My OperaMonty Hall Meets Deal or No Deal
Friday, January 2, 2009 2:04:02 PM
This article was brought on by the missing child problem in the last post. The Monty Hall problem is a a game show scenario where you are shown three doors. Behind one of the doors is a car and behind the other two doors are goats.
You must now pick one of the doors. Doesn't matter which as you have the same chance no matter what door you pick. So let's pick door #1.
Then the host will look behind door #2 and door #3 and open one of them where there is a goat. If both are goats, one door is picked at random. But you're not told that both are goats.
Now you are asked if you wish to switch doors with the remaining one. What do you do?
Switch? Or keep the door you have?
You must now pick one of the doors. Doesn't matter which as you have the same chance no matter what door you pick. So let's pick door #1.
Then the host will look behind door #2 and door #3 and open one of them where there is a goat. If both are goats, one door is picked at random. But you're not told that both are goats.
Now you are asked if you wish to switch doors with the remaining one. What do you do?
Switch? Or keep the door you have?
Switching is the correct move as this gives you 66% chance of winning the car. Sticking with the original door gives you 33% chance.
This, of course, is assuming that the host always shows a goat. For example, if he only lets you switch if you've chosen a car, then you'll lose for sure.
Why is it 66%? I've seen a lot of bad explanations. And I'm not sure I have a brilliant one, but I'll give it a shot.
When choosing a door at the beginning, what are the possibilities? And instead of using generic goats, we'll use B for a blue goat and R for a red goat.
CBR
CRB
BCR
RCB
BRC
RBC
Those are the six possibilities. As you can see, no matter what door you pick, you will have 2 out of 6 chances of winning the car. Or 1 chance out of 3. It'd be better if you could choose the other two doors. Those two doors have a probability of 2/3 of having a car behind one of the doors.
Consider the case where you now have the choice of keeping your ONE door or switching to the other TWO doors. TWO doors are better, so switching is obvious. You'll win 2/3 of the time.
What messes with people is that one of the doors is opened. If you have two doors, one of them MUST be a goat. So revealing that goat that must exist changes nothing.
What are the odds of showing a goat behind the first door?
CBR 100%
CRB 100%
BCR 0%
RCB 0%
BRC 100%
RBC 100%
What are the odds of showing a goat behind the second door?
CBR 100%
CRB 100%
BCR 100%
RCB 100%
BRC 0%
RBC 0%
What are the total possibilities if a goat is shown?
We add all the possibilities and subtract duplicates. 4+4=8. We counted ABR and ARB twice. So we must remove the duplicates. 8-2 = 6. If you don't understand this, there can only be 6 possibilities. 8 is impossible.
So when a goat is shown, we still end up with ALL SIX possibilities. But the odds of your door having the car is STILL 33% as shown in the table. Only two possibilities have a car behind door number 1. But the other four possibilities have the car behind one of the two doors EVEN AFTER ONE IS SHOWN!
Another way to look at it is this:
Pick red goat behind door #1, blue goat behind door #2 is revealed and you switch to car (WIN!)
Pick red goat behind door #1, blue goat behind door #3 is revealed and you switch to car (WIN!)
Pick blue goat behind door #1, red goat behind door #2 is revealed and you switch to car (WIN!)
Pick blue goat behind door #1, red goat behind door #3 is revealed and you switch to car (WIN!)
Pick car behind door #1, goat behind door #2 is revealed and you switch to the other goat (LOSE!)
Pick car behind door #1, goat behind door #3 is revealed and you switch to the other goat (LOSE!)
So 4 out of 6 times (2/3), you win if you switch.
The reason the goat color doesn't matter for the last two scenarios is that you can only have BR or RB for goats. Opening one door or the other won't change the ordering of the goats. It's still the same overall permutation. For example, for the permutation CBR, the host can pick the left door or the right. Either way, it represents the same permutation. But since there are two doors possible instead of one, each is half as likely to be picked (50/50 instead of 100%) for each permutation (CBR and CRB).
All this is explained multiple times all over the Internet, so I'm not going to bother with it any more.
What I'm interested in is if this applies to the game show Deal or No Deal. The consensus that I saw was that it was 50/50. In no event does it matter if you switch or not.
I ran simulations and these numbers work out.
To see the reason it's 50/50 for two cases where you're looking for the top prize, we must look at when there are three cases remaining. This looks similar to the Monty Hall scenario. You've chosen a box and two other boxes remain. You select one to open and then get to switch.
So what's the catch?
The catch is that in the Monty Hall scenario, the prize is IMPOSSIBLE to be disclosed. But in Deal or No Deal, you can very much pick the top prize and lose it. This causes different odds to occur. No matter what box you open, the top prize has equal chance to be in ANY remaining box and to consequently be taken out. This is true no matter the round. Only if you could guarantee that the top prize cannot be eliminated would the Monty Hall scenario apply. But if you can never eliminate the top prize, the show Deal or No Deal would be cancelled as you would have 96.2% chance of winning the million if you switch cases. The game wouldn't make much sense since only two boxes would remain no matter what.
Sean Connerspc476 # Monday, January 5, 2009 7:58:55 PM
As for Deal or No Deal, I thought part of the game was a counter offer from "The Banker" to cash out, which (I assumed) to be dependent upon both the case you have currently "picked" plus the value of the cases left unopened. Wouldn't that also provide some information? Or are you excluding that from your odds calculation?
Vorlath # Tuesday, January 6, 2009 8:00:56 PM
In the Monty Hall Problem, the host may not reveal your "box" and neither can he reveal the top prize. In Deal or No Deal, you can indeed reveal the top prize. That's what creates the excitement of the show (if you like that sort of thing).
What is our sample space? Easy.
Picked door, Car door
AA
AB
AC
BA
BB
BC
CA
CB
CC
Those are the only possibilities. And each is equally likely since they are unrelated events.
There is one more sample space. How likely is it to reveal a specific door in each of these cases?
For AA, odds of revealing A is 0%
For AA, odds of revealing B is 50%
For AA, odds of revealing C is 50%
For AB, odds of revealing A is 0%
For AB, odds of revealing B is 0%
For AB, odds of revealing C is 100%
For AC, odds of revealing A is 0%
For AC, odds of revealing B is 100%
For AC, odds of revealing C is 0%
You can continue on, but it's just a permutation of this.
Combine the odds...
AAA 1/9 * 0% = 0%
AAB 1/9 * 50% = 1/18
AAC 1/9 * 50% = 1/18
ABA 1/9 * 0% = 0%
ABB 1/9 * 0% = 0%
ABC 1/9 * 100% = 1/9
ACA 1/9 * 0% = 0%
ACB 1/9 * 100% = 1/9
ACC 1/9 * 0% = 0%
...
Again, you can go on, but they're just permutations of this.
To simplify, we can just say that we picked door A and keep the above table.
AAB 1/6
AAC 1/6
ABC 1/3
ACB 1/3
The odds of you winning where the door you picked is equal to the car door is AAB and AAC. So 1/6 + 1/6 = 1/3. You have 33% chance of winning.
If you switch, you must take whatever is not your original door and what is not disclosed.
AAB (switch door is C) 1/6
AAC (switch door is B) 1/6
ABC (switch door is B) 1/3
ACB (switch door is C) 1/3
So when is the switch door equal to the car door (second column)? ABC and ACB. Add those up. 1/3 + 1/3 = 2/3.
With 3 box deal or no deal, the third sample space is 100% for one specific box. So there is nothing to skew the results toward any of the boxes.
You'd have this sample space.
AAB 1/6 (Switch case is always C)
AAC 1/6 (Switch case is always B)
ABB 1/6 (Switch case is always C)
ABC 1/6 (Switch case is always B)
ACB 1/6 (Switch case is always C)
ACC 1/6 (Switch case is always B)
The times when the switch case equals the prize case is ABC and ACB. 1/6 + 1/6 = 1/3.
So switching doesn't help.
Even after the specific case is revealed. Since we picked case A, we need to reveal either B or C. Let's pick B just to see what happens.
AAB 1/2 (Switch case is always C)
ACB 1/2 (Switch case is always C)
Note that ABB is no longer possible since we picked B and can no longer switch to it.
You can increase the sample space to more boxes and it always ends up being the same thing.
Not only that, but even if B WAS the top prize, either of the two remaining cases will have to be the new top prize. And you're left with the same odds as before.
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