A die is rolled repeatedly until a three shows up for the first time. What is the probability that the three appears at the second roll, given that the first three appears at an even number of rolls?
I defined event A : the three appears at the second roll$=1/36$; and event B:the first three appears at an even number of rolls, which I guess it should be $=1/2$; but I didn't know how to proceed.
On
It is strange that odds are so often mistakenly used as synonymous to probability, and aren't used at all where their correct usage would greatly simplify computation:
If neither A nor B in sequence win on first try, the cycle repeats,
so the odds against $B$ are $\dfrac16:\dfrac56\dfrac16 = 1:\dfrac56 = 6:5$
Thus P(B wins on even try) $= \dfrac5{5+6} =\dfrac5{11}$
But P(wins on try $2) = \dfrac5{36}$
Hence P(wins on 2nd try | B wins on even try) = $\dfrac{(5/36)}{(5/11)} = \dfrac{11}{36}$
There's no need to do any infinite sums.
Let $p$ be the probability that a $3$ first appears on an odd roll. Then we will get a $3$ on the first roll with probability $\frac 16$, and with probability $\frac 56$ the likelihood of having the first $3$ on an odd roll will be $1-p$ (because odd and even rolls have now swapped positions). Therefore, $p=\frac 16 + \frac 56(1-p)$, so $p=\frac{6}{11}$. Thus, the probability that the first $3$ appears on an even roll is $1-p=1- \frac {6}{11}=\frac {5}{11}$.
The probability that the first $3$ appears on the second roll is $\frac 56 \cdot \frac 16=\frac{5}{36}$, and if that does occur, then the first $3$ necessarily occurs on an even roll. Thus, given that the first $3$ occurs on an even roll, the probability that it's precisely the second roll is $\frac{5/36}{5/11}=\frac{11}{36}$.