# Deal or No Deal



## cmhardw (Oct 28, 2008)

Hi everyone,

This is a math question about the game show Deal or No Deal. Please see here for a description of the US version of this game. http://en.wikipedia.org/wiki/Deal_or_No_Deal My question is this. I saw a particular episode one day where they were offering multiple million dollar cases, and I am struggling with a probability question.

Let's assume that out of the 26 cases to start there are 7 cases that contain 1 million dollars. I pick a case, and proceed to choose 24 cases after that one. After choosing 24 cases we are down to my case, and the final remaining case on the stage. I know that there is exactly one case containing a million dollars left in play and the other case is not a million dollar amount.

The probability question I am struggling with is this: *what is the probability that my case contains 1 million dollars in this situation?*

I've looked at this two ways. For the first method I imagine that I choose a case out of the 26. There are 7 million dollar cases, so it should seem pretty obvious that at this moment I picked a case containing one million dollars with probability 7/26. The rest of the show is just theatrics and regardless of what Howie Mandel says, or what the models do, or which cases are opened I am still stuck with the case I initially chose, with 7/26 probability of being 1 million dollars.

The second method for estimating the probability I used Bayes' Theorem to try to figure it out.

P(A|B) = P(A) * P(B|A) / P(B)

Let's define the events A and B as follows:

A = the event that my case is a one million dollar case
B = the event that I choose 24 cases (after my initial case choice), leaving me with exactly one case containing a million dollars and exactly one case containing an amount less than a million dollars. This is equivalent to saying that I choose 24 cases (after my initial case choice) and see exactly 6 cases containing one million dollars in the group of cases that I open.

P(A|B) is the probability that my case is a one million dollar case, given that I have opened 24 cases and know that exactly one of the remaining cases contains the amount of one million dollars.

P(A) = 7/26
P(B|A) = 19/25
P(B) = 7/26 * 19/25 + 19/26 * 7/25

so P(A|B) = (7/26 * 19/25) / (7/26 * 19/25 + 19/26 * 7/25) = 1/2

Now I am imagining the same situation in both cases, but I get different answers for the probability which should not happen. Howie Mandel always gives the Bayesian answer it seems, but I am hung up on the fact that it seems like the probability should remain 7/26 despite the additional information of opening cases. I know this may seem like the game show where you have a car behind one of three doors and you pick a door. After picking the door the host opens an empty door and asks if you want to switch your door, or leave it the same as your original choice. Here I understand that you should always switch to the other available door, since there was a 2/3 chance you initially chose the wrong door.

But for this Deal or No Deal situation where you are down to two cases with exactly one case being a million dollar amount, which is the probability that you are holding a million dollar case? You aren't allowed to switch cases, so you're stuck with your initial choice regardless, and at the time that you made that choice the probability was 7/26 that you chose a million dollar case. But using Bayes' Theorem the probability is 1/2, which is the chance that your case is the million out of two possible remaining choices. Which is the real answer?

Thanks for any help,
Chris


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## blah (Oct 28, 2008)

Actually, Chris, why don't you just run a test of say, 1000 trials, and find out what the answer is? Then figure out where the flaw in "the other hypothesis" was? (That's what I do most of the time anyway )


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## amateurguy (Oct 28, 2008)

Ok. I don't know much about Bayes' Theorem, but I hope I can answer your problem. 

This is definitely like a twisted version of the Monty Hall Problem (the one you referenced in the second-last paragraph). I guess information does really change your probability.

The probability should be 1/2 because:



> I know that there is exactly one case containing a million dollars left in play and the other case is not a million dollar amount.



Let's run through an example:

You're playing the game and you pick one case in the beginning (for simplicity's sake let's call it the 'red case'). Like you said, it has 7/26 chance of being a million-dollar case.

Elimination time. You reject a case... it is opened and... it's not a million dollars.

25 cases left in play. 7 cases still having a million dollars... This means your red case has a *7/25* chance of having a million dollars, from 7/26.

You reject another three cases. None of them have a million dollars. Your red case now gets a *7/22*. 

You reject another case. Not so lucky. It was a million dollars. So 21 cases left in play... 6 cases with a million dollars... your red case now gets a *6/21*.

This goes on until you are left with the ending situation, where two cases are in play (one of them being your red case) and you *know from the elimination round earlier* that there's still one million-dollar case left.

Therefore, your red case gets a *1/2*.

....

The first answer of 7/26 would be correct *only if the money in the rejected cases are not revealed*. i.e. the models on stage do not open the cases you choose. 

I really, really hope this helps! (And really, really hope I did not get it all wrong and embarrass myself!)


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## Mike Hughey (Oct 28, 2008)

Imagine there are parallel universes for each possible choice of case you had. In most of the cases where you didn't get down to the last two, you would have already been eliminated (or you would have chosen to stop), and the game would already be over (and you would not have gotten the million dollars). In order to come up with the 7/26 probability, you would have to consider all of those possible universes at once. Since you're in the "current universe", the only thing that matters is the current state, and the current information clearly shows that the probability now is 1/2.


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## brunson (Oct 28, 2008)

I got into the same sort of logic trying to estimate winning probabilities for poker hands, chasing myself through a maze of twisty little tunnels, but I read a book that explained the overall simplification and it saved my sanity. 

Even though someone else has seen their cards, you haven't, so when you calculate the odds of your draw you simply base it on the number of cards you haven't seen. In this case, you don't know what is in your case or the other case, so the chance that you have the million is 1/2. There's a name for this in mathematics, something like "Observer Unknowns" or some such, but essentially it says you can't base your probability on what someone else knows. (Unless they use that knowledge to alter the outcome, like the Monte Hall Problem.)

Reading that first paragraph just made me think of something else. Does the "Banker" know where the million dollar cases are? Would that change his offers of a deal? At first I thought it would, because he knows whether you have the million dollars or not, but he doesn't know the order in which you'll pick the other cases, so expected return stays the same for both him and the contestant. 

However, when you get down to the last two cases, and lets say there is a $1 case and a $1M case. Expected return is $500,000.50, so if he doesn't know what's in your case, his offer should be somewhere close to expected return. However, if he knows you don't have the million in your case, he could offer you $1 and you're screwed. On the other hand, if you have the million, he could offer you $750,000 to "save" $250,000. However, if he knows you *don't* have the million he could offer you $800K for your case, attempting to convince you that you do have the million, then open the case and lose big.

I think the greatest story surrounding the Monte Hall problem was in a follow up to Marilyn vos Savant's original column. This reporter for Parade went and met with Monte Hall on the set of "Let's Make a Deal" where Monte demonstrated his ability to influence the reporter to choose the *wrong* door 5 times in a row, which completely violates the predicted outcome of 2/3 if you change your doors.


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## Bryan (Oct 28, 2008)

With all the information known, the probability is 1/2.

For example, what's the probability that you went to the US Open? Now, earlier this year, it may have been up in the air, depending on schedules, location, etc. But with all the information we know now, the probability is 1.


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## Mike Hughey (Oct 28, 2008)

Bryan said:


> With all the information known, the probability is 1/2.
> 
> For example, what's the probability that you went to the US Open? Now, earlier this year, it may have been up in the air, depending on schedules, location, etc. But with all the information we know now, the probability is 1.



Are you sure? It's true that his wonderful 4x4x4 BLD result was something that very few others in the world could have been capable of, but honestly - he missed both his 5x5x5 BLD attempts! Could that really have been Chris? I think maybe it was a clone. Or perhaps I was just dreaming, and it hasn't even happened yet. I put the probability that Chris went to the US Open somewhere less than 1.


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## cmhardw (Oct 28, 2008)

I don't know why I have such a hard time understanding that the answer is 1/2, this really bugs me especially since I am studying probability right now, and want to eventually use probability in my career. I shouldn't have issues with a probability problem that is this simple :-(

Here is my thinking. What if the game was incredibly boring, and after you chose your case the model immediately opens it and you win that amount. By doing this, assuming there are 7 cases out of 26 to start, you win the million with probability 7/26.

However, I am also thinking of situations where you choose a case. After this you open 7 cases and each and every one of them is a million dollar case. After this information you know that you will win the million dollars with probability 0. Or also after you choose a case, you could open 19 cases and not see a million dollar case. This means there are 7 million dollar amounts left, and 7 cases left in play (including your case). So after this information you know that your case is a million dollar case with probability 1.

So I understand that the probabilities can change given new information of opening cases. I still feel though that if there is the possibility to win a million dollar amount, that the probability stays at 7/26 despite the fact that I've been proven wrong by Bayes' Theorem, and also by people's explanations here.

What if the player's strategy was simply to choose a case and say "No Deal" to the banker no matter what until they get to open their case? Isn't this exactly the same as my version of the game above where the model immediately opens your case and you win the million with probability 7/26?

If 26 people went on the show and played with this strategy, then 7 of them are expected to win the million this way.

Now the probability that someone gets down to 2 cases, with only one million left, is my P(B) mentioned above which is 133/325. So of the 26 people playing with my boring No Deal to the end strategy 10.64 of them or roughly 11 of them should make it down to 2 cases left with one of them being a million dollar case. Half of them win the million and half don't which is 5.32 people.

Now the probability that you get down to 2 cases, and 2 of them are the million is 7/26*6/25 = 21/325 so out of the 26 people playing 1.68 people should end up in this situation. Adding up both numbers, since these events are mutually exclusive gives you 7.00 people winning the million dollar cases.

Ok nevermind, I guess I can understand why the probability is 1/2 if you're down to 2 cases. The expected outcome still adds up to what it should be if you break it down into the two possible ways to be down to 2 cases and still win the million. It's just a result that completely violates my old intuition about the game. Dang.. math is cool, this blows my mind why it works this way.

Chris



> Are you sure? It's true that his wonderful 4x4x4 BLD result was something that very few others in the world could have been capable of, but honestly - he missed both his 5x5x5 BLD attempts! Could that really have been Chris? I think maybe it was a clone. Or perhaps I was just dreaming, and it hasn't even happened yet. I put the probability that Chris went to the US Open somewhere less than 1.



Haha Mike you could always use Bayes' Theorem. It helped me to understand my probability problem ;-)


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## brunson (Oct 28, 2008)

> So I understand that the probabilities can change given new information of opening cases. I still feel though that if there is the possibility to win a million dollar amount, that the probability stays at 7/26 despite the fact that I've been proven wrong by Bayes' Theorem, and also by people's explanations here.


Try simplifying it this way. Open your case and look inside. What's the probability that your case contains what you see in it? 

The probability is an *expectation* and that expectation has to be based on what you know, not what may have happened in the past. New knowledge changes expectation of what will happen, not what could have happened.

I think in your first post you correctly showed that Bayes does predict that given a 7/26 chance originally, if you take the conditional probabilities based on the outcome of each case opening, the probability goes to 1/2 by the time you hit your last two cases.

Another way of looking at it... what's the chance of flipping a coin and getting 100 heads in a row? Very slim, 1/2^100. Now say you flipped a coin and got 99 heads in a row, does that mean that that you have a 1/2^100 chance of flipping another head? No, you have a 1/2 chance.


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## Brett (Oct 28, 2008)

I'm not sure if I understand any of this, so ignore me if i'm wrong 

Wasn't this at the beginning of the movie 21? The kid is picking between 3 doors, then one is eliminated and he asks to change his answer. He choses yes, because his door has a 1/3 chance of winning, and the other has a 1/2 chance of winning.

I think your case remains as a 7/27 odds, not the 1/2 that makes you feel good inside.


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## Michel (Oct 28, 2008)

The other door has a 2/3 chance of winning...


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