# throwing a dice probability distribution

I was asked to find the probability distribution of a random variable, $X$, that counts the number of tries until getting the result $3$ and the result $2$.

I'm not quiet sure about my answer and I'd like to know if I'm on the 'right path' of solving the question, and if not then is there another way of solving it? Using Geometric distribution maybe?

$P(X=k) = \frac1{6}*\frac1{6}*(1-\frac4{6})^{k-2}*2*(k-1)$

$\frac1{6}$ = the probability of getting 3 throwing a dice

$\frac1{6}$ = the probability of getting 2 throwing a dice

$1-\frac4{6}$ =the probability of getting anything but 3 or 2 throwing a dice and since there are k tries, then we multiply the possibilities by $k-2$

$2*(k-1)$ = choosing on which try we got 3 (since the last will be 2) or the opposite.

• Does the $2$ have to occur after the $3$ or can it come earlier? For example, if the first three rolls are $2,3,2,$ is $X=2$ or is $X=3$? It sounds like you would have $X=2$ but it isn't completely clear. – David K Nov 17 '17 at 18:16
• Looks correct to me! – Karn Watcharasupat Nov 17 '17 at 18:53
• @DavidK what if the first three rolls are 2, 2, 2, and the forth is 3? how do I solve it then? Like, if the question is changed and that's what I'm asked to find – Lola Nov 17 '17 at 20:22
• That's a good point, it looks like that particular sequence would not be counted in your formula although it is a legitimate way to use $4$ rolls to get both of the results $2$ and $3.$ So you need a different way to compute the probabilities. – David K Nov 17 '17 at 20:53

I would try modeling it this way:

First you have to roll one of the two numbers $2$ or $3.$ Let's say this first happens after $X_1$ rolls; then $X_1$ is $1$ plus a geometric random variable with parameter $1/3$ (because of the six equally likely outcomes of each roll, two of them will end this sequence of rolls). That is, $P(X_1 = 1) = \frac13,$ $P(X_1 = 2) = \frac13\left(\frac23\right),$ $P(X_1 = 3) = \frac13\left(\frac23\right)^2,$ and so forth.

Then, after the first time you roll either $2$ or $3,$ you have to roll the other number. This takes some number of additional rolls. If the additional number of rolls required is $X_2,$ then $X_2$ is $1$ plus a geometric random variable with parameter $1/6$ (because now on each roll there is only one outcome that will allow us to end this sequence of rolls). That is, $P(X_2 = 1) = \frac16,$ $P(X_2 = 2) = \frac16\left(\frac56\right),$ $P(X_2 = 3) = \frac16\left(\frac56\right)^2,$ and so forth.

The total number of rolls is $X = X_1 + X_2.$

Example: Suppose the first roll is $2.$ Then you have already rolled one of the numbers $2$ or $3$ on the first roll, so $X_1 = 1.$ Next you have to roll the "other" number, which in this case is $3.$ Now suppose the next three rolls are $2, 2, 3.$ Since you got your first $3$ on the third roll, $X_2 = 3.$ The entire sequence of rolls was $2, 2, 2, 3,$ so it took four rolls to roll both of the numbers, and indeed $X = X_1 + X_2 = 1 + 3 = 4.$

• The answer doesn't include the example I gave in the comments? – Lola Nov 17 '17 at 21:05
• Thanks for the reply! – Lola Nov 17 '17 at 21:09
• Sorry again for the many questions, but why is X= X1+X2 and not X1*X2? – Lola Nov 17 '17 at 21:10
• For $X_1,$ on each roll there's a $2/6$ probability that you roll $2$ or $3,$ thereby ending the sequence of rolls that you needed to make in order to get one of those numbers, and there is a $4/6$ probability that you'll roll one of the other four numbers and that you'll have to roll again in order to try to get a $2$ or $3.$ I have added your example at the end of the answer, showing how $X_1$ and $X_2$ apply to that example, and I hope this also makes it clear why $X = X_1+X_2.$ – David K Nov 17 '17 at 21:24
• Brilliant! Thanks for the answer – Lola Nov 17 '17 at 21:29