Probability that no decision is reached in the first n trials 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?
 A: 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
$$
P_{decision}  = {{3\sum\limits_{1\, \le \,k\, \le \,5} {k^{\,2} } } \over {6^{\,3} }} = {{165} \over {216}} = {{55} \over {72}}
$$
A: 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 
$$
p_2=1-\left (\frac{15}{17}\right)^{n-1}.$$
