Goat or No Goat: applying the Monty Hall problem to Deal or No Deal

Can you apply the logic from the Monty Hall problem to Deal or No Deal? I’m no statistician, but after a conversation this week, I think you can…

For those of you unfamiliar with the Monty Hall problem, imagine a game show where you have three doors. Behind one door lies a shiny new sports car, while the other two doors hide goats (or, if you’re concerned about animal welfare, they can be empty). You’re asked to pick one door. The game show presenter will then open one of the doors that does not hide the car and then ask you if you’d like to change you choice.

Many people think that this is a 50/50 choice. In fact, if you change your choice, the odds are 2:1 in your favour of winning the car. Yes, it’s counterintuitive, but if you take the time to play out all the combinations you’ll see that this is correct.

Some people assume that you cannot apply the same principle of swapping your selection to Deal or No Deal. Firstly, the Monty Hall problem relies on a single item of value and multiple worthless items, and secondly, in the Monty Hall problem, the presenter always knows where the prize is located.

Although the Monty Hall problem features only one item of value and multiple low-value items, the game can be extrapolated. Provided there are always more low-value than high-value items to choose from, and the low-value items are always removed, you are more likely to win by swapping at the end of the game. Therefore, the Monty Hall problem can be used to consider games featuring multiple high-value items.


How can this be applied to Deal or No Deal? I suspect that most contestants on the game show don’t come on just hoping to win the top prize (£250,000 in the UK). Most people will be happy to win, or at least want to win, a range of values, from a minimum amount all the way up to the £250,000.

So the game will start and prize money will be lost as different boxes are selected by the contestant. Now, in the Monty Hall problem, the presenter knows the location of the high value prize(s) so these aren’t removed. This isn’t the case on Deal or No Deal – some of the bigger prizes will be removed by chance. However, some prizes in this range will still remain.

I wrote a short computer program to measure the likelihood of winning your minimum target or greater if you chose to swap when there was only one prize left in this range. As you can see, the greater your minimum amount, the more likely you are to win that value or greater if you swap; this mirrors what we see in the Monty Hall problem.

Minimum desired winnings Successes if swap Successes if stay
£1 13585 86415
£5 17947 82053
£10 22757 77243
£50 27148 72852
£100 31962 68038
£250 36342 63658
£500 40848 59152
£750 45572 54428
£1,000 50151 49849
£3,000 54796 45204
£5,000 59293 40707
£10,000 63494 36506
£15,000 67958 32042
£20,000 72867 27133
£35,000 77271 22729
£50,000 81821 18179
£75,000 86068 13932
£100,000 91092 8908

Obviously, you don’t have this option in the real game but you could apply its logic to your decision making when offered a deal by the banker: if there are fewer prizes in your target range than outside it, and the banker offers you a deal greater than your minimum amount, you should take it, as this is tantamount to swapping. From a purely rational point of view, you should take it anyway (and now you have a statistical reason why) although humans clearly aren’t rational and contestants clearly enter the show either with grandly over ambitious goals, or get overwhelmed with excitement (or possibly greed?) as the game progresses.

I haven’t quite worked out what to do if you get given an offer below you target prize, though. Statistically, the winning box(es) are less likely to be picked so by playing a little longer the deal could shift in your favour. This could be especially beneficial if there are some very high value boxes left, and anecdotal evidence suggest that the banker’s offers get more generous as the game goes on.

And, for the sake of completeness, if you came to the final two boxes, with one containing a penny and the other £250,000, you should swap (but don’t re-swap). However, I’d say you’d be very foolish not to take the banker’s offer at that point!


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12 Responses to Goat or No Goat: applying the Monty Hall problem to Deal or No Deal

  1. Pingback: Responding to the Monty Hall Problem | Eventually Almost Everywhere

  2. PalmerEldritch says:

    If you come down to 2 boxes, 1 with a penny and 1 with $250K, it makes absolutely no difference to your chances of winning the $250K whether you swap or stay.

    • Thanks for taking the time to read my post. I’d be interested to know why you think this is the case. Which bit of my analysis above do you disagree with?

      • PalmerEldritch says:

        You explained it yourself at the the beginning of the article “.. in the Monty Hall problem, the presenter always knows where the prize is located.”. In DoND this isn’t the case. DoND is like the MHP where Monty has amnesia: you pick a door, Monty randomly chooses 1 of the other 2 – it’s a goat fortunately – the 2 unopened doors are now equally likely to contain the car.. Why? In the 2 out 3 times that amnesiac Monty doesn’t open the car door , 50% of the time you’ve already picked it.

        It’s the same in DoND, say there’s 24 cases. You pick 1 , it’s 1/24 chance of it being $250K. There;s a 23/24 chance it’s in 1 of the other 23 cases. If you open 22 of those 23 cases at random it’s a 1/23 chance the 1 not opened is the $250K. 23/24 X 1/23 = 1/24 = same chance as you picking it to begin with.

        In DoND it makes no difference if you pick 2 cases to begin with and then open the other 22, open 22 to begin with, or pick 1 then open 22 cases., you’re still left with 2 cases both equally likely to contain whatever 2 prizes happen to be left.

  3. I thought the same thing which is why I modelled it in the second part of my article. This clearly shows better odds for swapping to win higher amounts and suggests it isn’t a two box game. I can’t explain the statistics but it doesn’t mean they’re not true. Would be interested to hear an alternative explanation for the stats, though. After all, I admit I’m no statistician.

  4. PalmerEldritch says:

    I didn’t understand what your model was trying to show. It’s not a 2 box game for sure since you can take the banker offer at any time. However, at the start of the game all boxes are equally likely including yours. As prizes get revealed all remaining boxes (including yours) stay equally likely – the odds of your box containing any of the remaining prizes increases as each box is opened, in line with all the other unopened boxes. If you’re down to 10 boxes then your box has 1/10 chance of containing any of the remaining prizes. If you’re down to 2 boxes it’s 50/50 irrespective of what prizes are left.

  5. TheBanker says:

    No, the odds are still 50/50 with a swap or a stick on DOND. The odds only increase where there are 3 boxes and the host knows where the car is and has to reveal a goat behind one of the 2 remaining doors (this increases your chances of picking the car from 1/3 to 2/3 when you swap). This scenario is not reflected in DOND where only 2 boxes are involved and the independent adjudicator is the only person knowing where the money is. It’s 50/50 whether you swap or not, odds remain the same.

    • Sorry for the slow response (I’ve been away) but thanks for your comment. As you can tell, you’re not the first person to say the odds are 50/50 for DOND but can you prove it? Or to put it another way: if the odds are 50/50, why does my table above show the opposite; namely that when you swap you are more likely to get your desired winnings or higher (provided it is still there for the taking)?

  6. PalmerEldritch says:

    Taking the banker’s offer is not “tantamount to swapping”. Why do you think that?
    As for a proof: say there’s 25 boxes, then:
    The probability that you pick the $1m box is 1/25. The prob. it’s in the remaining boxes is 24/25.
    The prob. it’s still in the remaining boxes after the banker opens 1 box at random is 24/25 x 23/24.
    The prob. it’s in the remaining boxes after the banker opens another box. is 24/25 x 23/24 x 22/23
    The prob. it’s in the last box after the banker has opened 23 of the 24 boxes is 24/25 x 23/24 x 22/23 x ……….3/4 x 2/3 x 1/2 = 1/25, i.e.the same as the prob. you picked it in the first place.
    So if you get down to the situation where the $1m is still out there (and 23 times out of 25 it won’t be) then it’s 50/50 between the last 2 boxes.

    “I wrote a short computer program”. it must have a bug in it.

  7. Phil says:

    I also wrote a program for this problem, and came out with a 50/50 split.
    Would you consider posting the code ?

    • Yeah. I can do that. Check back in the new year. It’s on my old laptop so I have to dig it out first.

      One possible reason for the difference might be when our respective bits of code choose to ‘swap’. If I remember correctly, I had my code swap when there was only one value greater than or equal to the desired winnings, even if that was half-way through the number of boxes; the argument being it was analogous to taking the banker’s offer (although I admit the banker’s offer does not come round every turn).

      Come to think of it, that almost certainly explains the difference: the higher the amount of money you want to win, the less likely you are to have a prize of that value in your box, and the more likely you are to up with a winning prize when you swap: I think the odds are stacked in your favour because statistically you’ll usually need to get rid of more lower-value boxes before most of the higher-value boxes are gone. Hmm…

    • Added the code to pastebin: http://pastebin.com/tGs2cBU2

      Sorry for the delay

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