Its easier to consider the chance that you are wrong: in MH you have a 66% chance of being wrong initially. When MH shows you that information, you know that 66% of the time you were going to be wrong. That 66% is now represented by the one remaining unopened box.

In the same scenario but with DOND you also have a 66% chance of being wrong. Unfortunately you must now guess again, and have a 33% chance THIS guess will be wrong as well – the host doesn’t spot you a freebie. So you guess randomly, 33% of the time you lose and open the 1 mil w/ 3 cases to go.

If u guess correctly u had a 1/3 chance you were initially right, 2/3 chance you were wrong (-1/3 from your latest guess) so 1/3 chance u were right 1/3 chance u were wrong. However only two cases left, so 1/2 right or wrong. AKA 50% chance whether u swap or not.

If we class the 250k as a “win” box, and ALL the rest as “lose” boxes, there are 3 ways a game of DOND can go, assuming we end with 2-box.
1) You choose the “win” box, open 20 other boxes. Left with the “win” and a “lose”.
2) You choose a “lose” box, open 20 other boxes including the “win” and are left with 2 “lose” boxes.
3) You choose a “lose” box, open 20 other boxes but not the “win” and are left with the “win” and the “lose” again.

BUT, and this is a HUUUGEE BUT, we’re not considering ALL possible games here. We’re only considering the ones that end with the “win” box in play.
Therefore you can COMPLETELY IGNORE the probability of having the “win” box left at the end of the game.

Therefore in this subset of games, there are only 2 ways the game can go:
1) You chose the “win” at the start and a “lose” is in the other box.
2) You chose a “lose” at the start and the “win” is in the other box.

These chances are clearly 1/22 and 21/22.

HOWEVER!

The information we have at the end of the game is not just that we have a “win” and a “lose” left.
This is where DOND differs from the MH problem. The “lose” boxes are NOT THE SAME.

We have some extra information.
This means, that out of all the 21 situations where we chose a “lose” and ended up with the “win” in the other box, this is reduced to just 1!

Therefore we’re left with two 1/22 chances, and thus it’s 50:50. Swapping makes no difference.

Where the analogy goes wrong, is where you simplify the problem to being 21 “lose” boxes, and 1 “win” box. If in DOND, all the boxes except the 250k were blank, then there should be a bonus to swapping, as in MH. But since we know which of the 21 “lose” boxes in particular we ended up with, there is no other options.

I think.

Conclusion: It’s wrong to say that at the end of the game your box still has a 1/22 chance of containing £250,000. This fails to take into account the fact that odds change when information becomes known. In this case, information becoming known (opening a box) reduces odds by reducing the choices available.

I’d be happy for anyone to point out flaws with this.

If you knew that every game you’d get to a situation with £250,000 and the rubbish, then clearly you should swap, aka the monty hall, and win 21/22 times. However, you don’t know this, and in fact, if you have picked the £250,000 box to start with, you are 21 times more likely to reach this situation than if you hadn’t picked it, which makes it kind of equal. I’ll try and explain why below.

So imagine a situation where you play 22 games. In 21 of those games, you don’t bring £250,000 to the table. Now the chances of getting to this final ’showdown’ are 1/21. So from the start of the game, the chances of getting to the showdown without bringing £250,000 to the table is 21/22 * 1/21 = 1/22. However, if you do bring £250,000 to the table, you are guaranteed to reach this showdown, so the chances of reaching the showdown bringing £250,000 to the table is 1/22. (For those who have made comments earlier about these not adding to 1, that’s because the specified showdown only happens 1/11 times). Hence, as the chances of reaching this showdown are equal, regardless of whether the £250,000 box has been brought to the table, neither is more likely than the others.

By the way, I’m sure this is adapted from other people’s comments I came to this page earlier today fully believing that you should swap …

In the same way that MH is scalable upwards (see wikipedia entry), DOND can be condensed to what happens when 3 (or fewer) boxes are left in play.
If you had space & time, you could create the same list of outcomes starting with 22 or more boxes if you like (you’d have a lot of iterations where desired winning box is eliminated early…) – the odds don’t change once you’re down to 2 boxes…

I’ve tried to illustrate this below, assuming M to be the value (or car) you want to win, and A & B to be lesser (or ‘goat’) values.

Let’s assume you’re down to the last 3 boxes, about to ‘remove’ one, in the same way as MH would ‘identify’ one of the goats.

You’re in one of 3 situations at this point – you either chose M (the high value box) at the outset, or you chose a goat value (A or B) – this is the left hand column below, with your selected box in brackets. The top 1-4 scenarios illustrate what happens if you brought (M) to the table – in 1 & 2, you remove box A from the game, and are left with (M) and B – you then can keep (M) for a win, or swap to B and lose.
In 3 & 4, you remove B from the game, leaving (M) and A – keeping (M) wins, swapping to A is a loss.

5-8 illustate what happens if you started with (A), and have M still in play with 3 boxes to go…
9 -12 illustrate what happens if you started with (B), and have M still in play with 3 boxes to go.

In the first 4 examples, this relates to what happens in MH when you picked the car door first.
Monty shows you a goat, and if you swap, you lose – this is your 1 in 3 odds-of-winning equivalent.

In the examples 5-12, this is where, in MH, you initially chose a goat – MH can then show you the other goat, and any swap here would give an increased chance of a win (2 out of 3 times).

Here’s where the fault lies in the comparison, and why MH is not compatible with DOND – in scenarios 5-12, you as the player, can remove the M value box from the game (specifically 5,6,9 & 10). These scenarios don’t occur in MH – it would be the same as MH not only showing you, but also removing the car – this doesn’t happen in MH, and it’s why you’re always left with a chance to win.

Even if you removed any iterations from DOND where the M box isn’t present in the final 2 box choice, you’ve still got 8 potential outcomes where you can chose between M and the remaining A or B box (asterisked *). 4 of these outcomes can win (**) – two if you swap, and two if you keep – and 2 chances in 4 is 1/2 or 50/50 – therefore a SWAP GAINS NO ADVANTAGE!

In other words, however you get to the point, at two boxes, where the high value M box is still in play (either at the table, or because it’s left in play) there’s the same chance that a swap will result in a win, as there is a loss…

1. (M) A B -> (M) B = (M) no swap wins**
2. (M) A B -> (M) B = swap B loses*
3. (M) A B -> (M) A = (M) no swap wins**
4. (M) A B -> (M) A = swap A loses*

5. (A) B M -> (A) B = no swap loses
6. (A) B M -> (A) B = swap loses
7. (A) B M -> (A) M = (A) no swap loses*
8. (A) B M -> (A) M = swap M wins**

9. A (B) M -> A (B) = (B) no swap loses
10. A (B) M -> A (B) = swap A loses
11. A (B) M -> (B) M = (B) no swap loses*
12. A (B) M -> (B) M = swap M wins**

Therefore in DOND if you say there are 21 boxes which are failures and 1 sucess it is modelled as a binomial, a fixed probability at the start and either a success or failure outcome. Let us say the success is the 250,000 prize. you have a 1/22 chance of picking it at the start, you then remove 20 more boxes and are told none of them contain the success. so you are left with the box you picked, which has a 21/22 chance of being a failure, and another box. As you almost certainly didn’t pick the 250,000 at the start, statistically it is probably not in your box hence you should swap.

The Monty Hall problem applies only to the situation given above, which is very unlikely to occur, because it involves choosing 20 boxes in a row which do not contain the success, but when it does occur you know to swap. As you have defined all other boxes as failures if you end up opening a box with the 250,000 (the success box) then you have lost as you are left with only failures which you cannot differentiate between. As DOND has varying prizes, not a simple win or lose scenario you can only apply the monty hall if you class 21 of the 22 boxes as failures. If you end up having to choose between two boxes which are known not to be the success then you are equally as likely to get a higher prize if you swap or not. The Binomial model no longer applies.

I hope this makes sense. I have tried to reason using logic and common sense but a mathematical proof is obviously more favourable.

MH –

1 in 3 times – car selected, Goat reveal, 2 options : SWAP = Goat NO SWAP = CAR

2 in 3 times – goat selected, Goat reveal, 2 Options : SWAP = CAR No SWAP = Goat

Therefore SWAP is twice as likely to get the car than no swap.

DOND –

1 in 22 times Top prize selected first

20 in 22 times Top prize not selected first but eliminated before only 2 boxes are left.

1 in 22 times Top prize left till last

So in 2-box finish, the middle 20 chances are ruled out and the 1 in 22 chances now both become 1 in 2. Therefore, swapping makes no difference.

If in DOND 20 non-top prize boxes were taken out by a host who knew where the top prize was, it WOULD be favourable to swap boxes as MH principle would apply.

DOND played like MH –

1 in 22 times Top prize selected first

21 in 22 times Top prize in last box [20 options removed by intelligent selection]

Therefore HIGHLY favourable (21 times) to swap in this hypothetical game (like MH)

To sum up, you’re most likely to NOT choose top prize at first. Swapping at this point makes no difference. Any information which reduces the number of non-top prize options in remaining set of options means this groups is more likely to contain the top prize.

The 2-box DOND offer when Top prize left is unlikely to happen – but equally likely to happen because of original choice being Top prize or Top prize not yet selected or opened. Hence equal probabilities in this situation.

That Monty can reveal a goat tells you nothing. You already know there’s at least one goat there somewhere. He opens a door and shows you one. No surprise. You learn nothing. Your odds of having picked the right door stay at 1/3.

In DONR every time you open a box randomly and it isn’t the top prize you are relieved. You are surprised. Why? Because you have gained information and your odds adjust accordingly.

Would a TV show format which is sold all over the world give unfair odds, even in unlikely situations? I find it hard to believe that a game show based on pure luck would do that.. the television companies would surely lose more money that way.. even if none of the contestants knew of/applied the MH idea, they’d still lose more money that way than by giving it an even chance.

However, I’m still unconvinced mathematically one way or the other.

I have to dissagree with you Chris. (I think!)

The comparison only applies when you get to the end and you have a swap choice between 2 boxes. In this case you do not have to take a chance on elimination as it has already happened.

Guy