This is the first time I've seen a proof that makes sense to me. All other times I've "heard" people state this solution, it always felt to me that "the mathematician" was "simply" saying something. . .without any actual proof. Even Bon on this same problem in another version she provides, it seemed like she "simply" said something that made her proof, not actually prooving her solution. Thank you, Bon, for this visual. Now it makes sense to me.
Here's a little history of the problem. And I _am_ trying to be brief. :) Most is in order, but I'll come back to the very beginning before I close. In the May, 1959 issue of Scientific American, Martin Gardner asked: "Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?" He answered "1/3" because there are four equally likely combinations. Ordered by age, they are: BB, BG, GB, and GG. Three have at least one boy, and one of those has two. So the chances seem to be one in three. But the next October, he retracted that answer. The event to condition on is not "there is at least one boy," but "what we learned is that there is at least one boy." Since you can learn that there is a girl in a BG or GB family, we really need to know how we learned it. The answer could be 1/2 (if it was random), or 1/3 (if we asked him "do you have a boy?"). (Spoiler: 1/2 corresponds to the the correct answer in the Monty Hall Problem, which has a different number of cases.) To illustrate, Gardner proposed the Three Prisoners Problem, which is mathematically identical to Monty Hall. To keep this brief, I'll translate the specifics he added, to how they apply to this problem: "If I pick the Steamroller, open the door with the Pink Car. If I pick the Pick Car, open the door with the Steamroller. But if I pick the Delorean, flip a coin to decide." Gardner provided the correct answer, that the chances of picking the Delorean remain 1/3. Because of how he explained it, there was no huge controversy. But Monty Hall never played a game like this on Let's Make a Deal (it's actually against regulatory mandates to alter the game based on what the contestant picks). What he would do, was _regardless_ of what the contestant picked, offer to buy the unknown prize back for cash. If the contestant declined, he might increase the offer or even open a losing door. In a 1975 letter to a statistics journal, a man name Steve Selvin suggested that after the reveal, a smart contestant should suggest the switch. This removes much of the ambiguity in the problem, that was introduced by Marilyn vos Savant in the 1990s. She didn't make the specifics explicit, and she never actually solved it. Which is why it remains controversial. But this all goes back to a man named Joseph Bertrand. In 1889 (I think), he proposed a problem that is now called Bertrand's Box Paradox. He was trying to warn people about why the seemingly-simple approach to such problems is wrong, and the paradox he intended was the proof: Instead of opening a door, suppose Monty Hall points to one and says "This door doesn't have the Delorean." It is tempting to say that, if he would then open it to reveal the Steamroller, the chances you picked the Delorean change from 1/3 to 1/2. But if that is true, opening it to reveal the Pink Car makes the same change. Since it has to be one or the other, and this change happens regardless of which it is, you don't need him to open it. The fact that he pointed is what makes the change. But it can't. This paradox disproves 1/2.
Your answer is correct if the cell with the prize is randomly chosen. In other words, if the prior probability is 1/3, this is the correct answer. But this is not necessarily true. If the prior is not 1/3, the correct answer can be different. If, for example, the car is in the leftmost cell more than half the time, your best bet is to pick the leftmost door and stick to it, regardless of what Monty does. h. This is the Bayesian analysis that someone asked for. It depends on the prior. Aviel Shapira Beer Sheva, Israel
the main problem that i can see with this paradox is that is all has to do with luck of the draw of the first selection i can understand the theory but in reality if you pick the car at first instead of the goat then you have no chance of getting it if you switch its only when you would pick the goat that your chances can be doubled henceforth if it still boils down to luck of the draw at the first selection in hopes of actually picking the goat as your first selection. nice video btw
Yes, its luck of the draw if its 50/50 that you will pick the car or a goat, but its highly probable (2 to 1) that you will pick a goat. So, with the odds in your favour that you have picked a goat, you can then swap to the car.
This is the first time I've seen a proof that makes sense to me. All other times I've "heard" people state this solution, it always felt to me that "the mathematician" was "simply" saying something. . .without any actual proof. Even Bon on this same problem in another version she provides, it seemed like she "simply" said something that made her proof, not actually prooving her solution.
Thank you, Bon, for this visual. Now it makes sense to me.
The best explanation I found on internet
Here's a little history of the problem. And I _am_ trying to be brief. :) Most is in order, but I'll come back to the very beginning before I close.
In the May, 1959 issue of Scientific American, Martin Gardner asked: "Mr. Smith has two children. At least one of them is a boy. What is the probability that both children are boys?" He answered "1/3" because there are four equally likely combinations. Ordered by age, they are: BB, BG, GB, and GG. Three have at least one boy, and one of those has two. So the chances seem to be one in three.
But the next October, he retracted that answer. The event to condition on is not "there is at least one boy," but "what we learned is that there is at least one boy." Since you can learn that there is a girl in a BG or GB family, we really need to know how we learned it. The answer could be 1/2 (if it was random), or 1/3 (if we asked him "do you have a boy?"). (Spoiler: 1/2 corresponds to the the correct answer in the Monty Hall Problem, which has a different number of cases.)
To illustrate, Gardner proposed the Three Prisoners Problem, which is mathematically identical to Monty Hall. To keep this brief, I'll translate the specifics he added, to how they apply to this problem: "If I pick the Steamroller, open the door with the Pink Car. If I pick the Pick Car, open the door with the Steamroller. But if I pick the Delorean, flip a coin to decide." Gardner provided the correct answer, that the chances of picking the Delorean remain 1/3. Because of how he explained it, there was no huge controversy.
But Monty Hall never played a game like this on Let's Make a Deal (it's actually against regulatory mandates to alter the game based on what the contestant picks). What he would do, was _regardless_ of what the contestant picked, offer to buy the unknown prize back for cash. If the contestant declined, he might increase the offer or even open a losing door. In a 1975 letter to a statistics journal, a man name Steve Selvin suggested that after the reveal, a smart contestant should suggest the switch. This removes much of the ambiguity in the problem, that was introduced by Marilyn vos Savant in the 1990s. She didn't make the specifics explicit, and she never actually solved it. Which is why it remains controversial.
But this all goes back to a man named Joseph Bertrand. In 1889 (I think), he proposed a problem that is now called Bertrand's Box Paradox. He was trying to warn people about why the seemingly-simple approach to such problems is wrong, and the paradox he intended was the proof:
Instead of opening a door, suppose Monty Hall points to one and says "This door doesn't have the Delorean." It is tempting to say that, if he would then open it to reveal the Steamroller, the chances you picked the Delorean change from 1/3 to 1/2. But if that is true, opening it to reveal the Pink Car makes the same change. Since it has to be one or the other, and this change happens regardless of which it is, you don't need him to open it. The fact that he pointed is what makes the change.
But it can't. This paradox disproves 1/2.
I would like you to solve it using Bayes theorem
Your answer is correct if the cell with the prize is randomly chosen. In other words, if the prior probability is 1/3, this is the correct answer. But this is not necessarily true. If the prior is not 1/3, the correct answer can be different. If, for example, the car is in the leftmost cell more than half the time, your best bet is to pick the leftmost door and stick to it, regardless of what Monty does. h. This is the Bayesian analysis that someone asked for. It depends on the prior.
Aviel Shapira
Beer Sheva, Israel
the main problem that i can see with this paradox is that is all has to do with luck of the draw of the first selection i can understand the theory but in reality if you pick the car at first instead of the goat then you have no chance of getting it if you switch its only when you would pick the goat that your chances can be doubled henceforth if it still boils down to luck of the draw at the first selection in hopes of actually picking the goat as your first selection. nice video btw
Yes, its luck of the draw if its 50/50 that you will pick the car or a goat, but its highly probable (2 to 1) that you will pick a goat. So, with the odds in your favour that you have picked a goat, you can then swap to the car.
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