Given four integers p, q, r and s. Two players are playing a game where both the players are hitting a target and the first player who hits the target wins the game, the probability of the first player hitting the target is p / q and as that of second player hitting the target is r / s. The task is to find the probability of the first player winning the game.
Input: p = 1, q = 4, r = 3, s = 4
Input: p = 1, q = 2, r = 1, s = 2
Approach: The probability of the first player hitting the target is p / q and missing the target is 1 – p / q.
The probability of the second player hitting the target is r / s and missing the target is 1 – r / s.
Let, the first player be x and the second player be y.
So the total probability will be x won + (x lost * y lost * x won) + (x lost * y lost * x lost * y lost * x won) + … so on.
Because x can win in any turn, Its an infinite sequence.
Let t = (1 – p / q) * (1 – r / s). Here t < 1 as p / q and r / s are always <1.
So the series will become, p / q + (p / q) * t + (p / q) * t2 + …
This is an infinite GP series with common ratio less than 1 and it’s sum will be (p / q) / (1 – t).
Below is the implementation of the above approach:
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