Deal Or No Deal vs Monty Hall
I had an interesting phone call from a friend who wished to settle a Christmas family argument. The argument was over Deal Or No Deal. Imagine the case where there are only two boxes left closed, one belonging to the contestant and the other to someone else, and the million is in one of these two boxes. Now Noel offers the contestant to swap his box for the other, should he?
My friend was instantly reminded of the Monty Hall problem and said he should swap. His family disagreed, but he saw this of ignorance of the lessons of MH. So who was right?
My first inkling was that the difference between DOND and MH is that in MH the person who reveals the goat knows where it is. Therefore, they are choosing to reveal a certain piece of information, and this choice is important. But in this case the choice is random so does the analysis still stand? I think the easiest way to resolve this is to consider the conditional probability of the car/million being in the chosen box given that you are offered a swap.
So in the case of DOND the swap doesn’t matter. And just to check, here is the numerical simulation in python:
edit: properly format code
My friend was instantly reminded of the Monty Hall problem and said he should swap. His family disagreed, but he saw this of ignorance of the lessons of MH. So who was right?
My first inkling was that the difference between DOND and MH is that in MH the person who reveals the goat knows where it is. Therefore, they are choosing to reveal a certain piece of information, and this choice is important. But in this case the choice is random so does the analysis still stand? I think the easiest way to resolve this is to consider the conditional probability of the car/million being in the chosen box given that you are offered a swap.
Let M be the event a million is in the original box Let S be the event we are offered a swap For MH, we are always offered a swap: P(M|S) = P(M and S) / P(S) = (1/3) / 1 = 1/3 For DOND with N boxes, contestant always keeps choosing boxes until the million is revealed or only two boxes left. P(M|S) = P(M and S) / P(S) Clearly, P(M and S) = 1/N Since if the million is chosen, a swap will always be offered And P(S) = 1/N + (N-1/N)*(1/N-1) = 2/N i.e. the probability of a swap is the probability you chose the million plus the probability you didn't choose it and didn't open it in the random choices. So, P(M|S) = (1/N) / (2/N) = 1/2
So in the case of DOND the swap doesn’t matter. And just to check, here is the numerical simulation in python:
import random size = 3 iters = 10000 choices = (['Goat'] * (size - 1)) + ['Car'] print "doors are", choices os = 0 ss = 0 for n in range(0,iters): random.shuffle(choices) if 'Car' in choices[0:2]: os += 1 if choices[1] == 'Car': ss += 1 print "offered swap", float(os)/iters print "should swap", float(ss)/iters print "should swap | offered swap", float(ss)/os
edit: properly format code
December 27th, 2006 at 2:34 pm DAMM YOU AND YOUR LOGIC!!!
But I do concede that you have a 50/50 chance in DOND when you get to the final two boxes and the large cash prize is still up for grabs.
Best to just make a deal with the Banker then…
January 4th, 2007 at 4:31 pm This is pretty stupid, you are answering the wrong question. There is no point working it out in terms of whether the swap is offered at that is random. Here is the interesting bit:
The problem for DOND is:
- You are down to the final 2 boxes
- the million is in one of them
- you are offered the swap.
Do you stick with the box you originally chose or do you take the swap? (Quite a dfficult question given the monty hall problem shows you should always swap)
The answer is, it doesn’t matter for deal or no deal because the other boxes are eliminated at RANDOM by the contestant. In the monty hall problem the wrong box(s) are eliminated by the INTELIGANT CHOICE of the quiz master, so you increase your chances if you swap.
Cheers
Tom Howard
January 4th, 2007 at 5:24 pm I worked this through with Tom from a different angle. Which I think is valid and agrees with the conclusion that the swap has no MH like advantage:
Given that you are in the situation of one box left plus the one you chose originally and are being asked if you want to swap. The top prize is still in play. Oh and the usual 22 boxes at the start.
There are only two possible ways this situation could arrive:
Path 1. You chose the right box at the start
Path 2. You chose a ‘wrong’ box at the start and have now randomly eliminated 20 other ‘wrong’ boxes leaving the top prize box as the swap you are being offered.
Which gives the following:
Path 1.
Probability of choosing the top prize box at the start = 1/22
Path 2.
Probability of choosing any other box at start = 21/22
Probability of randomly eliminating all 20 other boxes and leaving the top prize box = (20/21)*(19/20)*…*(1/2) = 1/21
Total probability of swap box being top prize = (21/22)*(1/21) = 1/22
So both boxes are equally likely to contain the top prize.
As James said the difference with MH is that in DOND the boxes are randomly eliminated. If as in MH Noel knew where the top prize was and intelligently selected to eliminate 20 of the other boxes to arrive at the situation then MH would apply.
January 14th, 2007 at 3:14 pm Tom, I’m not sure I follow your criticism. The point is I conditioned on whether the swap was offered. Conditional probability is pretty standard tool.
Your `proof’ is really just a claim as I see it.
January 18th, 2007 at 11:46 am When you say that the difference in this problem is that MH KNOWS where the prize is and in DOND it is random picks doesnt actually matter. The example STILL holds right but ONLY if you take ALL the other boxes as empty and ONLY want to win the 250,000. When you get to the 2 final boxes (an ‘empty one’ and 250,000) what has happened is that you have played out the role of MH BY PURE CHANCE. Hence you should swap. In any other given situation with 2 boxes in DOND swapping will make no difference as there is always a prize on offer. I am sure this is right!
January 18th, 2007 at 1:59 pm Dannie,
Your conclusion is fundamentally flawed. You say that the probability it is in your box is 1/22 and the probability it is in the sole remaining box is 1/22. What happens to the other 20/22 chances?……….
Part of the problem with this discussion is that the terms of reference are not clearly defined. The question, I believe, should be: “in DOND when there are two boxes left one of which contains nothing (or 1p) and the other contains £250,000, what is the probability that the box you chose contains the £250,000?”
The answer to that has to be 1/22. You chose it at random from 22 boxes. Nothing has intervened. The probability remains the same. Therefore it must be that the probability that it is in the other box is 21/22.
The key to the question is in the fact that you choose the box at the beginning of the game when there are 22 boxes.
Ollie is right that it is the same as MH albeit by pure chance and whether you have the knowledge that you’re taking out ‘empty’ ones or this happens by pure chance that you take out ‘empty’ ones makes no difference.
January 30th, 2007 at 5:00 pm I think you are all being misdirected…
The salient point here is that whether swapping the box or not, Noels house party will never again disgrace our screens.
Surely that makes everyone winners, whatever particular box they are perving over in the meantime.
Hirsty
April 3rd, 2007 at 6:11 pm Think of it that YOU eliminate 20 boxes, you gain no other information along the way. You actually pick one, eliminate 20 and are left with one more.
This is no different from picking just two boxes at the outset (other than Noel would not have a 1 hour show to fill).
So each box is 1/22 and thus a 50/50 probability.
May 1st, 2007 at 4:57 am Everyone appears to have mixed opinions here.. so I’m going to give a scenario -
You pick a box at the beginning of the game, 1/22. Let’s say you eliminate boxes one by one and come down to two boxes. Obviously one box contains a larger sum than the other. Statistically, should you swap or stay with the original box. Which, if either, option has more of a chance of providing the larger sum?
May 13th, 2007 at 8:42 pm Okay, so I’ve been turning this in my head over and over. RE: Dannie’s calculation, you have:
Path 1.
Probability of choosing the top prize box at the start = 1/22
Path 2.
Probability of choosing any other box at start = 21/22
Probability of randomly eliminating all 20 other boxes and leaving the top prize box = (20/21)*(19/20)*…*(1/2) = 1/21
Total probability of swap box being top prize = (21/22)*(1/21) = 1/22
Now, we’ve all been assuming that you’re down to a scenario where you’ve successfully eliminated 20 “empty boxes.” In DOND, you have to do it randomly, in MH, it’s by intelligent choice. This does not matter.
If we are to assume that you successfully eliminate 20 boxes, and are left with your box and another, and one of those boxes has the prize and the other is empty, you SHOULD SWAP. Why?
Because you have a 1/21 chance of going down path 1, but a 21/22 chance of going down path 2. Remember, we’ve already stated that you’re in a scenario where we happen to be able to eliminate 20 empty boxes. That is, we KNOW we are on path 2! Therefore, to win, you should swap!!!
May 13th, 2007 at 8:49 pm ^Okay, let me clarify it a little. We don’t KNOW we’re on path 2, but we KNOW we are comparing those two paths, because we’ve always talked about a scenario where there are only two cases left, and one has the prize money.
So if we know that, let probability of getting down to two cases where one has the money = 100%.
Prob(selecting the prize case & getting down to two cases) = 1/21
Prob(selecting an empty case & getting down to two cases) = 21/22
May 30th, 2007 at 5:39 pm Everyone here seems to be missing a key thing here. Let’s say that at the end of the game you’re left with 2 boxes, the top prize and the bottom prize. You can, of course, apply the reasoning that the odds were 1/22 for you to pick the top prize, having opened all the boxes but the remaining box you can say that it is therefore 21/22 that the other box contains the jackpot. However you could just as easily turn this reasoning on it’s head. For example, at the beginning it was also 1/22 that you chose the bottom prize, and when you reach the end of the game, according to the exact same conditional probability some of you are employing, it’s also a chance of 21/22 that the last remaining box contains the booby prize. This is obviously a paradox, hence the reasoning used to get here is false.
The correct solution is to consider the problem the following way. Let’s imagine you get to the final two boxes again, 1 box contains the jackpot and 1 contains another prize. There are 2 ways you could have got here, either by choosing the jackpot as your box, or by choosing to open every box but the jackpot.
The odds of choosing the jackpot are 1/22
The odds of leaving the jackpot until last are 21/22*20/21….*2/1 = 1/22
Hence this situation will occur equally often via both routes, thus 50% of the time you will have the jackpot and 50% of the time you won’t. It’s a coin toss.
June 19th, 2007 at 7:33 pm If you use the pre tag, perhaps around code tags just for fun, Wordpress (actually, the web browsers) will not lose your indents.
October 5th, 2007 at 5:15 pm This argument has been worked to death on a probability newsgroup - I used to be fooled into the MH analogy until I realised that the chance of your picking the £250K was 1/22 ONLY at the very begining - each time a non - £250k box is opened ( at random ) the odds shorten - until
ultimately you arrive at 50/50.
Swapping makes no difference to your probability of success.
January 30th, 2008 at 11:57 am Well… I’ve certainly enjoyed all these arguments for and against the equivalence of DOND (Deal or No Deal) and MH (Monty Hall), but let us look at the weightings of choice based on the values of boxes remaining and does that affect the choice. Strictly speaking, no… as it would appear that you always have a 50% chance of choosing a higher or lower box.
Nonetheless, I would like to see a probability tree of all the outcomes, with the option to swap at the end… and I think that would resolve it for all scenarios.
FINAL POINT: If it’s different to MH then with 50% probability, there’s no harm in swapping. If it’s the same as MH then swapping is better. As we have split opinions, we should swap anyway (even if correct in the opinion that MH is different)… Therefore I conclude that you should always swap!!!
June 27th, 2008 at 5:37 am I was just watching DOD and the scenario unfolded where the guy had three cases: $100, $10,000 & $750,000.
He took the deal, but when asked what he would done he said Case 3 ($10,000). He then said he would swap (which I thought he should of bc of the MH problem). Had he swapped he would have won the $750k.
I was trying to explain why he should have swapped to my roommate, but I’m not much smart in the maths, so came to the Internet. I was interested to see such debate and to see the consensus is a no swap.
I disagree and say they should swap.
Think of it this way:
There are 26 cases. You obv. have 1-26 of picking the $1M. But let’s not limit it just to that. Really, there are about 5 Big Money cases. So let’s say those are the cases you are after. The rest are Crap Cases
Your odds are 5-26 or picking a Big Money Case, and 21-26 of picking a Crap Case.
So from the start, the odds are far likelier that you picked a Crap Case vs. a Big Money Case.
If you have gotten down to 2 cases, one we know is a Crap Case and the other a Big Money Case, it doesn’t matter if it was revealed to you intentionally or by random, you are still at that same point. It is irrelevant how you got there.
I know the perception is it is 50/50 between the Crap vs. Big $ case because you see the two in front of you, but I refuse to believe it.
When you picked the cases there was a 21/26 chance you picked a Crap Case. Thus, when given the option, you should swap your Crap Case with what is likely to be a Big $ Case. You have now inversed your odds of being 5/26 that you now have a Crap Case. i.e. 21/26 of having a Big $ Case.
Again, I am not a statistician by any means, but this sounds pretty logical to me, and in spirit with the MH problem, which everyone seems to think doesn’t hold water in this case. (No pun intended.)
June 27th, 2008 at 12:59 pm Tom, you really haven’t thought that through. The probability that the OTHER box started with a big one is ALSO 5/26. Your own (flawed) logic could equally be applied from the perspective of the other box, to say you shouldn’t swap.
The odds do NOT remain the same as the start because every box that is opened change the odds of what is in your box and every other box. According to your logic, you would still have a 1/26 chance of having the sum that is in a box that has already been opened. Clearly, once a box has been opened, the probability that you chose that box at the start is no longer in existence, hence the probabilities continue to evolve rather than remaining consistent with the initial probabilities.
August 26th, 2008 at 8:24 am Hey idiots. Swap the case. Monty Hall DOES apply because you are given knowledge throughout the game - just as if Monty Hall was telling you a goat was inside of each case eliminated. It doesn’t matter if Monty Hall tells you this, or if you just get lucky on DOND. The fact remains that after your choice is made, you are given knowledge that $1 million still remains. This is new knowledge. By the time you get down to the end with only two cases remaining, you KNOW that one of the cases has $1 million, and the other case does not. This knowledge could have been given to you by Monty Hall, Helen Keller, Jack Benny, or by your stupid luck. Nonetheless, you are left with two cases and one has $1 million. You know that one of them has $1 million. If the total number of cases is 22, then the odds that your case has $1 million is 1 in 22 (these are the original odds when you picked your case). The odds that the remaining case has $1 million is 21 in 22.
Switch cases. The Monty Hall principle does apply.
It is a completely different story if Howie did not tell you that you were left with two cases, one with $1 million and the other with something else. Pretend that you started with 22 cases and just randomly eliminated 20 cases. The big information board isn’t working, and hot chicks holding the cases never open the cases. They just sit there looking stupid and so damn hot, and they forget to open the cases to show you what you just eliminated. So, you have just eliminated 20 cases. You don’t know what you eliminated. Your thumb is up your ass and you are now left with two cases. Should you switch? You should switch only if it gives you the opportunity to walk up to the hot model holding the case. It does nothing for your odds of winning. But, hopefully you were able to get a little sniff of her perfume. Odds are you already lost the $1 million with one of your earlier eliminations anyway. You don’t know anything about the last two remaining cases. If you switch or not, the odds are still 1 in 22 that you picked $1 million.
Take that stupid probability math and stick it where the sun doesn’t shine. It’s so pretentious. Dont’ let someone confuse you with flawed logic even if they use fancy numbers.
August 26th, 2008 at 10:58 am
Thanks, John for your analysis, and for the show Cops
That ‘pretentious’ ’stupid probability math’ is backed up by a simulation that requires no probability maths at all. Have you found a flaw in that as well?
August 28th, 2008 at 8:39 pm Thank you James Casbon. Now I’m the idiot because I thought I knew what I was talking about. I apologize for my abrasive language.
I’m back to believeing it’s a 50/50 split when we get down to the final two cases. In my scenario I assumed the $1 million took the path down to the final two cases. But, just as easily, the goat could have taken that path too. In the end, I am left with two cases that took a path. I don’t know which scenario happened.
Monty Hall knows which scenario happens. He eliminated the goat for you. So, in 21 of the examples, you are going to be presented with an ending scenario that involves $1 million and a goat. But, you really don’t desirve this because on DNOD, you would have arleady eliminated the $1 million earlier. You woudl never get to a Monty Hall scenario at the end. The only time you do get to the end of having $1 million in a case and a goat int he other case is the 50/50 case.
I’m not an Einstein. But, Einstein proved all of his theories through logic and reasoning. I don’t like numbers because I work with them as part of my job. I know how they can get manipulated. I value companies. OMG. You just wouldn’t believe the things that people do with numbers when it comes to valuing a company….
If I could delete my comment sent yesterday, I would. Not becuase I was wrong. But, because I called all of you nice people idiots.