[nixbits]

does the monty hall reasoning apply to 2 box deal or no deal decision to swap?

A 'monty hall' situation with 22 boxes is the following: only one box has a prize and 21 boxes are empty. After the player chooses a box the host opens 20 empty boxes (he knows where they are so is not opening boxes at random) leaving 2 boxes in the game. The player is offered the swap. Under these conditions they should always take the swap as the probability the prize was in the original box is 1/22 but the probability it is in the other remaining box is 21/22. But the deal or no deal 2 box situation is different and does have a probability of 1/2 for both boxes containing the big prize. This can be seen from the following consideration:
in the monty hall case the configuration : 'the win box is originally chosen alongside the 21 loose boxes' is counted once while if the original box chosen were a loose box then the prize can be in any of the 21 other boxes so the configuration in which the player holds the loose box is counted 21 times.This is what leads to the assymetry in the probability of win/loose for the swap/ no swop cases. In the deal or no deal case if the player originally picked the £250,000, then the 1p value can be in any of the 21 remaining boxes so this is the number of configurations which leads to the 2 box game with £250,000 and 1p with the player holding the big prize. If the player starts out picking the 1p the 250,000 can be in any of the 21 remaining boxes so the number of configurations leading to the same 2 box game in which the player holds 1p is also 21. So it is equally likely that the player is holding the 1p or the 250,000 prize and swapping boxes will not change the probability of winning on the average

[Hopple]

I'm struggling to read that, but if I get the gist of it, you're talking about what to do if you've opened all the boxes except for £250k and 1p?

I think the main difference between DOND and Monty Hall is that the contestant picks the boxes not knowing what'll happen. If we strip down DOND to three boxes, £250k, 5p (for example, I think that's the next one up) and 1p, the contestant is still choosing which of the three has £250k. With the same three boxes in Monty Hall, the host would open a box that he knew contained the 1p or 5p.

[Alexx53]

Deal or No Deal is 50/50 at the end. Say there's 1p and £250k left at the end. At the start, there is a 1/22 chance you have it in your box, but as you move through the rounds that probability changes (increases), 1/21, 1/20, 1/19...until the final round when it is 1/2.

[Alexx53]

(Original post by TenOfThem)
So, there is a 21/22 chance that the box is "out there" and only a 1/22 chance the contestant has it

Implying that swap is the best move

OK, say the two boxes are £250k and 1p.

There was a 1/22 chance of picking the 1p box and a 21/22 chance you didn't pick the 1p box, so should you always stick to avoid the 1p? Doesn't work like that.

At the start there would be a greater chance that you don't have it than that you have it.
But as you remove boxes that chance gets smaller - 21/22, 20/21, 19/20...1/2, whilst the chance of you having that box gets bigger - 1/22, 1/21, 1/20...1/2

Eg. At the start, there are 22 boxes, and each box would have a probability of 1/22 of containing the £250k. As the boxes get removed, the probability gets larger of each box containing the £250k...1/22, 1/21, 1/20...1/2, because the boxes are removed at random, instead of the host removing the boxes he knows don't contain the prize.

If it was played like the Monty Hall gameshow (£250k is the car, 1p being the goats - 21 of them), then you would always swap because that game is skewed by the host knowing where the car is.
You have a 1/22 chance of picking the £250k. The host then removes every 1p leaving one box, which is either £250k or 1p. Always swap, as the chance is 21/22 that the other box is the £250k.
However, DOND doesn't get skewed by the host.

[Hopple]

(Original post by TenOfThem)
So, there is a 21/22 chance that the box is "out there" and only a 1/22 chance the contestant has it

Implying that swap is the best move

Does it work like that though?

If we strip it down to three cases, one is the 'winning' £250k, the others are 1p and X, then we get the following possibilities:

Contestant chooses 250 initially (1/22), opens X (whatever the chances of opening all but 1 and 250) meaning swapping loses.
Contestant chooses 1p initially (1/22), opens X (same as above) meaning swapping wins.
Contestant chooses X initially (20/22) which means the OP's scenario cannot happen so we can ignore it.

So if you were to find yourself in the OP's scenario, you have the same chance of winning by swapping as losing. This would apply to all pairs of boxes.

[nixbits]

It is correct that the probability of winning the big prize £250,000, if it is present in an N box round, is equal to 1/N. So it starts out being 1/22 then assuming it is still in the next round (17 boxes) it will be 1/17 for that round and if it is still in the final round it will be 1/2. (Remember that this is a conditional probability: the condition being that the big prize is still in play. If we have a specific 2 box final round ( Eg 250,000 and 1p still in play) then the probability that we have a final round with these two = 2/(22x21) (number of equally probable ways of choosing 2 from 22 = n =22X21/2 so probability of a specific pair is 1/n ). so the probability of the player winning the 250,000 = prob of this 2 round game occuring x prob of winning this round = {2/(22 x21)} 1/2 = 1/(22 x21). The probability of the player having the 1p in this pair is also 1/(22x21). The probability of the 250,000 remaining in the last 2 box round is 21 times this as it can be in any of the 21 equally probable pairs i.e Prob {250,000 , any value} = 2/22 while probability of winning 250,000 in any game= 1/2 x {2/22} = 1/22. The probability of the 2 box {250,000, 1p}occuring = 2/(22x21)

So in 2200 games all played to the end with no deals 200 of the games will end in a 250,000 vs any value 2 box game and 100 of these games will end with the player winning the big one! The number of these 200 games which are 250,000 vs 1p will be 200/21= 10 games of which the player will win 5.

In thinking about these problems it helps to visualise the way the prize values can be put into the boxes: There are 22 distinct prizes. They can be put in the boxes as follows; first box has a choice of 22, second box 21, third 20 etc. So there are 22x21x20x...2x1 = 22! equally probable different permutations of the prizes in the boxes. Suppose now that the player picks a box at random and plays the game to the final 2 box stage where the box values are a or b. The number of the original permutations in which the player had a and the other boxes had any permutation of the remaining 21 values = 21! so the probability that the player picks a = (number of permutations with a in choosen box)/ (total number of permutations) = 21!/22! = 1/22 initially. Now the value b is in the remaining box; the other 20 values eliminated.The number of ways of arranging these = 20! This is the number of the original permutations which lead to the two box game in which the player holds a and the other box b. But if the player picked b at the start the same argument shows that 20! of the original permutations lead to this 2 box game {a,b} so in total the number of the original permutations which lead to the 2 box game where we dont know who holds the winning a is the sum of these = 2 x 20!. So the probability of ending up with the specific two box game {a,b} =
number of permutations leading to this game / total number of permutations=
2 x 20!/22! = 2/(22 x 21). The probability of the player holding the winning value in this 2 box game a = number of permutations in which player holds a/ number of permutations in which player holds a or b = 20!/2 x 20! = 1/2

[nixbits]

In response to Ten of tens point:

that the probability of having the 250,000 is 1/22 so the probability of it "being out there" is 21/22 ---this is true. But it doesnt follow that you must swap because

in the two box game the probability of the unchoosen box having a specific value is also 1/22. So the two boxes are equally probable in any specific 2 box game.

The monty hall case is like a game in which there is 250,000 and then 21 boxes with 1p. But even then if the host opened 20 of the unchoosen boxes randomly there would still be an equal probability of win or loose. It is only when the host opens the boexs only if they contain 1p that the swap advantage occurs.

[Hopple]

Yeah, I think the main point is that in Monty Hall the host treats the winning and losing boxes differently, whereas in DOND they're all treated the same so you can use symmetry arguments.

[Dog4444]

Guys, it's simpler.

In Monty Hall, your initial chances are 33% to win and 66% to lose. So, you must swap.

But, in Deal or No Deal you're as likely to pick 250grand as picking 1p.

[Hopple]

(Original post by Dog4444)
Guys, it's simpler.

In Monty Hall, your initial chances are 33% to win and 66% to lose. So, you must swap.

But, in Deal or No Deal you're as likely to pick 250grand as picking 1p.

The fact that the host makes an informed action does play a part though. If in Monty Hall it were just you picking a box, then opening one (which happened to be a bad/losing one) and then choosing to switch, then swapping wouldn't do any good there either.

[poohat]

(Original post by Hopple)
The fact that the host makes an informed action does play a part though. If in Monty Hall it were just you picking a box, then opening one (which happened to be a bad/losing one) and then choosing to switch, then swapping wouldn't do any good there either.

Yes exactly, its crucial that the host knows which box contains the money and chooses to open an empty box instead. If the host randomly opened a box and it was empty then it wouldnt matter whether you switched. This drops out very clearly if you do the maths.

Suppose we have 3 boxes, one contains money while the other two are empty. Let M be the event that you box you initially picked contains the money, and let E be the event that the host opens an empty box. We are interested in P(M|E) (ie the probability that the box you initallly picked contains the money, given the host opened an empty door). You should switch if P(M|E) < 0.5.

By Bayes theorem we have:

P(M|E) = P(E|M)P(M) / (P(E|M)P(M) + P(E|M')P(M'))

Now, P(E|M) = 1 regardless of what the host knows (if you picked the money, then every other box is empty so only an empty box can be opened) and P(M) = 1/3 (you have a 1 in 3 chance of having picked the money). But the P(E|M') term (which denotes the probability of the host opening an empty box given that you never picked the money) depends on whether the host knows where the prize is, or is guessing. If we let q=P(E|M'), then:

P(M|E) = (1/3) / (1/3 + 2q/3)

Now, if we assume that the host knows where the prize is and always opens an empty door, then q = 1, so P(M|E) = 1/3 and you should switch. But if the host doesnt know and opens a box randomly (as in Deal or no Deal situations) then q=0.5, so P(M|E) = 1/2, and switching makes no difference

[nixbits]

A quick visual summary of the situation:



We are agreed that swapping the boxes in deal or no deal makes no difference to the odds of winning.

Thanks for your comments. I now understand the problem and its solution.

[Venomilys]

in certain situations, yes. You could have opened 20 of the boxes and in the end it is either 1p or 250k. The 20 boxes could represent 1 box.

[nixbits]

In all situations in a deal or no deal game the probability of the player holding the larger value in any specific 2 box game {a,b} (where a and b are any of the prize values) is the same as the probability for the player holding the smaller of the two values in this two box game. So swapping never improves the odds of winning.

[nixbits]

No, it is never a monty hall problem provided the boxes are opened without knowledge of their contents.

[div curl F = 0]

No it isn't. I looked at this before last christmas with another teacher.

The Monty Hall setup has the host with knowledge of the layout of the prizes and directly influences the game whereas Deal or No Deal has the host with no knowledge of the layout of the prizes and has no input on the player's game.

This subtle difference makes all the difference.

[tommm]

If you got to the end and were left with 1p and £250,000 and offered the swap, it wouldn't make a difference.

However, if Noel Edmonds knew which box contained the £250,000 and wouldn't let you open it during the game, then it would be similar to Monty Hall and you should swap.

[nixbits]

This raises an interesting point, can we trust that the game truely is random? Yes if it strictly follows the rules since the rules of the deal no deal game set it up to be random. But We can only know that past games have been truely random by analysing the statistics of the outcomes and looking for correlation.

Suppose we were playing a monty hall game with 3 boxes but we didnt know if the host were opening boxes at random or not. The host opens a box to reveal a booby prize. What should we do if offered the swap?

From a statistical point of view always take the swap: for if the game were totally random there will be no statistical disadvantage to swapping but if there were any non randomness there is a statistical advantage to swapping!

Of course 1/2 the number of people who take the swap will loose if the game is random while 1/3will loose if it is non random.

[Dog4444]

Btw, we can expect 1 out of 22 participants to get 250k box, but in reality this number is significantly lower, what might imply that the boxes or players aren't completely taken by random. And the sample size is quite big, thousand games or smth I reckon. Maths got it busted!

[nixbits]

The reason there are many less big prize winners than statistically expected is because the contestants deal at some point in their game! (not because the maths got it wrong!) On average in the uk game the banker would be expected to pay out £26,712 per game if no deals were ever accepted. Recently it was announced that in 1967 games just over £30 million had been paid out which means the average payout per game is about £15000, significantly less than the £26k. This shows the banker is doing his job which is to make offers which minimise the payout but maximise the probability of the offer being accepted.