[MikeL05]

Compare your math to rolling a die. There are 6 sides, so chances of rolling a SIX is .166667. Now, if you roll it 6 times, it's not the case that you'll hit 1.0 SIXes. You'll hit on average 1.0 SIXes. Likewise, multiplying your percentage (which I believe is right) by 38, you've hit upon .59. This is not your chance of hitting one string of three in a row; this is the average number of 3-in-a-row sets you'll hit in one package.

What I think you need to do is take your probability and subtract it from 1.00. So 1.00 - .0156 = .984. Now raise this number .984 to the exponent "n", where "n" is the number of wafers.

So .984^38 = probability of NOT getting a set of 3 in a row.

1.00 - that = probability of hitting at least one set in 38 wafers.