In a certain game, players can do good (G), bad (B), or average (M). Players play one at a time, not at the same time. In way to win, a player needs to do better than the opponent. So G> B and M, M>B. If they do the same, then they need to repeat all the game.
Let's say that Player A does G or B with probability of $50\%$ while player B does G, B, or M with probability $\frac{1}{3}$.
You can't see the game but you know that one of the players between players A and B has finished playing and the table of winning probabilities says that Player A has now $75\%$ chance of winning. What has happened in the game?
$P(A wins) = \frac{1}{2}\frac{2}{3} + \frac{2}{6}P(A wins) = \frac{1}{2} = 0.5 $
$0.5(1+X) = 0.75 \rightarrow \frac{0.75}{0.50} – 1 = 0.5 $
So the probability of A has increased by 0.5. It means that A has done G. What do you think?
Best Answer
I will show that $A$ has done $G$ is not the correct answer. I will leave it for you to follow the same approach to get to the right answer.
At the beginning of a new game, the probability of $A$ winning is $0.5$ as you obtained. If $A's$ probability of win has increased to $0.75$, what are likely cases? It must be that either $A$ had the first turn and finished well or $B$ had the first turn and finished poorly.
Case $1$: player $A$ played first and did good ($G$).
$ \displaystyle P(A ~ \text{wins}) = \frac{2}{3} + \frac{1}{3} \cdot \frac{1}{2} = \frac{5}{6} \gt 0.75$
Explanation: $A$ has already finished with a $G$. Now there is $2/3$ chance that $B$ finishes with $B$ or $M$ ensuring a win for $A$ or there is a tie with probability $1/3$ in which case a new game starts afresh and from the beginning of a game probability of a win for $A$ is $1/2$.
Now what happens in case where player $B$ played first and did badly ($B$)? Can you find the probability of $A's$ win?