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Geek Culture / Deal Or No Deal (The Monty Hall Problem)

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Pincho Paxton
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Posted: 21st Feb 2006 12:27 Edited at: 21st Feb 2006 12:40
Suppose you’re on a game show, and you’re given the choice of three doors. Behind one door is a car, behind the others, goats. You pick a door, say number 1, and the host, who knows what’s behind the doors, opens another door, say number 3, which has a goat. He says to you, “Do you want to pick door number 2?� Is it to your advantage to switch your choice of doors?

Mathematically the answer is always swap doors.

Read this for more details.....

http://www.open2.net/moreorless/montyhallproblem.html

But what about 'Deal Or No Deal'?

Now you have 22 boxes, and at the end presume that you are always allowed to swap the last box. I would think that there is an advantage to swapping the last box, presuming that you have the £250000 remaining what is the mathematical sollution to this problem? How much advantage do you get?


Pincho.

BatVink
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Posted: 21st Feb 2006 14:57 Edited at: 21st Feb 2006 15:00
None. The first problem revolves around the fact that you know the box/door that the host discards is empty. In Deal or No Deal, this isn't a factor.

Pincho Paxton
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Posted: 21st Feb 2006 15:02
If you play Deal Or No Deal, you have discarded 20 boxes already. You are left with 2 boxes. I think it is the same situation. Should you swap from the remaining two boxes.

BatVink
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Posted: 21st Feb 2006 16:02 Edited at: 21st Feb 2006 16:03
The whole Monty Hall theory is seriously flawed. It bases the whole calculation on past results, when probability is not affected by what happened last time.

With the 3 doors thing, it implies that first time, you had a 1 in 3 chance of getting it right. So second time, with just 2 doors, your chances are 1 in 2, so therefore you increase the chance of winning. Which is complete drivel. The 2 doors have no concept of what you did when there were 3 of them.

Take it to it's simplest form. If you throw heads on a coin 100 times in a row, the next throw still has a 50/50 chance of being heads. The coin does not have a memory and decision making powers.

Don't confuse probability with chance.

Fallout
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Posted: 21st Feb 2006 16:25
Btw, this goat, car, stick or swap thing. I made a simple VB proggy using random choices of stick or swap that proved you have a 66% chance of winning if you swap.

However, I dont think it applies to Deal Or No Deal, as the remaining box that the player doesn't own hasn't been highlighted as potential. In the way that the swap box is left by the host opening all other remaining doors except one (that's possibly the win), it'd require good old Noel to open all other boxes except the £250,000 (assuming that isn't the one the player has) without the player eliminating any boxes. Only then would he/she have a 21/22 chance of winning the £250,000 with swap and a 1/22 chance of winning the £250,000 with stick.

dark coder
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Posted: 21st Feb 2006 16:44
@batvink the 3doors thing is actually more probable to switch to the other door as its not 50/50, im sure you can write a program to do this for yourself and find that by choosing the other door you increace your chances of it being right by around 30%?

Halowed are the ori.
Scraggle
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Posted: 21st Feb 2006 16:49
I'm with batvink on this ... I can't see any reason for there to be more than 50/50 when there are only 2 options.

If anyone has code that 'proves' this to be wrong then I would love to take a look at it to see why the results are showing what they are.


dark coder
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Posted: 21st Feb 2006 17:30
not the neatest code ever



Halowed are the ori.
Peter H
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Posted: 21st Feb 2006 17:39
Quote: "I'm with batvink on this ... I can't see any reason for there to be more than 50/50 when there are only 2 options."

Because it's not just that you have two options now...

the key is that the host picks a door that has a goat behind it... if he just randomly picked a door and removed it from the selection (without showing you what was there) then the chances would still be the same...

if you want we can look ar every single possible situation... (with one setup)
door 1 = goat
door 2 = car
door 3 = goat

if you pick door 1:
host removes door 3
swap = win
stay = lose

if you pick door 2:
host removes 2 or 3
swap = lose
stay = win

if you pick door 3:
host removes 1
swap = win
stay = lose

if you swap you win 66%... if you stay you win 33%

basically if you pick a goat the first time and swap, then you win, if you pick the car the first time and swap then you lose... and there are more goats then cars... (if you stay then you only win when you pick the car)

again, they key is that the host makes a discretion between removing a car and removing a goat...

"We make the worst games in the universe..."
Fallout
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Posted: 21st Feb 2006 18:23
Exactly, the host is the factor. Here's the other confusion ...

For someone that comes into the room when the two boxes remain and doesnt know which is which, he has a 50/50 chance of choosing the right box.

For the person who owns the box and has the decision of STICK or SWAP, his odds are improved if he swaps.

You can visualise this to an extent by asking someone to pick a number from 1 to 1000 and they'll tell you if its the one they're thinking of. You choose 562. You can see you'll have a 1 in 1000 chance of getting it right. If they then say it's not 1, not 2, not 3, not 4 and so on until they've said it's not everything single number between 1 and 1000, except 97 and the one you chose. Do you still honestly think you had a 50% chance of choosing the right number, or do you think it's probably 97 - the only number they left after they eliminated all the others?

It's the same principle, but reduced to such a simple example that it's hard to visualise the odds. In the example above, you have a 999/1000 if you swap and a 1/1000 if you stick. In the original example, you have a 2/3 chance if you swap and a 1/3 chance if you stick.

BatVink
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Posted: 21st Feb 2006 18:37
It all boils down to one thing...If I stick, or if I swap, will my tea be on the table and the children be in bed when I get home?

Fallout
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Posted: 21st Feb 2006 18:41 Edited at: 21st Feb 2006 18:41
Ahh. Well, the answer to that is no.

Edit: Actually, that depends on the wife!

BatVink
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Posted: 21st Feb 2006 21:40
The 66% figure is probability, based on 3 doors. The calculations above are based on 3 doors. At the point of making the decision, there are 2 doors. The chance is 50/50.

the_winch
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Posted: 21st Feb 2006 22:04
Quote: "At the point of making the decision, there are 2 doors. The chance is 50/50."


You are wrong. The first choice has a 1 in 3 chance of being correct. The second question is asking you if your first choice was correct. The fact there are two doors is irrelevent because they are nothing to do with the question.

By way of demonstration, he emitted a batlike squeak that was indeed bothersome.
Fallout
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Posted: 21st Feb 2006 22:33
You have a 50/50 if you choose A or B. But you have a 66/33 if you choose STICK or SWAP, as STICK or SWAP contain "knowledge" defined by previous events. STICK or SWAP is not the same as A or B, as A and B are arbitrary and can be assigned to either box, but STICK and SWAP are specific and are assigned based on previous probabilities and events.

Therefore, if someone enters the room and you say A or B. They have a 50/50. If they enter the room and you say STICK or SWAP, where the STICK box and SWAP boxes are defined by the previous rules of the game, they have a 66/33.

Pincho Paxton
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Posted: 21st Feb 2006 22:43
I made a computer program that proves that you should swap, but I still think that this applies to Deal Or No Deal as well, although I am not sure how to start the programming off yet.

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