Brain teaser for the week 20200815 每周烧次脑
▲Click above to subscribe 点击上方蓝字关注我们▲
Difficulty level: ✮ ✮ ✮ ✮ ✮
What is the value of the sum of the first 99 consecutive integers? (In other words, what is 1 + 2 + 3 + 4 +…+ 99?)
Welcome to try your answer
at the Comment Section!
Answer to be disclosed at our
next episode of brain teaser
答案下期公布
Answer to the last week 上期答案 Brain teaser for the week 20200808 每周烧次脑
He or she should open the remaining door.
The probability would be 2/3, not 1/2! Many people have thought that there is a 1/2 chance of having the car behind the door the contestant chose since there are two doors. However, because you have chosen a door first, out of three, there is a 1/3 probability there is a car behind it, and there would be a 2/3 probability any other door has the car behind it. So if you chose two of the other doors, you would have a 2/3 probability of the car. And if one door was eliminated by the host, since that door had the goat, if you switched doors, then it would be ineffect as if you were choosing two doors, which would give you the 2/3 probability that the car was behind the remaining door.
Anothe rexplanation:
Interestingly enough, the answer is that the contestant should switch doors, because there is actually a 2/3 chance of winning the car by switching, while there is only a 1/3 chance of winning the car if the contestant opens his or her original door!
This is how most people’s minds approach the problem: The host eliminates one door with the goat, so there is a goat or a car behind the contestant’s door. So the probability is 1/2. And it doesn’t matter whether the contestant opens his orher original door or changes doors. But let’s see what is really happening. The probability of having a car behind the contestant’s door when he or she originally chooses the door is 1/3, since there are three doors and only one door that has a car behind it. And there is a 2/3 probability that the car is behind one of the other two doors (since there are two ways a car can be behind one of the remaining doors [car-goat; goat-car] out of a total of three possibilities: car-goat; goat-car; goat-goat). Now, if the host eliminates a door, there would still be the 1/3 probability that the car was behind the contestant’s original door - so once a door is eliminated, there is a 2/3 probability the car was behind the remaining door.
I have pondered the Let's Make a Deal problem for some time, and I came to the conclusion that it is 2/3 when I took a more dramatic variation of the problem. Suppose I had 100 doors and 1 of them had a car behind it. The rest had goats. If I chose one door, there would no doubt be a 1/100 chance that there was a car behind it. So the remaining doors combined would have a 99/100 chance of the car. Now no matter what was done to the other doors, there would still be a 1/100 chance of the car’s being behind the door that was chosen. So if the host knew that 98 of the doors did not have the car, and then eliminated those doors, the remaining door then must have had a 99/100 chance that there was a car behind it.
Supposeyou had 100 doors. The chance of the car’s being behind the contestant’s chosen door is 1/100, obviously a very slim chance that won’t change even when the host eliminates 98 doors. When the doors are eliminated, and all but two are left, the contestant still has chosen a door with a very, very slim chance the car is behind it, namely 1/100. Because the host knows where the car is (and where the goats are), and because 98 of the goats were eliminated, the remaining door has a (98+1)/100 chance there is a car behind it.
Note that initial reasoning may indicate that when there is only one door left, since 98 doors have been eliminated, the total number of possibilities for what’s behind the contestant’s door is two, a car or a goat. The favorable number of possibilities is one, the car. So by strict definition of probability, the probability is 1/2 no matter what the host knows or did. But according to the previous paragraph, that is not the case.
I think probability has a somewhat ambiguous definition, especially if there is “conditional” probability, but it seems that if you tried the 100 doors out you would in fact find you’d get a 1/100 chance of getting the car.
Contributions will be appreciated！
欢迎投稿！
▲ QQ group
▲ Official account