Exercise C.4.6
⋆ Professor Rosencrantz flips a fair coin n times, and so does
Professor Guildenstern. Show that the probability that they get the same
number of heads is (n2n)/4n. (Hint: For Professor Rosencrantz,
call a head a success; for Professor Guildenstern, call a tail a success.)
Use your argument to verify the identity
k=0∑n(kn)2=(n2n)
Let's do what the hint says: call tails for Professor Guildenstern a success
and heads - failure. They have the same numbers of heads when they have
k+(n−k)=n successes out of 2n trials.
Pr{R=G}=b(n;2n;1/2)=(n2n)2n12n1=(n2)/4n
Alternatively, we can just calculate the probability and see what happens:
Pr{R=G}=k=0∑nPr{R=k}Pr{G=n−k}=k=0∑n(kn)⋅2n1⋅(n−kn)⋅2n1=4n1k=0∑n(kn)(n−kn)=4n1k=0∑n(kn)2
Those are two ways we can express the same value. This let's us verify the
identity.