Impossible Dice Sums – Mind Your Decisions

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.

Impossible Dice Sums

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“All will be well if you use your mind for your decisions, and mind only your decisions.” Since 2007, I have devoted my life to sharing the joy of game theory and mathematics. MindYourDecisions now has over 1,000 free articles with no ads thanks to community support! Help out and get early access to posts with a pledge on Patreon.

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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 p₁, …, p₆ and dice B have q₁, …, q₆. 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:

p₁q₁ = 1/11
p₆q₆ = 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.

p₁q₆ + p₆q₁ &leq; 1/11

From the first two equations, we can conclude none of p₁, q₁, p₆, q₆ 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 p₁q₆ and p₆q₁ must be positive numbers, and each is less than 1/11. That is:

p₁q₆ < 1/11
p₆q₁ < 1/11

Consider these four equations:

p₁q₁ = 1/11
p₆q₆ = 1/11
p₁q₆ < 1/11
p₆q₁ < 1/11

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

p₁q₆ < p₁q₁
p₆q₁ < p₁q₁
p₆q₁ < p₆q₆
p₁q₆ < p₆q₆

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

q₆ < q₁
p₆ < p₁
q₁ < q₆
p₁ < p₆

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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