What's your financial IQ?

[biaggio]

"Suppose instead there's 100 doors, 1 car and 99 goats.  Suppose you choose door 1.  Then the host opens doors 2, 3, 4 ... 56, 58, 59, ... 100.  So there's exactly 2 doors closed, door 1, which you picked, and door 57, which the host decided not to open for some strange reason.  Would you still claim that there's no benefit of switching from door 1 to door 57, that it's still 50/50?"

 

What would prevent the car from being behind door #1?

 

Say, you were the host, and  door #1 actually had the car, as the host you would still have to randomly choose one of the other doors even knowing that there would be a goat behind it (both doors would have a goat)

 

Before the host opens the doors you would have 1 in 100 odd of choosing the prize. After the door openings you would have a 50-50 odd of choosing the door with the car (ie door #1 + 57 would both have an equal chance of having the car..... ie  no benefit)

[RichardGibbons]

In that case, Biaggio, I'll propose a great deal for you then.  We'll play the game with a random number generator and 100 doors, 99 goats and 1 car.  The computer can hide the goat, and you can pick a door.  Then we'll have the computer open 98 of the other doors that contain goats.  At that point, you stick with your original pick, and I'll take the unopened door.

 

If I have the car, then you pay me $10.  If you have the car, I'll pay you $20.  If it's 50-50, then this should be a pretty lucrative game for you.

 

I propose that we both commit to playing this game 100 times, to attempt remove the effects of bad luck.  If the odds are 50-50, your expected return is $500.  If the stakes aren't big enough to make it worth your time, I'll commit to playing this game with up to 5,000 times, with you expecting to gain $25,000.  (My wife wouldn't be happy with me gambling more than $100,000.)

 

Just let me know.  Maybe Sanjeev would be willing to act as a intermediary to hold the cash for 5% of the winner's profits.  It could help pay for the website.  :)

 

Richard

[biaggio]

Sorry Richard I am not that confident.

 

With $$$ on the line, I thought it over it again...would the following make sense: The door I choose would have 1% chance of having the car...wheras the door the computer chooses after the other 98 door are opened  has a 50% of having the car?

[RichardGibbons]

The host will always open at least 98 of the doors that don't have goats.  Therefore, the probability of the car begin behind the other door is 1 - P(it's behind the door you pick).  In other words, when you start, there's a 1/100 chance that it's behind your door, so there's 99/100 chance that it's behind one of the other doors. 

 

After the other doors are opened (and the host will always open 98 other doors with goats), there's still a 99% chance that it's behind one of the other doors.  The only other door left is door 57, so there's a 99% chance that it's behind that door.

 

Richard

 

 

[biaggio]

makes sense

 

Thanks

[EdWatchesBoxing]

The 3 door problem can be verified by working it out yourself. Put the car behind the first door, the goats behind doors 2 and 3, and figure out all the scenarios. Repeat for the car being behind doors 2 and 3 and you'll find that switching improves your chances.

 

This problem came up in the blackjack counting movie, 21, so I already researched this.

[rkbabang]

The host will always open at least 98 of the doors that don't have goats.  Therefore, the probability of the car begin behind the other door is 1 - P(it's behind the door you pick).  In other words, when you start, there's a 1/100 chance that it's behind your door, so there's 99/100 chance that it's behind one of the other doors. 

 

After the other doors are opened (and the host will always open 98 other doors with goats), there's still a 99% chance that it's behind one of the other doors.  The only other door left is door 57, so there's a 99% chance that it's behind that door.

 

Richard

 

That is an excellent way to explain it and can also be used to explain the problem with 3 doors.

 

To rephrase what you said and apply it to the 3 door problem:

 

  The host will always open the door that doesn't have goats.  Therefore, the probability of the car begin behind the other door is 1 - P(it's behind the door you pick).  In other words, when you start, there's a 1/3 chance (33.33%) that it's behind your door, so there's 2/3 chance (66.66%) that it's behind one of the other doors. 

 

After the other doors are opened (and the host will always open the other door with goats), there's still a 66.66% chance that it's behind one of the other doors.  The only other door left is door 3, so there's a 66.66% chance that it's behind that door.

 

So, if you stay with your 1st pick you still have only a 33.33% chance of getting the car, whereas if you switch you have a 66.66% chance of getting the car.

 

--Eric