In the beginning of the movie the young math genius is in Kevin Spacy’s class. He’s figuratively offered to choose one of three doors, to try and find the car vs the goats. He chooses door #1.
Then Spacy as the game show host reveals that behind door three is a goat. Now he asks if he wants to change his mind and take door #2 or keep door #1. Genious takes two and when Kevin challanges his decision he says his choice is based purely on statistics rather than emotion.
When he first chose his chances of being right were 33 and 1/3%. He says that after knowing that door three has a goat, his chances increase to 66 and 2/3% by switching to door two.
My question, At first glance I might guess that his chances were now 50% since the choice is narrowed to two doors. Why are his chances now statistically 66 2/3 % by taking door two. Why wouldn’t his chances at door #1 be just as good as his chances at door #2 statistaicaly speaking? Isn’t staying with door 1 the same as choosing it with only two doors?
Because after you choose one door, the host gives you a chance to have one of the other two after he reveals a goat behind one of the doors. It’s essentially the same as if the host asked you, “okay, would you like to keep your door, or choose both of the remaining doors?”. The only difference is that he removes a goat from the two remaining doors first.
Repicking is 2/3 but only if the host always reveals a goat from a non-picked door. It’s like picking two doors and winning if the car is behind one of them.
g gC - Repick wins
g Cg - Repick wins
C gg - Repick loses
At the start you are offered the choice of three doors. There are two chances out of three that your choice will have a goat behind it.
The host then opens one door, that has a goat behind it. This does not affect the choice you have already made. There is STILL two chances out of three that you have a door with a goat behind it. Therefore, if you change doors, you improve your odds of getting the car.
okay that makes sense. So , you still might lose but at least statistically your chances are higher if you switch.
It’s hard to get my brain past the 50/50 two door idea. Also it seems a little like when you’re offered the 2nd choice it’s a new start. Now you have two doors and you have to pick one. Kevin did say in the movie that from the statistics mattered in a non emotional way.
One of the first things I learned when studying probabilities is that history is completely irrelevant. You might have gotten heads 15 times in a row and the chances you’ll get heads on the next coin toss is still 50-50.
In this problem, you are simply faced with two choices and it is completely irrelevant whether you chose door 1 or 2 in the past. Suppose the genius chose door 2 the first time around. Door 3 still has the goat. Does door 1 now assume a higher probability for the car?
That’s not the question. The question is will you get the car more often by switching or by not switching, and the answer is by switching (as has been amply demonstrated above).
The OP’s intuition was that switching would improve his probability of winning the car (although it was by an incorrect amount). Do you understand that the answer to the Monty Hall problem is to switch when given the choice and the probability of winning a car goes from 1/3 to 2/3?
In SmartAlecCat’s scenario, you pick one door out of 99. Your probability of winning is 1/99. You almost certainly don’t have the car. Assuming you don’t, the host knows which door has the car behind it so he won’t reveal the car behind that door, but reveals goats behind all of the remaining doors except for door #88. Since you know the probability that you first picked a door with a goat behind it is 1/99, the probability that door #88 has a car behind it is 98/99.
Just to make clear to myself. If I’m on Deal or no Deal and I’m down to two cases with one of them holding a million dollars, I can expect that switching cases is not going to increase my chances.
On Deal or No Deal all of the boxes are chosen randomly, so switching doesn’t matter. The probability of winning a higher amount doesn’t change if you switch. But, if Howie Mandel were the one to eliminate boxes and he didn’t do it randomly, he eliminated lower amount boxes first- definitely switch.
Also, in the Monty Hall problem, if Monty randomly showed you what is behind one of the two remaining doors and it just happened to be a goat and then offered you the chance to switch, it doesn’t matter if you do or not- your probability of winning is 1/2 either way.
It occured to me last night that if it’s presented as by switching your door to #2 it’s the same as being given two doors out of three, then why isn’t keeping door 1 pretty much the same thing?
The now open door is the 2nd door you’re given and you know it’s wrong. Whether you switch or not, that remains true. Don’t you now have a 662/3% chance whether you switch or not, because the open door is the other 331/3% for either of the closed doors. Then there’s the issue of considering the 1st choice. Why is the forst choice significant to the 2nd. If the 2nd choice stands by itself it’s 50/50.
Although I didn’t quite get the 99 goat example I think Cecil’s playing card example helped me out
I suppose I’d have to simulate it with three playing cards for it to really sink in. Interesting.