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?
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?
Chris and Jo put two red and four blue ribbons in a box. They each pick a ribbon from the box without looking. Jo wins if the two ribbons are the same colour. Is the game fair?
Here is another nut that has been waiting to
be cracked for months. Three solutions arrived simultaneously from
Bradley, Avery Coonley School, USA; Joe, Anglo Chinese School,
Singapore and Ling Xiang Ning, Allan, Tao Nan School, Singapore,
whose solution is given below.
As there are 9*8*7 possible choices of 3 numbers, and the order
of revealing the numbers does not matter, the number of choices of
the three numbers is (9*8*7)/6 = 84
The probability of winning is the number of winning combinations
divided by 84.
We have to count the number of winning combinations:
We start from 987, 986? and the first number chosen gets
smaller, and the other numbers must be smaller than the first so
that three numbers will not appear together more than once.
There are total 38 choices, as shown:
987, 986, 985, 984, 983, 982, 981, 976, 975, 974, 973, 972, 971,
965, 964, 963, 962, 961, 954,
953, 952, 943, 876, 875, 874, 873, 872, 871, 865, 864, 863, 862,
854, 853, 765, 764, 763, 754
So, the probability of winning a price is 38/84 or also,