Probability question: names in hats

[Aractus]

Here's the probability tree for 4 people:



So for Sn where n equals the probability that person n makes winning impossible for person 4, S1 + S2 + 2S3 = 1. S3 gives you both the probability of winning, and the probability that person 3 creates a loosing situation for person 4.

So:

S1 + S2 + 2S3 = 1
2S3 = 1 - S1 - S2

S3 = ½(1 - S1 - S2)

Okay, and here are the sums themselves.

S1 = ¼(1+1/3)
S2 = ¼(1/3)[3 + 3(1/3) + (1/3)½ + ½]
S3 = ¼(1/3)½[2 + 2(1/3) + (1/3)½ + ½]

Now do note that S1 only concerns itself with 4 being picked, whereas S2 requires that both 4 is not picked and 3 is not available to be picked next. So we could instead come up with a formula to compensates for this, we'll call it S2b that only cares if 4 is picked. It would look like this (I'll compensate for S3 while I'm at it):

S1 = ¼(1+1/3)
S2b = ¼(1/3)[2 + 2(1/3) + (1/3)½ + ½]
S3b = ¼(1/3)[2½ + 2(1/3)½ + (1/3)½2 + ½2 + (1/3)]