Best possible chance to succeed with free choice of how many dice to roll for 4 players

Question

Each of the 4 players can roll however many d10's as they like. Howerver, for any single player ,if any of their rolls come up as the same number, their total is treated as a zero. You are shooting for a total of 65 or more between everyone results How many dice should the survivors roll for the best chance of success? I had an idea of going for averages. An average roll of a d10 is 5.5. SO each survivor should roll 3 dice. However, this only gives about a 70 percent chance for each to succeed because (10*9*8/1000) I believe. Is this really the best chance to get over a 65 on the rolls? Or is there a better chance if one person were to roll 4 dice etc.?

Answer 1

According to my calculations, the four players' best chance of obtaining a total of $65$ or more is for all of them to roll four dice, even though the expected score of each, namely $11.088$, is slightly less than what it would be if they rolled only three dice, namely $11.88$. The probability that their total score would be $65$ or more for various combinations of dice is given in the table below.

\begin{array}{|c|c|} \hline \mbox{Combination}&\mbox{probability}\\ \hline 3333&0.1621\\ \hline 3334&0.1761\\ \hline 3344&0.1815\\ \hline 3444&0.1948\\ \hline 4444&0.2096\\ \hline \end{array}

The distribution of a player's score when he or she rolls $n$ dice is not difficult to compute. If the score is $\ S_n\ $, then \begin{eqnarray} \mathrm{Pr}\left(S_n=0\right)&=&1-\frac{10!}{(10-n)!10^n}\ , \mbox{ and}\\ \mathrm{Pr}\left(S_n=k\right)&=&\frac{\left\vert\left\{D\subseteq\left\{1,2,\dots,10\right\}\left\vert\,\left| D\right\vert=n, \sum_\limits{j\in D}j =k\right.\right\}\right\vert}{10^n}\\ &&\mbox{for $k=\frac{n(n+1)}{2},\dots,\frac{n(21-n)}{2}$} . \end{eqnarray} These distributions are tabulated below for $\ n\ $ from $1$ to $10$. These are the only values of $\ n $ that need to be considered, because throwing any more than $10$ dice is guaranteed to score zero.

If $\ \psi_n\ $ is the distribution of $\ S_n\ $, then the distribution of the players' total score when they throw $\ n_1, n_2,n_3,\ $ and $\ n_4\ $ dice, is the convolution $\ \psi_{n_1}*\psi_{n_2}*\psi_{n_3}*\psi_{n_4}\ $. To establish that $\ 4,4,4,4\ $ is the best combination, I evaluated $\ \sum_\limits{j=65}^\infty\psi_{n_1}*\psi_{n_2}*\psi_{n_3}*\psi_{n_4}(j)\ $ for all quadruples of integers $\ n_1, n_2,n_3, n_4\ $ with $\ 1\le n_1\le n_2\le n_4\le 10\ $ and $\ \sum_\limits{i=1}^4n_i \ge 7\ $, and found that this quantity was maximised by that combination.

Distributions of $\ S_n\ $

One die, $\ \psi_1\ $

Mean score = $5.5$ \begin{array}{|c|} \hline 1-10\\ 0.1\\ \hline \end{array}

Two dice, $\ \psi_2\ $

Mean score = $9.9$ \begin{array}{|c|c|} \hline 0&3,4,18,19&5,6,16,17&7,8,14,15&9,10,11,13&12\\ 0.1&0.02&0.04&0.06&0.08&0.10\\ \hline \end{array}

Three, $\ \psi_3\ $

Mean score = $11.88$ \begin{array}{|c|c|} \hline 0&6,7,26,27&8,25&9,24&10,23\\ 0.28&0.006&0.012&0.018&0.024\\ \hline 11,22&12,21&13,20&14,19&15,16,17,18\\ 0.03&0.042&0.048&0.054&0.06\\ \hline \end{array}

Four, $\ \psi_4\ $

Mean score = $11.088$ \begin{array}{|c|c|} \hline 0&10,11,33,34&12,32&13,31&14,30&15,29\\ 0.496&0.0024&0.0048&0.0072&0.012&0.0144\\ \hline 16,28&17,27&18,26&19,25&20,21,23,24&22\\ 0.0216&0.024&0.0312&0.0336&0.0384&0.0432\\ \hline \end{array}

Five, $\ \psi_5\ $

Mean score = $8.316$ \begin{array}{|c|c|} \hline 0&15,16,39,40&17,38&18,37&19,36&20,35&21,34\\ 0.6976&0.00012&0.0024&0.0036&0.006&0.0084&0.0108\\ \hline 22,33&23,32&24,31&25,30&26,29&27,28\\ 0.0132&0.0168&0.0192&0.0216&0.0228&0.024\\ \hline \end{array}

Six, $\ \psi_6\ $

Mean score = $4.9896$ \begin{array}{|c|c|} \hline 0&21,22,44,45&23,43&24,42&25,41&26,40\\ 0.8488&0.00072&0.00144&0.00216&0.0036&0.00432\\ \hline 27,39&28,38&29,37&30,36&31,32,33,35&34\\ 0.00648&0.0072&0.00936&0.01008&0.01152 &0.01296\\ \hline \end{array}

Seven, $\ \psi_7\ $

Mean score=2.32848 \begin{array}{|c|c|} \hline 0&28,29,48,49&30/47&31/46&32/45\\ 0.93952&0.000504&0.001008&0.001512&0.002016\\ \hline 33/44&34/43&35/42&36/41&37,38,39,40\\ 0.00252&0.003528&0.004032&0.004536&0.00504\\ \hline \end{array}

Eight, $\ \psi_8\ $

Mean score=0.798336 \begin{array}{|c|c|} \hline 0&36,37,51,52&38,39,49,50&40,41,47,48&42,43,45,46&44\\ 0.981856&0.0004032&0.0008064&0.0012096&0.0016128&0.002016\\ \hline \hline \end{array}

Nine, $\ \psi_9\ $

Mean score=0.1796256 \begin{array}{|c|c|} \hline 0&45-54\\ 0.9963712&0.00036288\\ \hline \end{array}.

Ten, $\ \psi_{10}\ $

Mean score=0.0199584 \begin{array}{|c|c|} \hline 0&55\\ 0.99963712&0.00036288\\ \hline \end{array}.