I'll formalise a bit the method I am proposing.
At a stage of the game where there are N ≥ 2 unopened boxes (including the contestant's one), a money offer is made, whose value M is smaller than the value of m of the unopened boxes, where 1 ≤ m ≤ N.
If the contestant takes the money, the game ends. If the contestant doesn't take the money, he must open another n boxes, where 1 ≤ n < N
How does the contestant decide whether to take the money or not?
Assume that all the boxes have different monetary value, and that every time a box is opened, the remaining N boxes are sorted by their ascending value and renumbered from 1 to N, so that vi < vj ∀ i < j ≤ N. From the above constraint, we also know that vN-m< M < vN-m+1.
The expected value E at this stage of the game is:
E = \frac{\sum_{i=1}^N v_{i}} {N}
Simply comparing E with M is not effective, as M < E in all cases known so far, therefore if the choice were based on getting at least the expected value, the money offer would never be accepted.
I propose the alternative approach of calculating the probability P(N,m,n) that all of the m boxes whose value is greater than M be opened in the next n choices.
If n < m, we know that this probability is zero, because of course we can't open all the better m boxes by opening a number of boxes smaller than m: at least m-n of the better boxes will remain.
If n ≥ m, the following formula should give the probability:
P(N,m,n) = \frac{n!} {m! \cdot (n-m)!} \cdot \prod_{i=1}^m \frac{m-i+1} {N-i+1} = \frac{n!} {m! \cdot (n-m)!} \cdot \frac{m! \cdot (N-m)!} {N!} = \frac{n! \cdot (N-m)!} {N! \cdot (n-m)!}
Numerical example.
N=10 boxes remain, prizes {0.01, 1, 5, 100, 500, 1000, 5000, 10000, 250000, 500000}.
Expected value = 85660.601.
A money offer M = 20000 is made, which is smaller than m=2 remaining prizes. If the money offer is refused, n=3 boxes must be opened.
P(10,2,3) = \frac{3! \cdot (10-2)!} {10! \cdot (3-2)!} = \frac{1} {15}
The contestant decides that this is a low risk, compared to his 1/5 chances of having one of the better prizes, and refuses the offer.
3 boxes are opened, and the new set N=7 is as follows: {1, 5, 500, 1000, 5000, 10000, 500000}.
Expected value ≅ 86643.71.
A money offer M=30000 is made, which is smaller than m=1 remaining prizes. If the money offer is refused, n=3 boxes must be opened.
P(7,1,3) = \frac{3! \cdot (7-1)!} {7! \cdot (3-1)!} = \frac{3} {7}
The contestant's probability of having the one remaining prize higher than M is 1/7, and the risk of losing it in the next round of box openings is 3/7.
He decides that this is too high a risk, and accepts the offer.
[Is this a good choice? Should he balance the probabilities against the actual values of the offer and of the better prize? E.g. (1/7)/(3/7)=1/3 but 500000/30000=50/3, so (50/3)/(1/3)=50, and taking the risk may be worth it? I'm not sure about this point].
Suppose instead that the following happened at N=7.
The new set N=7 is: {1, 5, 500, 1000, 5000, 250000, 500000}.
Expected value ≅ 108072.28.
A money offer M=100000 is made, which is smaller than m=2 remaining prizes. If the money offer is refused, n=3 boxes must be opened.
P(7,2,3) = \frac{3! \cdot (7-2)!} {7! \cdot (3-2)!} = \frac{1} {7}
In this case the contestant may decide to refuse the offer, because his probability of having either of the m better prizes is 2/7, whereas the risk of losing both of them in the next round of box openings is 1/7.
Another way of decreasing the contestant's chances of losing money would be to calculate the probability that the expected value decrease.
It is generally observed that the money offer M is somewhat proportional to the expected value, though always smaller.
So by minimising the chances of decreasing the expected value, the money offer should be accepted when the risk of it decreasing is too high.
A method similar to the one above should apply, with the difference that in this case it's sufficient to lose one of the prizes of value greater than E, for E itself to decrease.
Let's call r the number of prizes of value > E.
Then:
P(\text {at least 1 prize>E lost}) = 1-P(\text {no prize>E lost}) = 1 - \prod_{i=1}^n \frac{N-r-i+1} {N-i+1} = 1 - \frac{(N-r)! \cdot (N-n)!} {(N-r-n)! \cdot N!}
Example with the same numbers as above.
N=10 boxes remain, prizes {0.01, 1, 5, 100, 500, 1000, 5000, 10000, 250000, 500000}.
Expected value = 85660.601.
A money offer M = 20000 is made. If the money offer is refused, n=3 boxes must be opened.
r=2 prizes with value higher than E remain.
P(\text {at least 1 prize>E lost}) = 1 - \frac{(10-2)! \cdot (10-3)!} {(10-2-3)! \cdot 10!} = \frac{8} {15}
The contestant decides that this is a high risk, compared to his 1/5 chances of having one of the m=2 better prizes, and accepts the offer.
I think this strategy is a bit of a chicken-out one: the risk of decreasing the expected value is not as severe as the risk of losing the good prizes.
So in general I would prefer the other strategy.
I would like to know what people with experience in statistics and probability think about this, if they see errors in my reasoning, if they can suggest an even better strategy etc.
Thanks!
L
my favourite branch of maths, I need to study it better.
