A Question Regarding the Meaning of a Question About the Probability of a R-P-S Game The Question: Student A and B are playing rock-paper-scissors. What is the probability that A wins two times within the first 5 rounds played? (Assume both players play rock, scissor, or paper purely randomly.)
Rather than asking for the answer to the question, I'd like to ask for everyone's opinion on whether this means that A must win exactly 2 games or just 2 games or more and also whether the two players stop playing when A wins twice. The "within" in the question has me slightly confused about the exact nature of the question, and I'd like to ask the community what they think about it. Thanks in advance!
 A: Let $p$ be the probability that A wins a randomly-played round, and let $X$ be the number of times A wins in five independent played rounds. Then
$X \sim \mathsf{Binom}(n = 5, p).$ According to @callculus's interpretation, you seek $P(X = 2) = {5 \choose 2}p^2(1-p)^3.$ 
Any ambiguity must lie in whether 'A wins two times' means 'A wins exactly two
times' or 'A wins at least two times'. I think a carefully worded problem
should have been specific. 
However, as it stands, I vote in favor of the
second interpretation, so that you seek $P(X \ge 2).$ To a native speaker of English, I think the inclusion of the word within makes this clear.
If A were promised a prize for meeting the condition, and A wins the first two rounds, I think it would hard
be to argue that A should not get the prize. Indeed, if I were A, I would have my hand out to receive the prize immediately after those first two wins. (Certainly, two is 'within five'.)
A: Pretty sure it means two times exactly in 5 trials. Yes, the word "within" is unnecessary but this would be a bigger misfit if what they really meant was at least 2 times in 5 trials. 
So the answer is given by the Binomial distribution with N = 5 and p = 1/3 
If they are playing randomly, and given that rock > scissors ; scissors > paper ; paper > rock. A total of $3^2 = 9$ ways in which the players can play, of which there are 3 ways to win. Hence, P(success) = 3/9 = 1/3.
$P(X=2) = (^5C_2)p^2(1−p)^3 = 10 \times 1/3^2 \times (2/3)^3 = 0.329$
If what they meant was at least 2 times, then just calculate the probability of winning exactly 0 and 1 times and subtract from 1:
$P(X\geq2) = 1 - P(X=0) - P(X=1) = 1-(^5C_0)p^0(1−p)^5 - (^5C_1)p^1(1−p)^4  = 1 - (1\times1\times(2/3)^5) - (5\times(1/3)\times(2/3)^4) = 0.539$
