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The statistics of Deal or No Deal?

  1. Dec 13, 2015 #1
    Hi, I have a doubt on the statistics of this game, maybe more philosophical than mathematical.
    I assume most people are familiar with the game.
    Suppose you categorised the prizes into 'good' and 'bad'. The contestant plays a very lucky game, and halfway through the game is left with no bad prizes. To simplify, say that there are 20 boxes, 10 good and 10 bad, and the contestant opened the 10 bad ones. [BTW, I'm not sure how to calculate the probability of such a scenario. I wouldn't know how to account for the fact that the contestant's box is not included in the ones that can be opened during the game. But this is not my doubt: we know that the probability is quite low]. The point I don't get is, what happens if at this stage we forget about good and bad, and we use some other category to tell the prizes apart, say those that, expressed as numbers, are divisible by 6 as opposed to those that aren't? Now the distribution of what the contestant is left with and what is already revealed changes, and the probability of obtaining such a scenario has a different value.
    How is it possible that the same event has two distinct probabilities?
  2. jcsd
  3. Dec 13, 2015 #2


    Staff: Mentor

    If a prize distinguishable on multiple characteristics then you don't have a simple random variable with a single probability distribution, you have multiple random variables with some joint distribution. When you draw a sample from that joint distribution you can calculate joint probabilities, marginal probabilities, or conditional probabilities.

    You will often find that different marginal probabilities (what you described above) are different for the same sample from a joint distribution. Essentially this is the same as the fact that a single rectangle may have a different height and width.
  4. Dec 14, 2015 #3
    Thank you DaleSpam, I think I understand what you mean.
    Probability is probably:biggrin: my favourite branch of maths, I need to study it better.
  5. Jan 8, 2016 #4
    Coming back to the other part I was unsure about.

    In a national TV version of this game, there are 20 boxes at the beginning, with about 10 'non-prizes', 5 modest prizes and 5 rather good ones. The contestant has one of them and is asked to choose 3 boxes at a time, which are immediately opened. After each of these 'sessions', he's either offered a certain sum of money to stop the game, or the opportunity to swap his box. Toward the end the number of boxes to open at each session reduces to 2 or 1.
    We know that this is a 'non-Monty-Hall' game, so swapping the box should not change anything.
    And we know how to calculate the expected value (at each time, the sum of the values of all the unopened boxes divided by how many they are).
    The money offer is usually much lower than the expected value, so if one used that as a criterion to continue or not, he should never accept the offer, and that is equivalent to opening the original box, regardless of how the game went.

    So yesterday I was thinking, is there a better strategy?
    For instance, suppose at some point in the game there are 10 unopened boxes and you're offered a sum that is greater than 8 of them (and this smaller than 2 of them). You know that if you don't accept the money you'll have to open 3 boxes.
    What are the chances that you lose the two better prizes by opening 3 boxes?
    I tried to calculate that as follows.
    P(2 good and 1 bad) = (2/10) * (1/9) * (8/8) * (3! /(2!*1!)) = 1/15 ≅ 6.7%

    First question: is this a correct formula? I am not sure because the contestant's box can't be opened, so should I consider only 9 boxes? But then how would I know which numbers to use?

    Assuming the above is correct, say you refuse the offer and opened the next 3 boxes: you now have 7 boxes.
    The offer is again smaller than 2 of them, and you will open 3 boxes if you refuse it.
    P(2 good and 1 bad) = (2/7) * (1/6) * (5/5) * 3 = 1/7 ≅ 14%
    You consider the risk low enough, and open the next 3 boxes.

    You're left with 4 boxes, and the offer is smaller than 1 of them. You need to open 2 boxes if you refuse.
    P(1 good and 1 bad) = (1/4) * (3/3) * 2 = 1/2 = 50%
    You consider this risk too high and accept the money offer.
    Suppose instead that the offer was smaller than 2 of them.
    P(2 good) = (2/4) * (1/3) = 1/6 ≅ 16.7%
    In this case you think you still have a relatively low chance of losing both the good prizes, and refuse the money offer.

    So it looks like this strategy gives you an idea of the risk you run at each stage.

    Second question: do you think this strategy is effective in maximising the money you're likely to pocket? And if so, should the actual value of the offer be used somehow in the computation, together with the probability, to decide whether to continue or not?

  6. Jan 8, 2016 #5


    Staff: Mentor

    It is not a Month Hall game, but the additional information that you gain by seeing the contents of the opened boxes does change the odds.

    If you choose to swap the box, does that end the game immediately with opening the swapped box or does play continue. Especially, does the originally chosen box go back into the pile or is it opened to see "what you could have won"
  7. Jan 9, 2016 #6
    Sure, I realise that. The chances of the contestant having a 'good' box increase if the percentage of good boxes remaining in the game increases as the total number of unopened boxes N decreases.
    No, if you swap the boxes the game just continues as before, and the contestant's original box goes back into the pile and doesn't have to be opened immediately. That's why I thought this didn't change anything: in principle all the N unopened boxes have equal chances of containing any of the remaining prizes, so swapping any two of them is irrelevant - or am I wrong?
  8. Jan 9, 2016 #7


    Staff: Mentor

    The selected box has 0 chance of being opened, so I think that the information gained does not change the odds for the selected box.

    I am not very sure about this.
  9. Jan 10, 2016 #8
    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 < vji < 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:
    [itex]E = \frac{\sum_{i=1}^N v_{i}} {N} [/itex]

    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:
    [itex]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)!} [/itex]

    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.
    [itex]P(10,2,3) = \frac{3! \cdot (10-2)!} {10! \cdot (3-2)!} = \frac{1} {15} [/itex]
    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.
    [itex]P(7,1,3) = \frac{3! \cdot (7-1)!} {7! \cdot (3-1)!} = \frac{3} {7} [/itex]
    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.
    [itex]P(7,2,3) = \frac{3! \cdot (7-2)!} {7! \cdot (3-2)!} = \frac{1} {7} [/itex]
    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.
    [itex]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!} [/itex]

    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.
    [itex]P(\text {at least 1 prize>E lost}) = 1 - \frac{(10-2)! \cdot (10-3)!} {(10-2-3)! \cdot 10!} = \frac{8} {15} [/itex]
    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.

  10. Jan 14, 2016 #9
    I ran a simulation of 1000 games using the above strategy, i.e. accepting the offer if P(N,m,n)>m/N, continuing to play if not. I set the offer in each case at 1/2 of the expected value, which is more or less what happens on the TV show. I started with 20 boxes and then open 10, offer, 3, offer, 3, offer, 2. No swap offered.

    Result: in 61.5% cases the accepted offer was higher than the contestant's box, and the average of log(accepted_offer / contestant's box) was 1.15, i.e. the accepted offer was on average more than 10 times higher than the box assigned by chance.
    Not too bad, I would say.

    The strategy, however, is asymmetric: it gives more to those who have small prizes and it takes away from those who have big prizes.
    I suppose that's normal, because there are only a couple or really big prizes. So when the contestant has a big prize, it's quite likely to end up with a set of unopened boxes with only 1 big prize and for the rest small prizes. The expected value (and hence the offer) will be much lower than the big prize, and m=1, so P(N,1,n) will be large and bigger than m/N, causing the contestant to accept the offer.
    I'll keep thinking it over, maybe there is an even better strategy.
  11. Jan 14, 2016 #10


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    Isn't it more like a decision tree?
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