Expected number of rolls given that you roll the die till you get the same or higher than previous value

Question

Can anyone explain to me how the expectation of geometric distribution can be used here?

Answer 1

Think of scoring $U$ or above as 'success' in a series of trials. This gives us a geometric distribution and, if the probability of success on one trial is $p$, then the number of goes before success is achieved has expectation $\frac{1}{p}.$

As you realised, there are 6 equally likely outcomes for $U$ which we have to sum over. Furthermore, $p$ is given by $\frac{7-X}{6}.$

The sum is therefore the sum of six terms each of the form$\frac{6}{7-X}$ multipled by $\frac{1}{6}$ i.e. $\frac{1}{7-X}.$ This gives the sum which @Don Thousand gave an hour ago but which seems to have been questioned by other contributors.

I hope this helps.

Answer 2

At the demand of @Don Thousand

This is not an answer. It is only to show how I have simulated the issue and found a mathematical expectation of $\approx 1.52$.

Here is my Matlab program :

N=1000000;
for k=1:N
   r=ceil(6*rand);%initial roll 
   s=ceil(6*rand);nd=1;
   while s<r
      s=s+ceil(6*rand);
      nd=nd+1;
   end;
   T(k)=nd;
end;
mean(T)