6 cards labeled 1 to 6 are shuffled,
then turned over one by one.

What's the probability that the sum
of the first 2 will equal the sum of
the 2nd 2 and also the sum of the last 2?

I get 1 in 20.

For all three to have the same sum they must be the pairs 1,6; 2,5; 3,4. As the pairs can be in either order you can have any card in the first position of each pair. If you take the 1st, 3rd and 5th cards as the first card of the pair to represent the pair, you end up with a combination of choosing 3 cards out of 6 which gives 1/20.

Good try,Bob....but no ceegar !

Hint: if (as example) 1st card = 4, then there
is a 1/5 chance that the next one will be a 3;
then continue...

1/15, then.

The first card is anything, then the next card is correct with 1/5 probability. The third card can be anything not already chosen. The fourth will be correct match with 1/3 prob. The last two have to add up and can be in either order.

1/5*1/3=1/15

cute

Right on esube!

Thought twas cute too.