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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.
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 $\dfrac13$.
Suppose we have series of experiments, and we continue till event 1 occurs (with probability $p$, or the disjoint event 2 occurs (with probability $q$). If neither occurs (probability $1-p-q$) we throw again.
What is the probability that 1 occurs before 2? Call this probability $r$.
Now $r = P(\text {1 before 2}) = P(\text{1 before 2} | \text{first experiment event 1})P(\text{first experiment event 1}) + P(\text{1 before 2} | \text{first experiment event 2})P(\text{first experiment event 2}) + P(\text{1 before 2} | \text{first experiment not event 1 or event 2})P(\text{first experiment not event 1 or event 2})$.
Now this equals $1 \cdot p + 0 \cdot q + r(1-p-q)$, because if event 1 occurs in the first experiment, surely we have that 1 occurs before 2, and if event 2 occurs, it's certainly not the case that 1 occurs before 2, and if neither occurs, by independence of the experiments we have probability $r$ again, as we start over fresh.
So $r = p + r(1-p-q)$, so $(p+q)r = p$, or $r = \frac{p}{p+q}$.
Similarly we can reason that $P(\text {2 before 1}) = \frac{q}{p+q}$ and note that these sum to 1. So one of them happens, and the chances are in ratio to their respective success probabilities $p$ and $q$.
In your case $p = \frac{2}{16}, q = \frac{4}{16}$, so we get $\frac{1}{3}$ that sum 3 occurs before sum 5, and $\frac{2}{3}$ for the other way around.
$$ \left.\begin{array}{c|c|c|c|c} & 1 & 2 & 3 & 4 \\\hline 1 & 2 & 3 & 4 & 5 \\\hline 2 & 3 & 4 & 5 & 6 \\\hline 3 & 4 & 5 & 6 & 7 \\\hline 4 & 5 & 6 & 7 & 8\end{array}\right. $$ It is easy to tell various probabilities of the outcomes of this experiment using the table above. For example, the probability that we get $5$ in a single roll is equal to $4/16=1/4$.
Suppose that we roll the dice until we get either $3$ or $5$. The probability that we will eventually get either $3$ or $5$ is given by $$ \frac38+\frac58\cdot\frac38+\biggl(\frac58\biggr)^2\cdot\frac38+\ldots=\frac38\sum_{n=0}^\infty\biggl(\frac58\biggr)^n=\frac38\frac83=1 $$ using the independence and the geometric series (we can get either $3$ or $5$ on the first roll, on the second roll etc.). So this event happens almost surely, i.e. we will eventually get either $3$ or $5$. The question is what we are going to get first.
Hence, the the probability that a sum of $3$ is rolled before a sum of $5$ is $$ \frac18\sum_{n=0}^\infty\biggl(\frac58\biggr)^n=\frac18\cdot\frac83=\frac13. $$
One way to use conditional probabilities is to think about what would happen if you tried the dice-rolling experiment as described. How would you know which sum occurred first?
You would have to roll the dice until one of the sums occurred, and then see which one it was.
In other words, if $A$ is the event that the sum is $3$, and $B$ is the event that the sum is $5$, then you must wait for the first occurrence of $A \cup B$ and decide which of $A$ or $B$ occurred.
Looking forward to this experiment, you do not know how it will turn out but you know it will end with the event $A \cup B$. If the event $A$ also occurs at that time, then a $3$ was rolled first. So, how do you write the probability that $A$ has occurred when you know that $A \cup B$ has occurred, and how do you compute that probability?
When you are rolling the dice, three outcomes can happen:
With the first two outcomes, you're finished rolling, but with your third outcome you have to roll again. The probabilities are independent, so you can put together an equation that models this nicely. The probability of a 3-first-sum equals what I've calculated above plus this same probability times the probability of rolling something else:
$P(3 first) = P(3) + P(not 3 and not 5)*P(3 first)$
$P = \frac{1}{8} + \frac{5}{8}P$
$P = \frac{1}{3}$
