*q*_{6} = 1/11

The roll of 7 must also occur with chance 1/11, and two ways to get 7 are 1 + 6 and 6 + 1. So the sum of the probabilities of those pairs cannot exceed 1/11.

*p*_{1}*q*_{6} + *p*_{6}*q*_{1} ≤ 1/11

From the first two equations, we can conclude none of *p*_{1}, *q*_{1}, *p*_{6}, *q*_{6} is equal to 0, or else the product would have to be 0. Since each probability is also non-negative, each of these probabilities will be positive. Therefore *p*_{1}*q*_{6} and *p*_{6}*q*_{1} must be positive numbers, and each is less than 1/11. That is:

*p*_{1}*q*_{6} < 1/11*p*_{6}*q*_{1} < 1/11

Consider these four equations:

*p*_{1}*q*_{1} = 1/11*p*_{6}*q*_{6} = 1/11*p*_{1}*q*_{6} < 1/11*p*_{6}*q*_{1} < 1/11

We can pair each equation with two of the inequalities to give the following 4 inequalities:

We can pair each equation with two of the inequalities to give the following 4 inequalities:

*p*_{1}*q*_{6} < *p*_{1}*q*_{1}*p*_{6}*q*_{1} < *p*_{1}*q*_{1}*p*_{6}*q*_{1} < *p*_{6}*q*_{6}*p*_{1}*q*_{6} < *p*_{6}*q*_{6}

Since each probability is a positive number, we can cancel terms and preserve the inequality. This gives four inequalities:

*q*_{6} < *q*_{1}*p*_{6} < *p*_{1}*q*_{1} < *q*_{6}*p*_{1} < *p*_{6}

But now the 1st and 3rd equations contradict each other, as well as the 2nd and 4th equations.

Thus we have a contradiction.

The original premise of equal chance to all sums is wrong. It is impossible to roll two unfair dice and have an equal chance to all sums.

**Reference**

Math StackExchange solution by Tanner Swett

https://math.stackexchange.com/questions/4102888/is-it-possible-to-get-all-possible-sums-with-the-same-probability-if-i-throw-two

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