I don't see an easy way to do this, as there are 50 different numbers of heads.
Pr(1)=(.5)^1(.5)^49
and the Q(r) the second person is the same, so the joint probability that both will get only one H is
Pr(1,1)=(.5)^2(.5)^98 =(.5)^100
Now, Pr(2,2)=(.5)^4(.5)^96=.5^100
and Pr(3,3)=(.5)^6(.5)^94=.5^100
so you have to sum all these up, from 0.0 to 50,50
as in Pr(0,0)=.5^0*.5^100+(pr(1,1)=.5^100 + ....+Pr(50,50)=.5^100
That is 51 times, or
Pr(total, total same)=51*.5^100
check my thinking
Two People flip a coin, Each person flips it 50 times, Each person then counts the number of heads, What is the probability they both get the same number of heads?
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