Al tosses a fair coin three times, then Benjamin tosses it twice.
What is the chance, in per cent, that Al throws more heads than Benjamin?
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There is a 1/8 chance that Al throws 3 heads, in which case no matter what Benjamin does, Al throws more heads.
There is a 3/8 chance that Al throws 2 heads, in which case there is a 3/4 chance that Benjamin fails to match him.
There is a 3/8 chance that Al throws 1 head, in which case, there is a 1/4 chance that Benjamin fails to match him.
And there is a 1/8 chance that Al throws 0 heads, in which case no matter what Benjamin does, Al cannot better him.
So the chance that Al throws more heads than Benjamin is 1 / 8 + ( 3 / 8 ∗ 3 / 4 ) + ( 3 / 8 ∗ 1 / 4 ) = 4 / 3 2 + 9 / 3 2 + 3 / 3 2 = 1 6 / 3 2 = 50.00 per cent.