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Congratulations to Tom from Colyton Grammar 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.

This is how you can arrive at the 21 combinations:

Let's label the 6 positions on the outside ring A, B, C, D, E and F.

If 1 blue ball falls into the centre, the other blue ball can fall into any of the 6 positions on the outside ring.

If there is no blue ball in the centre, the
two blue balls can fall into the following 15 positions:

AB, AC, AD, AE, AF

BC, BD, BE, BF

CD, CE, CF

DE, DF

EF

This gives a total of 6+5+4+3+2+1 = 21
combinations

12 of these are winning combinations:

the first six (when one blue ball falls into the centre) and AB,
AF, BC, CD, DE and EF.

Congratulations also to Nick from Kegs 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,
so there is 1/7 of a chance of the blue balls touching with blue
ball 1 in the centre.

There are 6 positions left for blue ball 1 to finish. In each
of these positions there are 3 positions in which blue ball 2 will
be touching blue ball 1 and 3 positions in which it won't.
Therefore 1/2 of 6/7 is the chance of the two blue balls touching
when blue ball 1 finishes on the outside. This is equal to
3/7.

1/7 + 3/7 = 4/7 which is the probability of the two blue balls
touching. 4/7 = 0.5714.