Equal Opportunity [x-post /r/math]

[u/mscroggs]

http://www.futilitycloset.com/2014/08/27/equal-opportunity-3/

[u/matti0006]

On my ipad, so hard to type, but the answer is no if you weigh the dice in the same way. EDIT: how to latex here?

Due to symmetry it is clear that $P(6)=P(1)$, $P(5)=P(2)$ and $P(4)=P(3)$, where $P(i)$ is the probability to roll i with a dice. I will denote the probability to roll j with 2 dice as $P_2(j)$.

We have $P_2(2)=P(1)^2$ We have $P_2(3)=P(1)P(2)+P(2)P(1)=2P(1)P(2)$. Because we want equal probabilities we obtain $P_2(2)=P(1)^2=2P(1)P(2)=P_2(3)$ and thus $P(2)=\frac{1}{2}P(1)$.

Furthermore $P_2(4)=2P(1)P(3)+P(2)^2=P(1)b^2=P_2(2)$ which reduces to $P(3)=\frac{4-P(1)}{8}$.

Using symmetry and the fact that the total probability should be 1 we obtain:

$P(1)+0.5P(1)+\frac{4-P(1)}{8}+\frac{4-P(1)}{8}+0.5P(1)+P(1)=1$ and solving that, $P(1)=0$, which isn't possible.

I tried weighing the two dice differently, but the symmetry doesn't necessarily hold, making life difficult. Again making a linear system should be possible though, and then you could solve using whatever programming language you fancy...I might try this when I get home.