Monte Carlo Coin Simulation

Question

There are 30 coins. While 29 of them are fair, 1 of them flips heads with probability 60%. You flip all 30 coins 100 times. Use the Monte Carlo method to compute within .1% the probability that the unfair coin is in set of 10 coins that land heads most often? (Count the event, that the biased coin is tied with the 10th most common fair coin).

Hint: use np.random.binomial to speed up simulation. A binomial variable with parameters n and p is the number of heads resulting from flipping n coins, where each has probability p of landing heads.

Answer 1

NO!

I am getting convergence... 60% for the biased, 50% for the other 29.

Sorry, not a python fan.

//*******************************************************************************************************/

import java.math.*;

class Coin

{

public static final boolean HEAD = true;

public static final boolean TAIL = false;

protected double p;

public Coin() { p=0.5; }

public Coin(double p) { this.p = p; } // p is the probability of heads

public boolean Flip()

{

return (

( Math.random() ) >= ( 1 - p )

);

}

public static void main(String args[])

{

int numCoins=30;

Coin coins[] = new Coin[numCoins];

int numHeads[] = new int[numCoins];

int coinLoop;

for (coinLoop=0; coinLoop<numCoins; coinLoop++)

{

coins[coinLoop] = new Coin();

numHeads[coinLoop]=0;

}

coins[numCoins-1] = new Coin(0.6);

long numTrials = 1000000;

for (long iLoop=0; iLoop<numTrials; iLoop++)

{

for (coinLoop=0; coinLoop<numCoins; coinLoop++)

{

if (coins[coinLoop].Flip()) { numHeads[coinLoop]++; }

}

}

for (coinLoop=0; coinLoop<numCoins; coinLoop++)

{

System.out.println(" coin # " + (coinLoop+1) + " : " + numHeads[coinLoop]);

}

}

}