Congratulations to Tom, who sent in
the correct solution of 4/7
.
There are 12 combinations where you would win, 9 where you
would lose so there is a total of 21 combinations.
That means that there is a 12 in 21 chance of winning which
cancels down to 4 in 7 which is about 0.57 rounded up as a
decimal.
If you would like to have a go at
identifying all the possible positions yourself, you may find it
helpful to download the file on the problem page.
Congratulations also to Nick, aged 12 who
had a different approach.
There are 7 different places for blue ball 1 to finish.
If it finishes in the middle it must be touching blue ball 2,
1/7 of a chance of the blue balls touching.
There are 6 positions left for the blue ball 1 to finish. In
each position that blue ball 1 finishes there are 3 chances
blue ball 2 will be touching blue ball 1 and 3 chances that it
won't. Therefore 1/2 of 6/7 is the chance the 2 blue balls have
of touching if blue ball 1 finishes on the outside.
This is equal to 3/7.
3/7 + 1/7 = 4/7 which is the probability of two blue balls
touching. 4/7 = 0.5714.