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Three siblings cannot agree which TV programme to watch. They decide that each of them rolls a fair die and that the person with the highest number gets to choose the TV programme. If two or more persons get the highest number (e.g. numbers on dice are (5, 5, 2) or (4, 4, 4)) then they roll the dice again.

1) Let n ∈ N be fixed. How do I compute the probability that no decision is reached in the first n trials?

2) Let n ∈ N. How do I compute the probability that, if the decision is reached in the n trial, in at least one trial before the decision was reached all the scores were the same.

Could anybody help me with this problem?

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  • $\begingroup$ Welcome to MSE! What have you tried so far? $\endgroup$ – Stan Tendijck Feb 20 at 18:28
  • $\begingroup$ Binomial distribution for the first one but nothing for the second one, can't think of anything.. $\endgroup$ – The Poor Jew Feb 20 at 18:30
  • $\begingroup$ Try to search for geometric distribution ;) $\endgroup$ – Stan Tendijck Feb 20 at 18:30
  • $\begingroup$ For the second one, a quick hint would be to look at the probability of getting no events where all scores are the same in $n-1$ tries given that all scores come from the set of undecidable options $\endgroup$ – Stan Tendijck Feb 20 at 18:33
  • $\begingroup$ It would be sensible to start by calculating (a) the probability in a single trial that not decision is made and (b) the probability in a single trial all values are the same. Incidentally, in your $5,5,2$ example, how many people throw again, two or three? $\endgroup$ – Henry Feb 20 at 18:36
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We can approach the problem geometrically.

Each triple toss $\left( {x_{\,1} ,x_{\,2} ,x_{\,3} } \right)$ corresponds to an integral point in a cube with side $[1,6]$: wlog we can take the die to be numbered $0,1,\cdots , 5$, being more convenient to place the cube with one vertex at the origin.
Let's make it general and take a cube with side $[0,r]$.
Being the die fair, all the $\left({r+1} \right)^3$ points are equi-probable.

Consider the region made by the points points which obey to $$ {0 \le x_{\,1} < x_{\,2} < x_{\,3} \le r} $$ The integral volume of such a region is clearly $$ \binom{r+1}{3} $$ (choose three values from the $r+1$, and arrange in order).
There are six of such regions, corresponding to the permutations of the $x_k$.

The diagonal of the cube instead, $x_1=x_2=x_3$, has a volume of $r+1$.

Let's make it even more general, in the case of $m$ siblings, note that the expansion of the binomial in terms of [Falling Factorials][1] via the Stirling N. of 2nd kind $$ \eqalign{ & \left( {r + 1} \right)^{\,m} = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} { \left\{ \matrix{ m \cr k \cr} \right\}\left( {r + 1} \right)^{\underline {\,k\,} } } = \cr & = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,m} \right)} {\ underbrace {\left( {k!\left\{ \matrix{m \cr k \cr} \right\}} \right)}_{N.} \underbrace {\left( \matrix{ r + 1 \cr k \cr} \right)}_{Vol.}} \cr} $$ represents the splitting of the cube into the regions $$ \left[ {x_{\,1} < x_{\,2} < \cdots < x_{\,m} } \right],\left[ {x_{\,1} = x_{\,2} < \cdots < x_{\,m} } \right], \cdots , \left[ {x_{\,1} = x_{\,2} = \cdots = x_{\,m} } \right] $$ and in the formula above $k$ represents the number of $<$ signs + 1.
This can be demonstrated resorting to the meaning of the Stirling N. 2nd kind$\left\{ \matrix{ n \cr k \cr} \right\}$ as the number the number of ways to partition a set of n objects into k non-empty subsets

The decision will be taken when $$ \left[ {x_{\,1} \le x_{\,2} \le \cdots \le x_{\,m - 1} < x_{\,m} } \right] $$ or any permutation of it.
Fixing the value of $x_m$, then the volume of such region will evidently be $x_m^{m-1}$, so that
the total volume of such a pyramid is $$ V = \sum\limits_{1\, \le \,k\, \le \,r} {k^{\,m - 1} } $$ The m-tuples in which $x_m$ is strictly higher than the other components are of course different (non-overlapping) with those in which the highest component is $x_1$ or $x_2$ and so on. Therefore we have $m$ of such regions.

Now we have the basic elements to solve your question.

In your particular case, $m=3$ and $r=5$, the probability that a decision be taken is $$ P_{decision} = {{3\sum\limits_{1\, \le \,k\, \le \,5} {k^{\,2} } } \over {6^{\,3} }} = {{165} \over {216}} = {{55} \over {72}} $$

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  • $\begingroup$ A decision can be made when the two lesser values are equal: $0\le x_1\le x_2 < x_3 \le r$ $\endgroup$ – Daniel Mathias Feb 20 at 21:15
  • $\begingroup$ @DanielMathias: you are right: thanks for signalling!. I amended the answer $\endgroup$ – G Cab Feb 20 at 23:39
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The probability that a round ends without desicion is $$ \underbrace{3\left(\frac16\right)^2\left(\frac56+\frac46+\frac36+\frac26+\frac16\right)}_{\frac{15}{72}}+ \underbrace{6\left(\frac16\right)^3}_{\frac{2}{72}}=\frac{17}{72}. $$ Here the first summand stands for two equal scores with a value larger than the third one. The factor $3$ stands for the number of ways to choose one person from three people. The values $\frac56,\frac46,\frac36,\frac26,\frac16$ in the parentheses are the probabilities of the third score to be less than $6,5,4,3,2$, respectively. The second summand stands for the case with three equal scores.

Thus, the answer to the first question is $$p_1=\left (\frac{17}{72}\right)^n .$$

To answer the second question recall that probability to get all three score equal is $\frac{15}2$ times less than to get only two equal scores. Therefore $$ p_2=1-\left (\frac{15}{17}\right)^{n-1}.$$

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  • $\begingroup$ See the answer I linked in comment above. The probability that a round ends without a decision is $\frac{17}{72}$ $\endgroup$ – Daniel Mathias Feb 20 at 21:21
  • $\begingroup$ @DanielMathias Thank you for catching this. Indeed I did not notice the requirement that two scores have to be higher than the third one. Now my result agrees with yours. $\endgroup$ – user Feb 20 at 22:06

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