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.
You and I play a game involving successive throws of a fair coin.
Suppose I pick HH and you pick TH. The coin is thrown repeatedly
until we see either two heads in a row (I win) or a tail followed
by a head (you win). What is the probability that you win?
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?
Assume that removing a ball from bag A or bag
B is equally likely, the probability of choosing a
blue ball from bag A is 3/10 and the probability
of choosing a blue ball from bag B is 4/14 so the
probability of choosing a blue ball is 3/10 + 4/14 = 41/70.
Theoretically in 70 trials, 41 of the outcomes would give a blue
ball and 21 of these would of been drawn from bag
A . Given that the ball drawn was blue the
probability that it came from bag A is 21/41.
The only correct solution to date came from
Matthew (Smithdon High School)