Rolling a die until the sum of all numbers rolled is greater than 6

Question

I'm trying to solve a problem in which I'm asked the probability mass function of $X$ where $X$ is the number of times a dice is rolled until a sum greater than six is encountered. I was able to deduce that the possible values that $X$ can take are $2,3,4,5,6,7$ (at most we keep gettings $1$s each time), however deducing how often does each of them occurs wasn't easy for me to conclude without drawing the sample space tree, but the problem is that it gets out of control (in terms of size) quickly and I'm not able to find the pattern that would make me solve it quicker otherwise.

Thank you, have a nice day.

Answer 1

Recursion will do:

If $f(x,n)$ is the probability you need $x$ throws until you exceed $n$, then

• $f(0,n)=1$ for $n<0$ and $f(x,n)=1$ for $x \not=0$ and $n<0$
• $f(x,n)=\frac16(f(x-1,n-1)$ $+f(x-1,n-2)$ $+f(x-1,n-3)$ $+f(x-1,n-4)$ $+f(x-1,n-5)$ $+f(x-1,n-6))$ for $n\ge 0$

giving the table

   x  0   1    2      3      4       5       6       7
n   
<0    1  
0     0  6/6  
1     0  5/6  6/36
2     0  4/6 11/36  6/216
3     0  3/6 15/36 17/216   6/1296
4     0  2/6 18/36 32/216  23/1296  6/7776
5     0  1/6 20/36 50/216  55/1296 29/7776  6/46656
6     0   0  21/36 70/216 105/1296 84/7776 35/46656 6/279936

and you want the bottom row

Answer 2

Many parts of the tree are the same. There are only seven probability mass functions you need: If you have already reached $n$, how many more rolls will you need?
Work backwards,, starting with $n=6$ and ending with $n=0$.

Answer 3

Let $P (X) $ be the probability that the sum does not exceed 6 after $X $ throws. The number of such events is the number of positive integer solutions to the inequality: $$k_1+k_2+\cdots+k_X\le6, $$ which by stars and bars is $$ \binom 6X. $$

Hence $$P (X)=\frac1 {6^X}\binom6X.$$

Now observe that the probability you are looking for is $$P (X-1)-P (X). $$