Jane Street Puzzle Solution - Mar 2023

[Giovanni Crocioni]

Thanks for the post, its very well written and I had no problem understanding it. Only thing, isn't P(S>S') = 1/6(x^2+x+1) rather than 1 - 1/6(x^2+x+1)?

[Herbping]

An alternative way to relate p0 and X.

Prob(Eventually not blow up when the current score = 0) = \sum_{n = 0}^{\infty} Prob(Land in [X,1] after the n'th decision)

Prob(Land in [X,1] after the n'th decision) = Prob(Land in [0,X] after the (n-1)'th decision)\times Prob(U(0,1)\in [X,1])

According to the PDF of Irwin–Hall distribution,

Prob(Land in [0,X] after the (n-1)'th decision) = \int_{y=0}^X \frac{y^{n-1}}{(n-1)!} = \frac{X^n}{n!}

Therefore

Prob(Land in [0,X] after the (n-1)'th decision)\times Prob(U(0,1)\in [X,1]) = (1-X)\frac{X^n}{n!}

At last,

Prob(Eventually not blow up when the current score = 0) = \sum_{n = 0}^{\infty} Prob(Land in [X,1] after the n'th decision) = \sum_{n = 1}^{\infty}(1-X)\frac{X^n}{n!}=(1-X)e^X

Prob(Eventually blow up when the current score = 0) = 1-(1-X)e^X