[Probability] Derangement / gambling problem

[goraemon]

Homework Statement

Alice proposes to Bob the following game. Bob pays one dollar to play. Fifty balls marked 1, 2, . . . , 50 are placed in a big jar, stirred around, and then drawn out one by one by Zori, who is wearing a blindfold. The result is a random permutation (let's call it s) of the integers 1, 2, . . . , 50. Bob wins with a payout of two dollars and fifty cents if the permutation s is a derangement, i.e., s(i) =/= i for all i = 1, 2, . . . , n. Is this a fair game for Bob? If not, how should the payoff be adjusted to make it fair?

Homework Equations

Derangement formula: ∑(k=from 0 to n) [(-1)^k * C(n,k) * (n-k)!]

The Attempt at a Solution

I got this far:
Probability that any given random permutation is a derangement:
(num of all derangements) / (num of all permutations) = ∑(k=from 0 to n) [(-1)^k * C(n,k) * (n-k)!] / 50! ≈ 1/e

Then, computing the expected win/loss, given that he gains $2.50 (for a net profit of $2.50 - $1 - $1.50) if he wins, and is out $1.00 if he loses:
1.5(1/e) - 1(1-1/e) ≈ -0.08. (so on average, he'll lose roughly 8 cents per play)

Solving for the adjusted payment amount so that he'll break even in the long run...
(p-1)(1/e) - 1(1-1/e = 0
Solving for p, we get: p ≈ e

First, is the above correct? Right now I settled for an approximate answer (e) as the derangement equation seemed too messy to compute an exact answer. Is there a simpler way to get an exact answer? Thanks.

 

[Ray Vickson]

Homework Statement

Alice proposes to Bob the following game. Bob pays one dollar to play. Fifty balls marked 1, 2, . . . , 50 are placed in a big jar, stirred around, and then drawn out one by one by Zori, who is wearing a blindfold. The result is a random permutation (let's call it s) of the integers 1, 2, . . . , 50. Bob wins with a payout of two dollars and fifty cents if the permutation s is a derangement, i.e., s(i) =/= i for all i = 1, 2, . . . , n. Is this a fair game for Bob? If not, how should the payoff be adjusted to make it fair?

Homework Equations

Derangement formula: ∑(k=from 0 to n) [(-1)^k * C(n,k) * (n-k)!]

The Attempt at a Solution

I got this far:
Probability that any given random permutation is a derangement:
(num of all derangements) / (num of all permutations) = ∑(k=from 0 to n) [(-1)^k * C(n,k) * (n-k)!] / 50! ≈ 1/e

Then, computing the expected win/loss, given that he gains $2.50 (for a net profit of $2.50 - $1 - $1.50) if he wins, and is out $1.00 if he loses:
1.5(1/e) - 1(1-1/e) ≈ -0.08. (so on average, he'll lose roughly 8 cents per play)

Solving for the adjusted payment amount so that he'll break even in the long run...
First, is the above correct? Right now I settled for an approximate answer (e) as the derangement equation seemed too messy to compute an exact answer. Is there a simpler way to get an exact answer? Thanks.

As Feller (Introduction to Probability Theory, Vol. I) points out, the probability that in a permutation of ##n## numbers exactly ##k## of the numbers match (that is, we have ##s(i) = i## for ##k## of the numbers from ##1## to ##n##) is
[tex] P(k \; \text{matches}) = \frac{1}{k!} \left( \frac{1}{0!} - \frac{1}{1!} + \frac{1}{2!} - \cdots \pm \frac{1}{(n-k)!} \right). [/tex]
In particular, for ##k = 0## and ##n = 50## we get your probability of a derangement (0 matches) as
[tex] P(\text{derangement}) = 1 - 1 + \frac{1}{2!} - \frac{1}{3!} + \cdots + \frac{1}{50!} [/tex]
This, of course, is the first 51 terms in the series expansion of ##e^{-1} = 1/e##, and because you have an alternating series, the error made in stopping the computation of ##1/e## at ##1/50!## is smaller than ##1/51! \approx 0.6446959640\, 10^{-66}##. So, your error in replacing the answer by ##1/e## is not too bad. Admittedly, this is an approximation, but the error would only show up at the 66th or 67th decimal place, if that is what you were worrying about.