Gameshow - Easy Money - Solution
You should choose the door the game host is offering you.
Then you will get this ferrari ! No need for car hire any more! YOU are the lucky winner of the following FERRAI ! (you should get cheap insurance for the car).
If you have NOT choosen this door, you will get the following:
Pimp-My-Ride Car !! (no external link, page on this site)
Why his door ?
Well, ’cause there’s a 10% chance the door you picked first had the ferrari.
The game host then had to open 8 empty doors and had to leave one door closed. That door will probably (90% chance) hide the ferrari.
Scott Deen has sent in the following:
‘About the gameshow problem…
It is wrong. Both of the final doors have an equal probability of holding the car. If the person had picked the other door before the host opened the eight other doors, would that mean that the door he didn’t choose now had the car? Each door had a 10% chance of having the car. Take away 5 doors and each door has a 20% chance. Take away all but two doors and they each have a 50% chance of having the car.
Are you ever going to attempt correct this incorrect solution?’
Well, the solution is correct in my opinion. The chance that you pick the right door at first is 10%. The chance one of the other 9 doors is hiding the ferrari is 90%. If the host opens the doors of which he knows do not hide the ferrari, the last unopened door has still a 90% chance of containing the car.
More people have mailed me on this. Does anybody else have a better explaination ?
Finally, Mark Jahnke mailed in this explanation:
‘If, instead of opening the eight doors, the host offered to exchange the contents of the nine doors you did not choose for the one door you did choose you would be crazy not to take the nine doors. This is exactly what the host offers in Gameshow.
I realize I am preaching to the choir, but if the analogy helps you explain the solution please reply.’
I don’t think I could come up with a more elegant explanation !
I keep getting e-mail from people who think this answer is wrong. Tom Breedlove sent me the following. although he arguments for a wrong solution, I still think it is right. The argument he provides explains everything … it is just how you tell the riddle. The record was set with the proposition “the game host does know where the ferrari is“.
‘I’ve enjoyed the mind breakers, but…
On the “Gamehost” puzzle: The solution ~is~ wrong.
The explanation of Scott Deen is correct. Yes, you start with a 10% chance of picking the right door, and yes after opening 8 doors, the host knows which of the remaining doors hides the car: either the one you picked or the one he is offering as an alternative. Only two doors left so at this stage of the game you have a 50/50 shot at it. The probabilities at this point are no longer based on 10 closed doors as your numbers are suggesting. The probability of picking the correct door is 50% since there are only two in question. The other opened doors are no longer part of the equation. Your chances of picking the right door are increased from 10% to 50%.
The explanation of Mark Jahnke does not apply as the puzzle is currently worded. This explanation trades you 9 out of 10 choices (90%) for the 1 out of 10 choice (10%) the host originally offered. This is ~not~ the same as starting with a selection set of 10 doors, then revising the selection set to only two doors.
If you really need proof, have a friend sit down with 10 cards from a deck, and make sure one of the cards represents the prize. Make sure the person laying the cards face down knows where the “prize” is, then duplicate your puzzle as it is written. Do this enough times and you’ll see that your chances for picking the right card will always be 10%, not 90%.
Hope this helps clear things up.’
Okay, here is the LAST argument on this case from Jason Boomer.
‘On the “gameshow” question, you could tell scott dean this to explain his wrong answer. If you chose the wrong door the first time, then you are guaranteed to get the right door if you switch. There is a 90% chance you get the wrong door the first time, so if you switch you do have a 90% chance of winning. So you’re answer is correct.’
Guys, I suggest you go (back) to university before claiming this sollution to to be right. The eight eliminated doors no longer effect the odds of picking the correct door once they are opened, and the contestant indeed has a 50/50 chance of getting the correct door.
The solution is in fact correct, although the question is a very “nice” one. Yes the host opens 8 doors and there are 2 doors left which seems to be 50/50 probability. But don’t forget that host opens the doors after you pick your guess, which effects the probability of the left doors. The explanation from Jason Boomer is a very good one to present some intuition on this.
The percentages never change per door. The game show hosts door orignally has a 10% chance of containing the car, and your door has a 10% chance. Once the other doors are opened, the percentages dont change, except for the fact that you know it has to be on of the two doors…in essence there are only two outcomes, and they each have an equal chance of containing the prize. Their percentage does not necessarily increase to 50% and definitely one does not increase to 90%. They are both still part of the orginally 10, so they are 10% each, but since you only have two outcomes they each have an equal chance, so in essence, while the doors aren’t 50%, you would have a 50% chance of picking the right one.
As the puzzle is stated, you should absolutely trade doors. There really is a 9/10 chance that the host’s door holds the prize.
Reduce the puzzle, and you’ll see why. There are three doors, one with a car and two with nothing. When you choose one, you have a 1/3 chance of winning. There’s a 2/3 chance that the car is behind a door you didn’t choose. OPENING AN “EMPTY” DOOR DOESN’T CHANGE THOSE ODDS.
Don’t believe me? Get a piece of paper and write down numbers 1 through 100. Get a deck of cards and pull out the Ace of Spades and two Jacks. Get a friend to help by mixing the three cards face down. You pick one and slide it toward you with lifting it. Now your friend turns over one of the two remaining cards. If it’s an Ace (the door with the car), the round doesn’t count, because it doesn’t match the conditions of the puzzle. If it’s a Jack (an empty door) then this round counts. According to some of the respondents here, with your card down and a Jack up, you now have a 50/50 chance of having the Ace.
So turn over your card. If it really is an Ace, mark an X next to the number 1. If it’s a Jack, mark down an O. Repeat this until you’ve done 100 rounds where the first card your friend turns over is a Jack. If the odds really are 50/50 in these conditions, you’ll have about the same number of Xs and Os.
But guess what. You won’t. You’ll have many more Os. And if you did 1,000 or 10,000 or 100,000 rounds, you’d find you’d get closer and closer to twice as many Os as Xs. That’s because your chances of choosing the Ace are 1 out of 3. Always. Despite the upturned Jack.
Go ahead. Try it.
This logic posted above follows through perfectly… if you were asked to start with those two doors, then you would have a 50/50 chance, but becuase you are guaranteed a win for switching if you chose wrong originally, and you had a 90% chance of choosing wrong, if you switch, there is a 90% chance you will win.
“If you chose the wrong door the first time, then you are guaranteed to get the right door if you switch. There is a 90% chance you get the wrong door the first time, so if you switch you do have a 90% chance of winning. So you’re answer is correct.”