JoeSapphire: So everybody who enters has as much chance of fixing Vitek's mess as they have of ruining everything.

Lifthrasil: Interesting that one doesn't even need much maths to solve this seemingly difficult problem. :-)

Lifthrasil: I neglected one case in my above formula, also it contains a mathematical error. There are actually two special seats one has to consider. The chance that 'Vitek' sits in his own seat was covered. But there is also the chance that he sits in the seat of the last person. In which case the chance to get it correct is 0.

So the correct formula would be: P(n) = (1/n)*1 + (1/n)*0 + ((n-2)/n)*P(n-1). Sure, the (1/n)*0 falls out of the equation since it's 0. But the -2 instead of the -1 makes a big difference. It allows to solve by forward induction instead of working backwards over sums (which I neglected to do formally correct).

We know that P(2)=1/2
Now let's take the above equation while assuming that we know that P(n-1)=1/2

P(n) = 1/n + ((n-2)/n)*(1/2)
= 2/2n + (n-2)/2n = (2+n-2)/2n = n/2n = 1/2

So, if any P(n) is 1/2, then all higher P(n+m) are also 1/2 and we know that for n=2 the assumption P(n)=1/2 is true.

ZFR: And now I'm embarassed for notchecking your solution thoroughly...

JoeSapphire: and you call yourself the king of maths... tut tut tut...

yogsloth: It's legitimately distressing to me that I can no longer follow the math. I feel like less of a man.

I brute-forced n=2, 3, and 4, and they all came up 50%.

yogsloth: It's legitimately distressing to me that I can no longer follow the math. I feel like less of a man.

JoeSapphire: (you can make yourself feel better by claiming to have checked someone's working and telling them it all adds up. Makes you look very smartbrains.)

ZFR: Viiiiiiiiiiiiiiiiiteeeeeeeeeeeeeeeeeeeek

SirPrimalform: