[Math] probability that a sum of 3 is rolled before a sum of 5 is rolled in a sequence of rolls of the dice from a four sided die.

bayes-theorempermutationsprobability

I am stuck at this problem-

A pair of four-sided dice is rolled and the sum is determined.
What is the probability that a sum of 3 is rolled before a sum of 5 is rolled
in a sequence of rolls of the dice?

What I tried-

Let $A=$ sum of $3$;
Let $B=$ sum of 5;

then

$P(A \mid \text{not}~B)=$ i.e probability of A given that B has not happened.

$P(A \mid not B)= \dfrac{P(A~\text{and not}~B)}{P(\text{not}~B)}$;

But using this approach I am not getting answer.

Best Answer

There are $2$ ways of making $3(12,21)$, $4$ ways of making $5 (14,41,23,32)$ and $10$ ways of making anything else.

Represent these possibilities by $1,2,5$ - we wish to determine the probability of $1$ occurring before $2$.

We can ignore all initial throws of $5$, and as each throw is independent, the problem reduces to which occurs first of $1$ and $2$.

As $2$ is twice as likely to happen as $1$, the answer is $\dfrac13$.

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