This is a wonderful book which I've just bought. It presents 79 riddles in probability, organised into 11 chapters on different topics, building on and extending the material collected on the cut-the-knot website.
I've only dipped into the book so far but in Chapter 3 Likely Surprises I came across this gem.
A flips a fair coin 20 times and B flips another 21 times. What is the probability that at the end B has more heads than A?
Bogomolny gives five solutions but my favourite is the last one. To read my version of the argument scroll down.
Solution
Suppose A has HA heads and TA tails, and B has HB heads and TB tails.
We know HA + TA = 20 and HB + TB = 21, so HB + TB > HA + TA.
But this would not be possible unless either HB > HA or TB > TA: B must either have more heads than A or more tails than A (it can't be both because B only did 21 coin flips).
But the riddle is completely symmetric in heads and tails and therefore we have
prob(
B has more heads than
A) = prob(
B has more tails than
A) =
.
The same argument applies for any numbers n and n+1 in place of 20 and 21. We can easily check this by hand for n=1
| A |
B |
| H |
HH, HT, TH, TT |
| T |
HH, HT, TH, TT |
You can see that B has more heads than A exactly 50% of the time.
Tags: Bogomolny, coin flips, cut-the-knot
