Uva 10056 Solution

Solution : What is the Probability?

In This Problem Gives,

        1. The total number of people in this game.

        2. Probability of each person winning = Probability of the happing of a successful event in a single throw.

        3. Probability of the required number of people to win.

We assuming that all have N people everyone's chances of winning are p, the probability of losing is q = (1 - p)

# First Round( Each Person Winning Probability ) :

             ➤ Probability of the first person winning = p

             ➤ Probability of the second person winning = q . p

             ➤ Probability of the third person winning = q² . p

             ➤ Probability of the k^th person winning = q^(k-1) . p

# Second Round( Each Person Winning Probability ) : Assume everyone loses in the First round.

             ➤ Probability of the first person winning = q^N . p

             ➤ Probability of the second person winning = q^N . q .  p

             ➤ Probability of the third person winning = q^N . q² . p

             ➤ Probability of the k^th person winning = q^N . q^(k-1) . p

#............................................................................. Going On......

We don't know which round a person will be win. For that, we count each person probability in each round until it greater than 10^-7. Suppose this round number will be R.

# R^th  Round :

            ➤  Probability in R^th  round of the k^th person winning = q^(R-1).N . q^(k-1) . p

Coding Part : 

#include<bits/stdc++.h>

using namespace std;

int main()

{

int t;

cin >> t;

while(t--)

{

double total_player, probability, player_number;

cin >> total_player >> probability >> player_number;

double result = 0;

while(probability * pow( (1.0 - probability), (player_number - 1) ) > 1e-7)

{

result += probability * pow( (1 - probability), (player_number - 1) );

player_number += total_player;

}

cout << showpoint;

cout << fixed;

cout << setprecision(4);

cout << result << "\n";

}

return 0;

}