Latest video and the Monty Hall Problem

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    argyl53 WANTED $419

    In his latest video, while playing Deal or No Deal, the Bandit is offered the chance to swap his box with the one remaining box and although he doesn’t swap, he comments that you should. The rationale behind this is that if you are left with a high prize and a low prize, at the start of the game you had only a 1/22 chance of picking the high prize and therefore after eliminating all the others, it’s more likely to be in the one remaining box you didn’t pick.

    It may surprise you, particularly if the title of this post was familiar, that this is not correct and – assuming the game is fair and random – you actually have a 50/50 chance with either box of getting the higher prize.

    The reason people think otherwise is because they confuse this situation with a seemingly very similar situation known as the Monty Hall Problem, which describes this scenario:

    You are on a game show and are asked to pick 1 of 3 doors. Behind one of the doors is the prize. After you pick your door, the host opens one of the two remaining doors to reveal it does not contain the prize and then asks you if you’d like to stick with your door, or switch to the other remaining unopened door.

    Now, in the above scenario, you should swap and by doing so will double your chance of winning from 1/3 to 2/3.

    But there is a crucial difference between the Monty Hall Problem and a Deal or No Deal type game – in Monty Hall, the host’s choice of which door he opens is not random. He’s always going to open a losing door and if as would happen 2/3 times, you picked a losing door, that means there is only one door he can open. In other words, every time you play Monty Hall, there is a 2/3 chance the host opened the door he did because it was the only door which both wasn’t the one you picked and wasn’t the one with a prize.

    In Deal or No Deal, you are simply opening boxes at random and that top prize could be revealed at any time, so by the time you’ve narrowed it down to only two boxes, there is a 1/2 chance the reason it hasn’t been revealed yet is because it’s in your box and a 1/2 chance it’s because it’s in the other box.

    We can easily demonstrate these differences via probability tables.

    We’ll call the box/door you picked number 1.

    P = prize, X = lose (or lowest prize in case of Deal or No Deal)

    DoND (if you’re offered the swap, you can open either box):


    Box 1 – Box 2 – Result Stick – Result Swap

    P – X – Win – Lose

    X – P – Lose – Win

    As you can see, you will win/lose in 50% of those scenarios.

    Monty Hall (1 is the door you picked, host will automatically open a losing door from the two you didn’t pick)

    1 – 2 -3 – Result Stick – Result Switch

    J – X – X – Win – Lose


    X – J – X – Lose – Win

    X – X – J – Lose – Win

    Best way to understand this key difference is that in the case of Monty Hall, you’re effectively being offered one door you’ve chosen at random, or all the other doors together, whereas in DoND, you’re just being offered either of two doors.


    The Bandit WANTED $5,362
    El Bandito

    Yeah it was the monty hall one i was thinking of, providing the game is fair and random…. a much debated point, then you are correct. Do you believe the game is fair and random? It is an interesting one, i liked the side bets even though i didn’t quite understand them most of the way haha

    argyl53 WANTED $419

    The Bandit wrote:

    Do you believe the game is fair and random?

    Haha now that’s the question! I’d like to think so, but as we know from for example the Extra Chili gamble wheel, games can visually mislead you. Back when I played FOBTs, I used to get the feature a lot on Deal or No Deal virtual scratchcard game and the number of times I’d deal at around £100 only for it to show me I had £500 in my box was way more than I’d expect off fair and random probability. And of course I never had the £500 if I went all the way.

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