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If you can design two dice that not necessarily fair and not necessarily weighted in the same manner, then roll twice. How can you minimize the probability of getting the same sum of the two rolls?

My thought is since the probability of rolling the same sum of two fair dice is $\dfrac{6^2+2\times(1^2+2^2+3^2+4^2+5^2)}{36^2}$. Our goal is to minimize the numerator by changing the weight of each side of the dice. If we make one die with side $1$ weighted more and the other die with side $6$ weighted more, then we may reduce the probability of getting the same sum. Am I right? I have no idea what's next. Can someone help me please? Thanks.

jacky123456789
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  • You can work with the lateral areas of die – Martín Vacas Vignolo Apr 03 '16 at 12:59
  • Maybe you should make a general expression of what the probability of rolling the same sum of two dice is given that the probability of rolling $i$ for $1 \leq i \leq 6$ is $p_i$. – Noble Mushtak Apr 03 '16 at 13:01
  • Formally you can set up a 10 variables (as they are constrained by the sum of probabilities equal to 1) optimization problem, but here you can easily see that if you have set $\Pr{X_1 = 1} = \Pr{X_2 = 2} = 1$ respectively, then $\Pr{X_1 = X_2} = 0$ which is a minimum. So the minimum occur at the boundary and the fair dice case is probably the maximum case, which is the interior extremum. – BGM Apr 03 '16 at 13:24
  • Because the distribution of sums of dice have a bell shape the best thing you can do is try to flat the distribution all that you can. Then notice that $\Pr[X(\omega)\in{1,6}]>\Pr[X(\omega)\in{2,5}]>\Pr[X(\omega)\in{3,4}]$, and you must try to force $\Pr[S=2]\approx\Pr[S=7]$, where $S$ is the sum. – Masacroso Apr 03 '16 at 13:36
  • So, you roll both dice once, you note their sum and then again both dice and you compare? Do I get it right? – Jimmy R. Apr 03 '16 at 17:11

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From numerical experiments, it seems that the optimum is to have one die only show $1$ or $6$ with probability $\frac12$ each, and to use probabilities $\left(\frac18,\frac3{16},\frac3{16},\frac3{16},\frac3{16},\frac18\right)$ for the other die. The probabilities for the sums are then $\frac1{16}$ for $2$ and $12$, $\frac18$ for $7$ and $\frac3{32}$ for all other values. I can show that this is a local optimum, but I don't know how to show that it's a global optimum. The probability to get the same sum twice using these probabilities is $\frac3{32}=0.09375$, pretty close to the lower bound $\frac1{11}=0.\overline{09}$ that would be achieved if all sums had the same probability. This is to be compared to $\frac{73}{648}\approx0.11265$ for fair dice, so the distance to the lower bound has been reduced to $13\%$ of its value for fair dice. Interestingly, if we allow negative probabilities, the optimum seems to be $\frac{11}{120}=0.91\overline6$, which just so happens to be a number I recently came across at Project Euler.

Also interestingly, the optimal symmetric solution isn't nearly as good. In a fully symmetric solution, both dice would have the same distribution, and the distribution would be invariant under the inversion $x\to7-x$. That plus normalisation leaves only two degrees of freedom, $p_1$ and $p_2$, with $p_3=\frac12-p_1-p_2$ determined by normalisation and the others by symmetry. Then the probability to get the same sum twice is

$$ 2p_1^4+2(2p_1p_2)^2+2(2p_1p_3+p_2^2)^2+2(2p_1p_3+2p_2p_3)^2+2(2p_1p_2+2p_2p_3+p_3^2)^2+(2p_1^2+2p_2^2+2p_3^2)^2\\ =36 p_1^4+88 p_1^3 p_2-36 p_1^3+108 p_1^2 p_2^2-72 p_1^2 p_2+15 p_1^2+48 p_1 p_2^3-48 p_1 p_2^2+18 p_1 p_2-3 p_1+28 p_2^4-24 p_2^3+9 p_2^2-2 p_2+\frac38 $$

(Wolfram|Alpha calculation), and minimizing this yields

$$ p_1\approx0.243883\;,\\ p_2\approx0.137479\;,\\ p_3\approx0.118638\;,\\ $$

with probability $\approx0.104303$ of getting the same sum (Wolfram|Alpha calculation), which reduces the difference from the lower bound only to about $62\%$.

P.S.: I did the same numerical investigations for three dice, and it seems that these are optimized if two of them show only $1$ or $6$ with probability $\frac12$ each and the third has probabilities $\left(\frac3{26},\frac5{26},\frac5{26},\frac5{26},\frac5{26},\frac3{26}\right)$. In this case the probability for coinciding sums is $\frac{15}{208}\approx0.07212$, compared to the lower bound $\frac1{16}=0.0625$. The probabilities for the sums in this case are $\frac1{104}(3,5,5,5,5,9,10,10,10,10,9,5,5,5,5,3)$.

joriki
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