Deal Or No Deal/Monty Hall Problem

Factual Questions
1

Hello,

Long time lurker, first time poster here. I was arguing with some friends over whether Deal or No Deal is essentially the same probabilities issue as the Monty Hall problem. Although there are some fundamental differences, our basic question was: “When it is down to your case and only one other case, with $1 million and $0.01 on the board, are you better off switching your case?” We were also talking about it for when there’s 3 cases left, but that part doesn’t really matter.

I continually switch off in my own thinking, completely convincing myself back and forth that it is, or isn’t, the same. Also, don’t let this thread turn into a Monty Hall problem argument. It’s generally known that you are better off switching-- if you disagree, I ask you to do some other research.

Here’s my thinking both ways- $1 million is the car, $0.01 is the goat:

Yes, it’s the same: At the beginning, you pick one case, hoping for $1 million. You have a 1/26 chance of getting it. Just like in Monty Hall, other cases (that you KNOW aren’t $1 million because of the ground rules I’ve laid out) get eliminated, and in the end, you’re offered the switch. As far as I can tell, this is exactly the same as Monty Hall.

No, it’s different: In the sake-of-argument-style rules of Monty Hall, Monty always has knowledge of which door has the car, and purposely chooses a goat. This knowledge is a fundamental difference in Deal or No Deal and Monty Hall, and affects the probabilities.

So, tell me dopers: You’ve rejected all the deals, and are down to $0.01 and $1 million. You are offered the chance to switch. Should you?

Thanks,
Matt

2

No, no goat is revealed so there is no advantage to either switching or not switching.

3

In Deal or No Deal, you first choose one door and then choose 24 doors to open. So you’ve essentially chosen both of the doors yourself. In Monty Hall, you choose one door, and the host chooses one of the two remaining doors.

4

To make it more explicit, imagine a setup with 26 doors. You choose one, and then 24 more doors are opened and their contents revealed. In the Monty Hall scenario, Monty chooses which door remains unopened. In the Deal/No Deal scenario. you make the choice. Does this help?

5

You can switch or not switch; it makes no difference at all.

You’ve already illustrated why. The assumption in the Monty Hall Problem is that Monty knows where the goats are and deliberately reveals a goat, which is what makes it logical to switch. But if you were to consider a scenario where Monty DOESN’T know where the goats are, switching would be pointless; if Monty pulls back Door #2 to reveal a goat, he’s simply hit his 66.6% change of hitting a goat and the odds are now 50/50 for either remaining door.

Assuming no knowledge is involved in the reveal, there’s no advnatage to switching. In “Deal or No Deal,” there’s no knowledge involved; the host, Howie Mandel, is not picking any cases or you and anyway he doesn’t know where the million dollars is, either.

The “I might have a million bucks in my briefcase” idea in “Deal of No Deal” is purely illusory. It would make no difference at all in the way the game is played if you didn’t choose your briefcase to start the game, but instead just picked 25 straight cases to eliminate and were left with the last one.

Personally, I’d be the worst Deal or No Deal contestant ever.

“Okay, Rick, pick six brief…”

“Gimme one through six.”

“Uh, okay. (Reveals cases) Now, the banker is calling…”

“Tell him to go to hell, no deal. Now open cases 7 through 11.”

6

It doesn’t make one bit of difference. Both cases which are unknown have a 50/50 chance of being the $1million case.

If anyone could explain why it would be a good strategy to switch or not switch besides some supersticious reason I’d like to hear it,

7

RickJay: I just love your strategy. It made my day.

8

Lets say that you and I are playing the same game, and picked the same case. Lets say that I want the $.01 one, and you want the $1 million dollar one. By the logic you presented here, you have a 1/26 chance of originally picking the $1 million dollar case. By the same logic but replacing 1 million with .01, I have a 1/26 chance of having originally picked the $.01 case. Now, since we have picked the same case there is a 1/26 chance of it being the 1 milllion dollar case, and a 1/26 chance of it being the .01 case. Since those are the only two options for the case, one must wonder what happens for the other 24/26 chances. Obviously the answer is that this type of logic is wrong. There is a 50% chance of it being the 1 million dollar case, and a 50% chance of it being the .01 case.

9

This thread appears quite dead but I feel compelled to offer an alternative explanation to the author’s question for anyone who stumbles on this as I did. I would argue that deal or no deal and Montey hall are essentially the same. People seem caught up on the idea that the host’s knowledge of which door the car is behind fundamentally differentiates this game from the Monty Hall problem and I have yet to see a good explanation here of why. For me the most important thing regarding the last 3 briefcases/doors is that one is revealed and that that one is not the million dollars/car. It does not matter that the revelation of the goat or non-mollion dollar briefcase was revealed by chance or because of the host’s knowledge. What is most important is simply that it is revealed. Regarding the last three brief cases, you know you have a 1/3 chance of holding the million dollar briefcase, and a 2/3 chance of holding a briefcase that does not contain the million dollars. 2/3 of the time, the revelation of a non-million dollar third briefcase will simultaneously reveal the million dollars. There is only a 1/3 chance that of the remaining three you hold the million dollars, and thus 1/3 of the time it is in your best interest not to switch after the 3rd briefcase. 2/3 of the time, however, you will be better off switching, simply because of the greater chance of having chosen a non-million dollar briefcase in the first place. Perhaps this helps explain it. Just think that if in the montey hall game the goat WERE revealed by chance, it would still be a good idea to switch.

10

If amongst the last three briefcases one is revealed and it NEVER has the $1 million dollars in it, then it cannot be a random reveal since logically a random pick should have the $1 million one-third of the time. So if this describes the show (and I’ve not seen it), then it is not a random reveal and it conveys exactly the same information that Monty Hall does. On the other hand, if the reveal is random (and therefore it should show the $1 million one-third of the time – oops probably bad for the shows ratings), then it conveys no information and switching is irrelevant.

11

But the whole point of the Monty Hall game is that the goat ISN’T revealed by chance. It’s purposefully giving you the “wrong” choice of your 2/3 chance when the odds are still 1 or 2 out of the three.
In the Deal or No Deal situation there is always a chance the “goat” reveal may be the $1million box. In the Monty situation that chance does not exist.

12

If a goat is revealed by chance and two doors remain, each has an equal chance of being the winner. What asymmetry can exist here that would make one more likely than the other?

13

$1 million dollars turned out to be a goat, you’d be gutted.

14

Imagine a game with one thousand boxes to choose from. 999 contain goats. The other one contains a million dollars.

If the host is Monty, you pick one box to hold on to, Monty opens 998 boxes that he knows are goats, and gives you the option of switching.
Obviously, you’re going to switch. The remaining box has to be the one with a million dollars, because there’s only a slim chance that the first box you opened was the correct choice. In fact, you have a 99.9% chance of getting the million dollars if you switch.

On the other hand, if the host is Howie, you pick one box to hold on to, and Howie makes you open 998 boxes. Most of the time, you’re going to open one with the million dollars some time during that 998 run, and the game will be over- you’re going to go home with a goat no matter what. But in the unlikely (0.2% chance) event that you managed to open 998 boxes at random and every one happened to be a goat, it doesn’t matter if you switch or not. What’s the difference between the two boxes? Both of them you decided to hold off from opening, for whatever reason. What difference does the order make?

To sum up, with Monty as the host, you always have a 99.9% chance of winning as long as you switch.

With Howie, you have a 99.8% chance of losing during the initial opening phase, and then 50/50 chance of winning if you manage to get past that phase through sheer luck. Overall, you only have a 0.1% chance of winning no matter what you do with Howie.

I’d much rather be playing with Monty.

15

I didn’t know how Deal or No Deal worked, so I looked up the Wikipedia article on it, and when I saw the picture they chose to illustrate the article, I was pretty sure that the editor of that article is a marketing major who has studied the art of website promotion. :slight_smile:

16

Here’s the simplest possible explanation:

  1. Does Monty EVER open the door with the prize? Or does he ONLY open doors with goats?

  2. On Deal or no Deal, does the contestant EVER open a case with $1 million? Or ONLY cases with lesser amounts?

So, the difference is simple. In one game, the contestant can randomly reveal the big prize (thus eliminating it from play). In the other game, the host deliberately avoids revealing the big prize.

Of course, I like me some goat meat, so the Monty Hall problem is a no lose proposition.

17

It is the same asymmetry that exists in the original Monty Hall problem. With three doors, you know that you have 2/3 chance of having chosen one of the two goats, and so if a goat is revealed, even by an oblivious Monty, you know that there is a 2/3 chance that the door that was not revealed and that you did not choose is the car. 2/3 of the time after the goat is revealed the car will be behind the other door, regardless of who revealed the goat. This is the simplest way I can explain my thinking on this aspect of the problem. Yes, the odds of a goat being chosen are 100% with Monty, and everyone is totally right that this is a key difference between the two games, but among the last three briefcases, having revealed the goat by chance, it is still a better option to switch.

So yeah, amongst the last three briefcases (and definitely yamatotwinkie’s 999 boxes also), Monty’s knowledge is significant because it prevents the contestant from eliminating the prize. Definitely, without a doubt. Perhaps though the revelation of the goat by chance still gives one sufficient reason to switch from their 1/3 chance of winning the prize to a 2/3 chance. Someone try understand what I’m saying so I don’t feel so completely insane :confused:. I’m damn sure I’m right, though ;).

18

or, better yet, perhaps someone could try to explain why Monty’s knowledge versus his ignorance of the what lies behind the doors alters the original Monty Hall probability after the third door has been shown to conceal a goat. How does the “asymmetry” disappear? I am referring only to the last three doors or briefcases and not to the monumental feat of probability that you would have to achieve to make it to the last three briefcases with Howie as the host.

19

No. Go back to my example above of 1000 boxes and Howie as the host, where you keep one box, open 998 others, then choose between the remaining two. Before the game even began, you could have decided:

“I’m going to set aside box #12 first, then set aside #752 as my second box”

or you could have decided:

“I’m going to set aside box #752 first, then set aside #12 as my second box”

Either way, the game proceeds identically. In both events you result in two unopened boxes that you could have chosen randomly from the get-go. The order makes no difference. In the unlikely event that you didn’t reveal the million dollars in the 998 other boxes (which seems to be the part you’re still stuck on), it doesn’t matter which of the two boxes you set aside “first”. Why would it? You could have chosen the two at the exact same time!

20

IMHO I always thought that if it came down to the $1 million case and the $0.01 case you were better off switching. My reasoning was like the OP’s first thought; that the odds that you picked "correctly” the first time was 1/26 so the odds that the $1 million case was still on stage were 25/26.

I think the key is that if through chance it comes down to the 1cent and 1m cases you still know what the odds originally were. If Monty didn’t know where the prize was but got lucky and revealed a goat, the odds are still in your favor to switch, right?

The OP set the condition that chance played a major part and got you to this point; you were lucky and opened 24 “wrong” cases. So what are the odds that you picked the right case at the start? Still 1/26.

Of course I only took one class in statistics and I’ve only had 1 cup of coffee so far today.