Intersting probability situation - NOW with script

[coder1]

I have been learning probability and statistcs lately and was asked a simple question/riddle/puzzle or whatever you want to call it.

You are on a game show ("Lets Make a Deal") and are shown 3 doors in front of you. The host tells you that behind the one door is a check for 5 million dollars. The other 2 doors have nothing but a bag of coal behind each of them. You start of by choosing door number 2. Now the host opens one of the doors that he knows has a bag of coals behind it. He ask you if you want to switch to a new door. You can either stay with the original door you chose or choose the other unopened door. What is the best way to go?

I'm sure many of you who have taken math have seen this example, and I know it's all over the Internet, but my original thought was that it did not really matter if you switched or not, at that point it was 50/50. But probability shows that if you switch you have a 2/3 chance of winning the prize.

Here is a script that test this theory 200,00 times

LINK


another view

The 2 out of 3 explained

Case 1:
(Odds 1:3)
You pick the correct door.
1 Empty door is revealed.
1 Empty door is left

Case 2
(Odds 2:3)
You pick an empty door
1 Empty door is revealed
1 Correct door is left

 

[Syringer]

Yup, really interesting stuff.

Although I think the last time this was posted it went on to about 300 posts or so..

 

[Yax]

interesting. But common sense would say to stick with door number 2. Opening one of the other doors means that door is now a given and should not affect the outcome anymore. Now the new senerio should be:
1 bag of charcoal and 5mil, behind 2 doors.

With the new setup, I'd say its now 50/50. How can you get 2/3 out of that?

Its like looking at it this way:
What is the probability you will get heads or tails on a fair coin if you flip it? Heads you win 5 mil and tails you lose. Given that you just saw tails, doesn't do you any good since the results are independent from previous results. Looks like 50/50 to me.

 

[kranky]

That's one of the best probability problems I've ever seen. The reason it draws such discussion is that the common sense answer is wrong. When I first ran across this problem it took me a long time to understand why it's better to switch.

 

[mugs]

Link

 

[gopunk]

Originally posted by: cheapbidder01
interesting. But common sense would say to stick with door number 2. Opening one of the other doors means that door is now a given and should not affect the outcome anymore. Now the new senerio should be:
1 bag of charcoal and 5mil, behind 2 doors.

With the new setup, I'd say its now 50/50. How can you get 2/3 out of that?

Its like looking at it this way:
What is the probability you will get heads or tails on a fair coin if you flip it? Heads you win 5 mil and tails you lose. Given that you just saw tails, doesn't do you any good since the results are independent from previous results. Looks like 50/50 to me.

when you pick your door, it has 1/3 chance of being right. the other 2 have combined 2/3. when he eliminates one of the other 2, the remaining door you did not choose now has 2/3.

 

[coder1]

I wrote a quick VB program here at work that ran this simulation 150,000 times through a loop, and sure enough, if you switch doors you roughly have 66% of getting that prize. I guess the best way to think about it (for me that is) is that in the beginning you had 2/3 chance of getting the wrong door. When the host opens the door and shows you one of the wrong choices it does not change that 2/3 probability, so it's mathematically wise to change. This whole probability stuff is very interesting.

 

[coder1]

cheapbidder01, thats exactly what I thought at first, but remember that your original choice was based on a three door possibility. If someone came in your spot after the host opened the door and did not know what door you originally chose than yes it's 50/50 at that time.

 

[Yax]

Originally posted by: coder1
cheapbidder01, thats exactly what I thought at first, but remember that your original choice was based on a three door possibility. If someone came in your spot after the host opened the door and did not know what door you originally chose than yes it's 50/50 at that time.

Yeah, I can see the argument now, 1/3 vs 2/3.

Then there's the human factor. What if the host was showing you the wrong door because he knew you picked the right door and he wanted to sway your judgement? Well, that's not math, but what if. Oh well, that's not an argument, just some psychology thrown in.

Edit: No, I can't convince myself of the 1/3 vs 2/3 argument. I'm thinking, you have 1/3 chance of being right and 2/3 chance of being wrong. Now just because they exposed 1 of the 2/3 chances of being wrong doesn't automatically change it so that if you switch doors you'd get a 2/3 chance of being right. I still think they're independent trials and it will become 50/50.

 

[coder1]

LOL, thats true. I guess you would want to know if he showed the door for the last 5 people and if they were correct when they switched.

I just got done asking this question to a coworker and he swore that there was no way it made difference if the person switched doors. I then showed him the results of my VB program (this time I had the loop run 5,000,000 times) and he was in shock of the results.

 

[Yax]

Here's another senerio to think about using that 1/3 vs 2/3 argument:

You pick door 2, now they bring someone else in and he picks door 1. The host opens door 3 which has nothing. Now, if you two decide to switch doors, you'd each have 2/3 chance of getting the right door which makes it 4/3? That doesn't make any sense at all.

 

[RayH]

Bayes and Bayes Theorem

 

[kranky]

Originally posted by: cheapbidder01
What if the host was showing you the wrong door because he knew you picked the right door and he wanted to sway your judgement?

That's exactly what the host does. You pick one door. Of the other two doors, you know without a doubt that one of them is wrong (after all, there's only one prize). The host will always show you a wrong door, whether you chose right or wrong initially.

 

[Mallow]

I think the two trials are independent. I can logically work out how switching would increase your chances even after reading everything in this thread. Guess I'm a moron

 

[Syringer]

Originally posted by: cheapbidder01

Originally posted by: coder1
cheapbidder01, thats exactly what I thought at first, but remember that your original choice was based on a three door possibility. If someone came in your spot after the host opened the door and did not know what door you originally chose than yes it's 50/50 at that time.

Yeah, I can see the argument now, 1/3 vs 2/3.

Then there's the human factor. What if the host was showing you the wrong door because he knew you picked the right door and he wanted to sway your judgement? Well, that's not math, but what if. Oh well, that's not an argument, just some psychology thrown in.

Edit: No, I can't convince myself of the 1/3 vs 2/3 argument. I'm thinking, you have 1/3 chance of being right and 2/3 chance of being wrong. Now just because they exposed 1 of the 2/3 chances of being wrong doesn't automatically change it so that if you switch doors you'd get a 2/3 chance of being right. I still think they're independent trials and it will become 50/50.

The host will always pick a door that doesn't have anything behind it. You're not being GUARENTEED that you'll win, but only that your chances will be increased from 1/3 to 2/3.

 

[TallBill]

Originally posted by: kranky

Originally posted by: cheapbidder01
What if the host was showing you the wrong door because he knew you picked the right door and he wanted to sway your judgement?

That's exactly what the host does. You pick one door. Of the other two doors, you know without a doubt that one of them is wrong (after all, there's only one prize). The host will always show you a wrong door, whether you chose right or wrong initially.

yeah., thats the key to it.. hes not gonna randomly open the right door

 

[kranky]

Another view: we all agree that the initial selection is a 1/3 chance of being right. Would you trade your one door for the other two? Of course! That's the equivalent of switching. It's just that before you choose to switch, the host shows you that one of the other doors is empty, which isn't exactly news. You already knew ONE of the other two would be empty.

 

[Yax]

I guess the 1/3 to 2/3 argument assumes that if the wrong door of the other 2 doors were opened, then the probability of that door being the one alutomatically transfers to the other unpicked door. That's not correct. It should be distributed between the two remaining doors. It becomes 50/50.

 

[Toasthead]

But by eliminating one door you now have a 50% chance of being right by doing nothing. This is flawed statistics.

 

[Looney]

Classic probability problem. One of my best classes were statistics and probabilities, even though they were one of my hardest.

 

[Yax]

Originally posted by: mugsywwiii
Link

The argument from the link is flawed. After one door is exposed, the probability can be recalculated which brings it to 1/2 for each of the remaining doors. 2/3 for one door vs 1/3 for the other door is not correct.

 

[Looney]

Originally posted by: Toasthead
But by eliminating one door you now have a 50% chance of being right by doing nothing. This is flawed statistics.

No you don't... your chance of being right is still the same as your original action since it was based on 3 choices.

Take an extreme example. There are 100 doors, and you pick one... then 98 of them are opened, leaving you with 2 doors left. Do you stay, or do you switch? If you stay, your chances are 1/100.. but if you switch, your chances switches to 98/100.

 

[Looney]

Then there's the human factor. What if the host was showing you the wrong door because he knew you picked the right door and he wanted to sway your judgement? Well, that's not math, but what if. Oh well, that's not an argument, just some psychology thrown in.

He ALWAYS opens the wrong door... that's the point to the game.

 

[AntaresVI]

Originally posted by: Moralpanic

Originally posted by: Toasthead
But by eliminating one door you now have a 50% chance of being right by doing nothing. This is flawed statistics.

No you don't... your chance of being right is still the same as your original action since it was based on 3 choices.

Take an extreme example. There are 100 doors, and you pick one... then 98 of them are opened, leaving you with 2 doors left. Do you stay, or do you switch? If you stay, your chances are 1/100.. but if you switch, your chances switches to 98/100.

That's not true, though. Because the total number of doors was reduced to two, it does not matter that you had an initial guess in. Essentially, if you stay, you're "choosing" that door. If you move, you're choosing the other. There is a 50/50 chance that you are right either way.

 

[FuzzyBee]

Originally posted by: LeRocks

Originally posted by: Moralpanic

Originally posted by: Toasthead
But by eliminating one door you now have a 50% chance of being right by doing nothing. This is flawed statistics.

No you don't... your chance of being right is still the same as your original action since it was based on 3 choices.

Take an extreme example. There are 100 doors, and you pick one... then 98 of them are opened, leaving you with 2 doors left. Do you stay, or do you switch? If you stay, your chances are 1/100.. but if you switch, your chances switches to 98/100.

That's not true, though. Because the total number of doors was reduced to two, it does not matter that you had an initial guess in. Essentially, if you stay, you're "choosing" that door. If you move, you're choosing the other. There is a 50/50 chance that you are right either way.

Considering it's a "redeal" of opportunities at this point, and you aren't dumb enough to pick a door that has already been opened, you will have a 50/50 chance no matter which door you choose. Any other answer is bunk. None of the prior decisions affect the fact that there are 2 doors, and you have to choose 1 of the two.