probability question

[tania]

Q- Player 1 has 5 coins and Player 2 has 4 coins. Both players toss all the coins together and observe the number of times heads come up. Assuming the coins are fair, find the probability that player 1 has more heads than player 2.

What is the answer?

[Granpa Simpson]

I am getting answer as 0.5

[Granpa Simpson]

The possible cases are:
i) 1 head for player 1 and 0 head for player 2.
ii) 2 heads for player 1 and (0 or 1) head for player 2.
iii) 3 heads for player 1 and (0 or 1 or 2) head/heads for player 2.
iv) 4 heads for player 1 and (0 or 1 or 2 or 3) head/heads for player 2.
v) 5 heads for player 1 and (0 or 1 or 2 or 3 or 4) head/heads for player 2.
Now use binomial distribution and find:
Required Probability=Pr(X1=1)*Pr(X2=0)+Pr(X1=2)*Pr(X2<=1)+Pr(X1=3)*Pr(X2<=2)+Pr(X1=4)*Pr(X2<=3)+Pr(X1=5)*Pr9X2<=4).
Here X1: Random Variable denoting the number of heads for player 1.
       X2: Random Variable denoting the number of heads for player 2.
Consider getting of head as a success and compute the given probability by using Binomial Dist...i got the answer as 0.5.

[Amit Goyal]

Player 1 has 5 coins and Player 2 has 4 coins. Both players toss all the coins together and observe the number of times heads come up. Assuming the coins are fair, find the probability that player 1 has more heads than player 2.

The problem can be done by using the following symmetry argument:
Note that entire sample space can be partitioned into the following two events:
A = {Player 1 has more heads than Player 2}
B = {Player 1 has more tails than Player 2}
such that A U B = S and A ∩ B = ∅
Thus, Pr(A) + Pr(B) = 1.
And by symmetry, Pr(A) = Pr(B)
So, Pr(A) = 0.5