Deal Or No Deal/Monty Hall Problem

Factual Questions
21

What is possibly being missed in the random reveal scenario is that as goats are randomly removed from the “stage pool”, the probability that your original choice is correct increases. This is usually the intuitive thought - it’s often harder to grasp that when one of the stage pool goats is revealed intentionally, your odds of winning don’t improve.

Another way of looking at it is this: If Monty intentionally reveals a goat, he is selecting from all goats excluding your choice if it is a goat. That skews the results toward the winner being on stage once more and more goats are revealed. Once the number of goats left gets down to 2, it’s more likely than not you’ve got one.
On the other hand, a random selection that reveals a goat excludes your case as well. This time, your case is as likely as any unopened case to contain a winner. Since only goats are revealed in this scenario, the likelihood that any unopened case (including yours) contains a winner is equal. Once it gets down to two, there’s no reason to switch as you can’t do any better.

22

No. Again, with Howie, you could have decided, even before the game began, to hold those two cases aside. They have equal 1/26 probabilities of holding the million at the game start. Every case that gets opened and doesn’t reveal the million increases both case’s odds. When there’s two cases left, they have 50/50 odds.

Imagine you’re playing the lottery. You and your friend have the exact same ticket, except the last number is different (you picked 0, he picked 1, and there were only two options). The radio station starts slowly reading out the numbers, and both of you have matched the first set of numbers with only one number left to be read. Do you switch tickets with your friend before the last number is read?

According to your logic, since you only had a 1 out of ten million chance of winning the lottery when you first bought the ticket, your friend’s ticket should have a 99.999% chance of winning. But then wouldn’t your friend think the exact same thing about your ticket? You both can’t be right.

23

I think I see what you’re getting at, but it still makes my head hurt. I’m going to go take a nap first . . .

24

The entire point of the Monty Hall problem is that Monty has told you something you didn’t know at the start (or, another way, he is constrained in his actions during the reveal - your initial choice dictates his choice).

If Howie chooses the cases for you, always avoiding the $1 mill, then you should switch at the end. But since you choose the cases yourself, each time risking the 1/26 chance of opining the $1 mill, it makes no difference if you switch or not.

25

And, like RickJay pointed out five years ago, in the game of Let’s Make a Deal, if Monty didn’t know which door had the prize and which had the goats, but just got lucky on this one event, you have no advantage by switching. The “2/3 probability of winning if you switch” answer applies only in the case where Monty knows where the prize is and avoids opening that door.

26

If Monty does not know what is behind each door, the chances are:
1/3 You have picked the prize.
1/3 Monty opens the prize door.
1/3 The remaining door has the prize.
If Monty opens a door that reveals a goat, one of these possibilities has been excluded, leaving 2 possibilities with equal probibility. He could have just as easily opened the prize door, ending the game. It’s this new information that changes the odds of your door winning.

If Monty knows what is behind each door and only opens a goat door, the chances are:
1/3 You have picked the prize door.
2/3 One of the other doors is the prize door.
When Monty opens a goat door in this situation, you have learned nothing new. He always opens a goat door, so you have no new information. The odds of your door winning are still 1/3, and the odds of the other door(s) winning is 2/3.

27

That’s a sure sign. :slight_smile:

Let the doors be numbered 1, 2, 3. The prize is behind one of them, call it A, and the others B and C. You can choose A, B, or C. Then, a second one is choosen (by you in Deal or No Deal; by Monty in Let’s Make a Deal), and the last is left. So, our options are ABC, ACB, BAC, BCA, CAB, CBA, which exhausts the possibilities, and all of which are equally probable.

In Deal or No Deal, you win by staying with the two cases ABC or ACB, but you can also win by switching in the two cases BCA or CBA. Your chances are the same.

However, in Let’s Make A Deal, Monty would never choose the winning door, so of the six possibilities, what’s left is A(B/C), A(C/B), BA, BA, CA, CA, after Monty’s door is removed. You can see that your chances are improved by switching.

28

RM Mentock, among others states it clearly. Monty Hall will never reveal the grand prize, only a lesser prize. I don’t know much about statistics or probability, so to resolve this for my own understanding, I wrote a program to simulate the circumstances. It was immediatly obvious that the selection of the door to reveal was not random, it had to involve knowledge of the prize behind the remaining two doors. On Deal or no Deal, that isn’t a factor. First of all, the choice is not made when there are three cases left, just two, when you have a choice between your initial pick and the only remaining case still held by a model. Secondly, on Deal or No Deal, the choice to switch between the final two cases isn’t based on any knowledge of the contents of either case.

In layman’s terms it’s easy. On Let’s Make a Deal, you have a one in three chance of picking the right door. One out of three times you will pick the winner, just by chance. When Monty reveals one of the other doors has a goat behind it, he is telling you that there is a one in two chance that the door you didn’t pick contains the grand prize. Why? Because two out of three times the door you didn’t pick the right door contains the grand prize.

Here are your choices:

  1. You picked the right door, and the other two don’t contain the grand prize.
  2. You picked the wrong door, and the other door A that Monty has not revealed contains the grand prize.
  3. You picked the wrong door, and the other door B that Monty has not revealed contains the grand prize.

So two out of three times, the other door that did you didn’t pick contains the grand prize. One out of three times the door you picked contains the grand prize. So if you switch, you change from a one out of three chance, to a one out of two chance, to win the big one.

On Deal or No Deal, you get the opportunity to switch only when there are two cases left. So here are the choices:

  1. You picked the right case, and the other one is the wrong case.
  2. You picked the wrong case, and the other one is the right case.

It doesn’t matter whether you switch or not, the chances of winning are the same. One out of two.

Keep in mind, on Deal or No Deal you are only given the choice to switch when two cases are left. On Let’s Make a Deal, there are still three choices when you are given a chance to switch, and Monty only reveals the door which is not the grand prize. It’s not a random selection. Monty isn’t going to show you the door with grand prize, because then the show would be over, because everyone would know you lost, and there’s no point in revealing what’s behind the door you picked.

On Deal or No Deal, if they did something similar when there were three cases left, what would they do? If they reveal the contents of case you didn’t pick, and it’s the big prize, the show would be over, nobody would care what was in the case you picked. So they would have to reveal a case based on the knowledge that it didn’t contain the big prize.

29

If Monty Hall is too easy for you, here’s a generalization that may be a good exercise in applying symmetry principles to find optimal mixed strategies.

Montgomery and Sally play a game with K closed boxes. Montgomery hides a million-dollar jewel in one box (the other boxes are empty) and proceeds to open the empty boxes one by one as Sally plays (see below). When there are only two boxes left, Sally opens either box and wins if it contains the jewel. Prior to each of the K-2 box openings Sally chooses one box and locks it, preventing Montgomery from opening it next. That box is unlocked after the opening and cannot be so locked twice in a row.

What are the optimal strategies for Montgomery and Sally and what is the fair price for Sally to pay to play the game?

30

The third sentence contradicts the first two.

It should read: “So if you switch, you change from a one out of three chance, to a two out of three chance, to win the big one.”
Note that the MH problem is exactly equivalent to this:

Behind 3 doors are 2 goats and one big prize. You have the choice of choosing one door and keeping what’s behind it, or two doors, keeping your choice of what they conceal.

31

Yeah, poor writing.

32

It’s been a while and Deal or no Deal isn’t making new episodes, but I wanted to try my hand at this.

  1. A lot of people here seem to be saying the whole POINT of the Monty Hall problem is that the host KNOWS where the goat is and picks FOR the contestant. That it’s NOT done at random.

No. This is not at all the point.

The host knowing ONLY simplifies things and gets us to the ACTUAL point of the MH problem. Imagine if he didn’t know. Or you choose. Or your overweight nearsighted autistic swimming instructor chooses. Or a butterfly sneezed in the studio and accidentally knocked down a door revealing a goat. The host knowing is just a vehicle to quickly and simply get us down to the scenario where a goat is shown and you have the option of sticking with your door or switching to he other. How we get there really does NOT matter.

3 doors
1 car
2 goats

You are down to a prize and a lesser prize

  1. Yes, technically the $1,000,000 is the car and every lesser amount would be a goat. We’re using “goat” here as a variable. Now THIS is where the randomness and NOT KNOWING DOES MATTER. It determines what amounts are goats and what amount will end up being the “car”. This is the source of the confusion.

At the end of DEAL or NO DEAL you will always have two cases. One with a higher amount and one with less. ALL the OTHER cases have been opened and revealed an amount that is not the prize. The prize is the highest amount you can get. Say it’s $.01 vs $50,000 or $200 vs $100,000. We’ll go with the latter two amounts. It does not matter what amounts were actually in all the other cases.

We are standing on stage with $200 and $100,000 still on the board. All the other cases are our third door. Many of those cases revealed higher amounts than our $100,000 and many revealed less. But $100,000 is NOW what we stand to win or miss. Since you didn’t pick any of the other cases you couldn’t have won them… they may as well all have contained goats at this point. Or they may as well all have been one door containing a goat (or in this case $200)

What are the odds that you chose the $100,000 at the beginning? What are the odds that it’s still on stage and you have another $200 in your case? Monty Hall problem. 2/3 odds again if you switch.

  1. THE ONLY THING THE RANDOMNESS of the game does is gradually determine by chance just how much the “car” is worth when we get to the MH Problem at the end.

  2. This thread was frustrating to read

33

I thought Monty Hall posts were automatically locked and thrown into the deepest dungeon possible?

34

I should change the wording at the end bit. It’s not what the odds are that you chose $100,000 at the “beginning” (beginning of the game). It’s as if you only had two rounds to “begin with” knowing the options were two “not $100,000” (or in our hypothetical scenario two goats… or $200) and one “$100,000” (car). The entire game thus far has revealed all the cases that are “not $100,000” and can be consolidated into opening one door with a goat behind it. You’re left with two doors and the option to switch.

35

Only in the best of all possible worlds

36

You’ve completely misunderstood the problem, and your answer is wrong on nearly every level.

37

You could not be more wrong.

If Monty doesn’t absolutely know where the goat is, he opens the door with the car some fraction of the time (actually 1/3 of the time).

But this never happens, because Monty KNOWS where the car is and NEVER opens that door.

So, yes, the point IS that Monty knows where the car is, and that affects the probability.

38

: popcorn :

39

Please note that the “Monty Hall Problem” is not how the show actually worked on TV. In the Problem, it’s a given that the host is constrained to always open a goat door before offering you the choice to switch. In the actual TV show, Monty was not predictable like this.

If Monty is forced to always show a goat before offering the choice, then the answer to the problem is that you should switch. In the real TV show, Monty was capricious and there is not a clear answer about what’s best to do.

40

Since you’re new here, FeelScknd, you should know that the Monty Hall problem has been discussed in great detail here in previous threads, such as this one and this one, which contains links to yet others and to the Straight Dope column(s) in which Cecil Adams gave us the Straight Dope on the problem. The chances of your bringing us any new enlightenment on the problem are small.