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?
You have two bags, four red balls and four white balls. You must
put all the balls in the bags although you are allowed to have one
bag empty. How should you distribute the balls between the two bags
so as to make the probability of choosing a red ball as small as
possible and what will the probability be in that case?
The gambler will have less money than he started with.
Suppose the amount of money before a game is $m$, then:
$m \to 3m/2$ for a win and $m\to m/2$ after losing a game.
After $n$ wins and $n$ losses he will have $(3/4)^n$ times the money he started with, irrespective of the order in which his wins and losses occur. Eventually he will run out of money as what he has left will be smaller than the smallest coin in circulation.
The diagram was suggested by Roderick and Michael of Simon Langton Boys' Grammar School Canterbury who pointed out that if the gambler went on indefinitely he would, in theory, end up with an infinitely small amount which would be represented by nothing.