I have proceeded as follows: We need exactly 3 +1s, which could be in any order. So permutation comes into mind. What is wrong, and how should I proceed next?
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3$$\frac{\text{number of outcomes with exactly three "$+1$"s}}{\text{number of outcomes}}$$ – angryavian Aug 28 '17 at 03:19
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As noted in a comment above, the computation is $$\frac{\text{number of desirable outcomes}}{\text{number of possible outcomes}}.$$
First, we can ask how many desirable outcomes there are, which can be done via a "Stars and Bars" style of argument: you have 6 slots to fill with 3 ones (the stars) and 3 negative ones (the bars). There are 6 choose 3 ways to pick slots for the 3 ones, hence the negative ones would have to go in the other slots. Thus there are $$ \binom{6}{3} = \frac{6!}{3!(6-3)!} = \frac{6\cdot 5\cdot 4}{3 \cdot 2} = 5\cdot 4 = 20$$ ways of getting zero.
On the other hand, there are $2^6 = 64$ possible outcomes. Thus the probability of adding to zero is $$\frac{20}{64} = 0.3125,$$ or about 31%.
Xander Henderson
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I don't under the "stars and bars" reference. We are not placing identical objects in distinct boxes. We are selecting three of the six positions in the sequence for the positive ones, as you calculated. – N. F. Taussig Aug 28 '17 at 09:37
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As I said, it is a stars and bars style argument. The way that I usually such arguments introduced is to say that we have $k$ identical slots, into which we must place $k$ identical bars, which partition the $n-k$ stars into groups. This is the same kind of argument. – Xander Henderson Aug 28 '17 at 13:23