 Okay, this video will be about the Monty Hall problem, which was a somewhat notorious problem in probability theory that was popular about 30 years ago. So the name comes from a game show called Let's Make a Deal run by the host Monty Hall. And during this game show, what would happen is that there would be three doors and behind one of the doors, there was a car and behind the other two doors, there is a goat. Except you don't know which door has a car and which door has a goat. And what happens is the game show contestant picks one of the doors, say the first one. What happens is the game show host may then open one of the other doors and show that there is a goat behind it. And the question is, if you pick this door, do you then switch to the second door or does it make no difference? And there are two arguments. The first argument says it makes no difference that these two doors are equivalent. So there's a 50, there's a one and two chance the goat is behind first door and a one and two chance that is behind the second door. The second argument says that opening the door that does not change the probability that the first door is correct. So the probability that goat is behind the first door remains at one third and the probability that the goat is behind the second door is now two thirds so you should switch. And people get extremely excited about which of these two is correct. Most people think answer one is correct because these two doors are equivalent. In fact, given some assumptions that I will state later the second argument is actually correct. So I want to, I will explain why. Before explaining why I would give a bit of history of the problem. The problem was actually first introduced by Martin Gardner in his Scientific American Mathematical Games column in 1959. He republished it in his book, the Second Scientific American Book of Mathematical Puzzles and Diversions. And you can see his original version of the problem on page 226 if I zoom in so that you can actually read it. So he says here there's this wonderfully confusing little problem. He's not stated in terms of goats and cars. He's stated in terms of three prisoners and the sentence of death. And one of them was reprieved. Only they didn't know which and you had to had to calculate the what is the chance of being reprieved under some conditions. Incidentally, Martin Gardner probably didn't invent this problem. As you see, he says this little problem is now making the rounds, which suggests the problem was kind of circulating among people at that time. And it's not really clear who first invented it. Martin Gardner gave a very brief correct solution. You can see it here. He says, regardless of who is pardoned, that means regardless of where the car is in the Monty Hall version of the problem, the warden can give A the name of a man who will die. Well, that means Monty Hall can open a door that has a goat behind it. And Martin Gardner then points out that opening the door has no influence on the door you chose having a goat behind it. So Martin Gardner gave the problem and gave a correct solution back in 1959. The problem became really popular when it was appeared in a newspaper column by Marilyn Vosavent, who again gave the correct answer. And what is rather interesting about this is, let me zoom back out again. Marilyn Vosavent received enormous numbers of letters from professional mathematicians, basically most of which said something along the lines of I'm a professional mathematician with the brain the size of the planet and your solution is completely wrong. And most of the mathematicians who wrote to her were in fact wrong. So this is a really tricky problem. Even professional mathematicians get the answer wrong. Incidentally, the problem, as I stated, is a little bit ambiguous. So that I should really add two assumptions. The first assumption you need is that a car is more valuable than a goat. There is a school of thought that says that actually goats are worth more than cars. So when the host opens the door with a goat behind it, you should take the goat and well. So we're going to assume that you actually want to get a car rather than a goat. The second assumption is that the host always opens a goat door. In the actual game show run by Monty Hall, Monty Hall did not always open a goat door. Sometimes if the game show contestant selected a goat, he would immediately open the door that the game show contestant had selected. So the game show contestant would immediately lose. Now, if the host has the strategy that he only opens a door to show you a goat if you already picked the car, then of course you should never switch. So if the host is trying to trick you, then this modifies the problem quite a lot. And as I said, in the actual game show host, Monty Hall did sometimes try to trick the person. So we're going to assume a host is not malicious and always opens a door with a goat behind it. Now, the problem with probability is that it is really tricky. So I've given you Martin Gardner's solution which is in fact correct. If you're feeling rather doubtful about this then you're rather good instincts because you should always distrust all probability arguments unless you have had considerable training and probability and even if you have, you should be extremely cautious. So I'll now give you a few more arguments why it is correct to switch. One thing you can do is you can play this game 600 times and see what happens. So let's suppose we're going to play this game 600 times and let's say you always pick the first door. It doesn't really matter which door you always pick. So 200 times there will be a car behind this door and goats behind the other two. And so we're going to pick this door and what's the host going to do? Well, he's going to pick one of these other two doors. So they're going to be 200 times this happens. So they're going to be 100 times when the host opens this door and 100 times when he opens this door so this will happen a hundred times and this will happen a hundred times and then there will be 200 times when you pick a goat and the car is behind the second door and there is another goat behind the third door. And in this case, the host will always open this door. So we have 200 times when the host opens that door and you pick this door and finally there are 200 times when you have a goat and the car is behind the third door and there is another goat behind the second door and 200 times the host will open this door. So we're playing this game 600 times, 100 times 600 times, 100 times that will happen, 100 times that will happen, 200 times that and 200 times that. So let's see what happens if you don't switch. Well, if you don't switch, you're going to win when you're going to win in these two cases and you're going to lose in these cases. So here we lose 200 times, we lose 200 times and we win 100 times there and 100 times there. If you switch, then we lose 100 times here, we lose 100 times here, but you can see if we switch we would be getting this door. So we win 200 times here and we win 200 times here. So you can now add a stop if we switch, we win 400 times and if we don't switch, we win 200 times and lose 400 times and here we lose 200 times. So you can see switching is clearly better. If you switch, you have a two thirds probability of winning and if you don't switch, you have a one third probability of winning. There's a psychological problem going on here and if you switch, there's a danger that you might actually have picked the car and then lose it by switching and people are very, very, very reluctant to do things that might end up losing something they already possess. So there's a sort of big psychological incentive not to switch because people are afraid that they will lose the car and end up somehow losing something counts psychologically much more than the chance of gaining something for some reason. So that's a computational method of calculating what the best answer is. What you can also do, you can just get a computer to simulate this if you distrust this calculation. The only method that I've found that ever actually convinces people is to offer to gamble with them with this problem. So you can play this game with people a lot of times. So you will choose where the car goes and open a goat door and I will try and guess where the car is and if I get the car, I win $100 and if I get a goat, I pay you $110. So if you believe it doesn't make any difference whether I switch or not, you should be willing to play this game with me and I will be very, very happy to play this game with you. And after playing it a few times, you will notice that you're losing lots of money. So if you find someone who doesn't believe this solution just offer to play them for money and you will make a lot of money before they catch on. There's a final argument that it is best to switch. Suppose you're unsure whether this argument is correct or not. Well, if the argument is correct, it's best to switch. Suppose the argument is not correct. Then it makes no difference. And suppose you're unsure which of these are whether the argument is correct or not. Well, then you may as well switch because if the argument is correct, then you gains and if the argument is not correct, then you've lost nothing. So if you're unsure whether the argument is correct or not, you should still switch. Assuming of course that the host is not malicious.