A man went to Monte Carlo to try and make his fortune. Whilst he
was there he had an opportunity to bet on the outcome of rolling
dice. He was offered the same odds for each of the following
outcomes: At least 1 six with 6 dice. At least 2 sixes with 12
dice. At least 3 sixes with 18 dice.
Two bags contain different numbers of red and blue balls. A ball is
removed from one of the bags. The ball is blue. What is the
probability that it was removed from bag A?
A gambler bets half the money in his pocket on the toss of a coin,
winning an equal amount for a head and losing his money if the
result is a tail. After 2n plays he has won exactly n times. Has he
more money than he started with?
Again we had the solution to this question
from Joel, well done Joel.
The solution is to put all the balls into one bag, then (for any
number of balls, say n red and n white)
P(red) = 1/2 (1/2) + 1/2 (0) = 1/4.
The solution here is to put no white balls and at least one red
in one bag, then the probability of choosing that bag and a red
from it is 1/2, which is maximal. In order to make the probability
of choosing a red from the other bag as large as possible we put as
many reds in that bag as possible, which means putting exactly one
red and no whites in the first bag and all the remaining balls in
the second bag. With 4 red and 4 white balls the probability of
selecting a red ball is then 5/7.
The case for n = 100 is left to the reader. In general,
with 1 red in one bag and all the remaining balls in the other bag,
the probability of drawing a red ball tends to a limit as
n tends to infinity. You may like to find this limiting