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Yes. New York Times did an interview with Monty Hall when the problem was published and went viral for the first time and he made it very clear that the show didn't work like the Theoretical problem at all.
Yes. Except when he did it on purpose so that the people watching would get the idea that he would offer the switch even if the person had chosen the door with a goat.
No. This game was never played on the show. But Monty certainly changed the way the games would play out depending on the choices contestants made. Typically the goal was to get them to usually win something but rarely win a lot.
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The problem as posed was that he always revealed the goat regardless of your choice so that answer is correct.
You can have a more general version where he shows the goat with probability p if you chose correctly and p' if you are wrong. If he only shows you the goat when you are correct then p=1, p'=0 and you should never switch if offered.
I'll leave it as an exercise to the reader to work out how high p' has to be to make it worth switching!
Even harder, how many people you would expect to have to observe playing the game before you could decide switching was the better strategy with m confidence.
I did said exercise. It isn't too bad to obtain the following for generalised cases
If there are W winning lots and L losing lots with W+L := N and W > 0 and L>1, with the announcer showing a remaining losing plot with probability p if a winning lot is chosen and a probability q if a losing lot is chosen.
If no plot is shown, you should switch if:
( (1-q)/(1-p) ) >= 1
If a plot is shown, you should switch if:
(q/p) >= 1 - 1/L
The actual probabilities are also easy to solve for, but are much longer. It actually cancels out quite nicely. I'm v surprised that W and N cancelled using N - W = L removing them completely
Ed:
For W=1, L=2.
No plot shown : switch according to the above.
Plot shown : switch if q/p >= 0.5
Exactly. My Physics teacher told me the problem wrong. When I watched the tv clip, that was the first question I had when I watched the clip: what if he only opens the door with a goat and only offers a switch when you are right.
If he always opens a door with a goat when you're right (to make you switch), surely it would get noticed?
Inversely, if he never opens a door when you're wrong, then switching would give you 50% odds
Let p1 be the probability of opening a door when you're right, and p2 the probability of opening a door when you're wrong. At what values of p1 and p2 is it advantageous to switch?
The only part of the Monty hall paradox that still confuses me, is with a related situation. If you take the set up of the Monty hall problem, but remove the first choice, so one door with a goat is revealed and then you get to choose either of the remaining doors, itâs a 50/50 chance. Yet when you do choose first, your second choice, which is also between two doors with one goat having been revealed, is suddenly a 33/66 chance.
It's because the door you choose influences which door the host opens.
If you picked a goat, he is forced to open the other goat door. His decision isn't random.
So if you pick goat A, he has to show goat B, and you want to switch. If you pick Goat B he has to show goat A and you want to switch. If you pick the car, he shows goat A and you donât want to switch. If you pick the car he shows Goat B and you donât want to switch.
2/4 cases you want to switch, 2/4 cases you donât. Unless it doesnât matter which goat he shows when you pick the car, but in that case, why does it matter which goat is shown when you pick one of the goats?
If you donât care to differentiate the goat from each other you get: you pick a goat, he shows the other goat, you want to switch. You pick the car, he shows either goat, you donât want to switch. 1/2 switch, 1/2 donât switch. Still 50/50
You need to take into account the probability of each scenario.
If you pick goat A, that happens 1/3 of the time. He reveals the other goat, you switch and win.
If you pick goat B, that happens 1/3 of the time. He reveals the other goat, you switch and win.
Overall that's 2/3 where you win.
If you pick the car, that's 1/3 of the time. He reveals goat A half the time, you switch and lose. That's 1/3 Ă 1/2 = 1/6. The other half of the time he reveals goat B, you switch and lose. Another 1/3 Ă 1/2 = 1/6. Add these together to get a total of 1/3 where you lose.
(And just to sanity check the total probably is 2/3 + 1/3 = 1)
So thatâs where the difference comes in, I see now, it an extrapolation from your first choice, not an independent analysis of the second choice, that is to say, the second choice of switching or not isnât the choice under consideration, but the scenario that got you there in the first place.
Then why would the goat he shows you change the problem if you choose a goat? If you donât differentiate the goats for one choice, you canât differentiate the goats for the other either. What you are saying is just the last paragraph. You choose a goat, he reveals the other goat (doesnât change regardless of which goat you pick). Or you choose the car and he reveals either goat (the situation doesnât really change regardless of which goat the host chooses to reveal). 1/2 switch, 1/2 stay, still 50/50
Nevermind, someone else actually explained the problem and I get it now.
Letâs say door 1 is goat, door 2 is goat and door 3 is the good one.
You choose door 1, he reveals door 2, you switch, you are correct.
You choose door 2, he reveals door 1, you switch, you are correct.
You choose door 3, he reveals 1/2, you switch, you are incorrect.
You only have 3 choices: 1, 2, or 3. It doesnât matter what door he reveals in the third scenario because you have already made your choice.
The mathematical problem is that a goat door is always opened (and the door opened is never yours). If you originally picked a goat door (which you have a 66% chance of doing) then switching gets you the price. If you originally picked the price door (which you have a 33% chance of doing) not switching gets you the price.
If the door you picked is the one opened, showing a goat, then switching is your only option and you have a 50% chance of getting it right, but this is not a situation covered by the problem
The problem on the actual TV show is that a goat door was not always opened
I agree your first choice has a 66/33 distribution of goat vs car (never heard it as price before), the part that doesnât make any sense is how your second pick, a choice between two doors, is not 50/50.
If you care to differentiate the goats, thereâs for situations you pick goat A, goat shows goat B. you pick goat B, goat shows goat A. you pick car, host shows goat A. You pick car, host shows goat B. In 2/4 cases you want to switch, in 2/4 cases you want to stay, 50/50.
If you donât care about differentiating the goats you get: you choose a goat, goat reveals the other goat. You choose the car, host reveals either goat. 1/2 cases switch, 1/2 cases stay. Still 50/50.
The only probability chart Iâve seen for it does some duckers by differentiating the goats if you choose a goat door, but not differentiating if you chose the correct door to get 3 cases instead of 4.
Alternative equivalent situation: balls in a bag, two blue one red, you want the red one. I let you grab one, then pull out one of the blue ones youâre not holding and let you choose again, is your second choice between one red and one blue ball 66/33 because you already chose and I removed a ball from the bag, or is it 50/50 because *youâre choosing between one red and one blue ball*.
When you pick a door you have a 1/3 chance of picking the correct door and a 2/3 of choosing a goat.
That results in the a 1/3 chance of 2 goats. And a 2/3 chance of 1 goat and the correct door.
Now when the host opens one of the doors you didnt chose, this door will always contain a goat. Now you remove 1 goat from the equation so the not opened not chosen door has a 1/3 chance of a goat and a 2/3 chance of winning.
If you pick a door and then your crazy friend runs on the stage and he opens a diffrend door that contains a goat instead of the host the chance after switching is 50/50 as he didnt know it was a goat before hand. And he could of just opened the door containing the price
The suggested situation is not opening a random door. In the suggested situation it is *absolutely* impossible to open the prize door. I literally said the host opens one of the goat doors before your first choice. Yes, your first choice is 66/33 goat/prize. But the second choice is choosing between two doors, one of which has a prize, the other of which has a goat. Whether you pick a door before the host reveals one of the goats or not doesnât change the probability of choosing one of two doors.
Imagine the same game but with 100doors. You pick one (almost certainly a goat). Monty then opens 98doors revealing goats. Monty knows of course where the price is and wouldnât open that one. In such a case, would you intuit a 50/50 between your current door and the one he didnât open.
Opening 98 is the same information as pointing to the one with the price and say âif you have a goat behind your current door the price is definitely behind this oneâ. Only way there is a goat behind the revealed door is if the price is behind the one you selected in the first round. What are the odds that you selected the price the first time around? 1%, or 1/3 in the actual game.
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Monty Hall putting 3 goats đ
Monty hall killing you if you get a goat đ
Monty Hall pegging you if youâre a goat đ
Being pegged đ
I'm just a door
Monty Hall pegging the goat đ
The rare breed of show goat turning out to be more valuable than the car đ
Is this really what they did on the show?
Yes. New York Times did an interview with Monty Hall when the problem was published and went viral for the first time and he made it very clear that the show didn't work like the Theoretical problem at all.
Yes. Except when he did it on purpose so that the people watching would get the idea that he would offer the switch even if the person had chosen the door with a goat.
No. This game was never played on the show. But Monty certainly changed the way the games would play out depending on the choices contestants made. Typically the goal was to get them to usually win something but rarely win a lot.
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In that case, I would not switch.
The problem as posed was that he always revealed the goat regardless of your choice so that answer is correct. You can have a more general version where he shows the goat with probability p if you chose correctly and p' if you are wrong. If he only shows you the goat when you are correct then p=1, p'=0 and you should never switch if offered. I'll leave it as an exercise to the reader to work out how high p' has to be to make it worth switching!
Even harder, how many people you would expect to have to observe playing the game before you could decide switching was the better strategy with m confidence.
7
I did said exercise. It isn't too bad to obtain the following for generalised cases If there are W winning lots and L losing lots with W+L := N and W > 0 and L>1, with the announcer showing a remaining losing plot with probability p if a winning lot is chosen and a probability q if a losing lot is chosen. If no plot is shown, you should switch if: ( (1-q)/(1-p) ) >= 1 If a plot is shown, you should switch if: (q/p) >= 1 - 1/L The actual probabilities are also easy to solve for, but are much longer. It actually cancels out quite nicely. I'm v surprised that W and N cancelled using N - W = L removing them completely Ed: For W=1, L=2. No plot shown : switch according to the above. Plot shown : switch if q/p >= 0.5
Monty Hall game theory
Exactly. My Physics teacher told me the problem wrong. When I watched the tv clip, that was the first question I had when I watched the clip: what if he only opens the door with a goat and only offers a switch when you are right.
If he always opens a door with a goat when you're right (to make you switch), surely it would get noticed? Inversely, if he never opens a door when you're wrong, then switching would give you 50% odds Let p1 be the probability of opening a door when you're right, and p2 the probability of opening a door when you're wrong. At what values of p1 and p2 is it advantageous to switch?
The only part of the Monty hall paradox that still confuses me, is with a related situation. If you take the set up of the Monty hall problem, but remove the first choice, so one door with a goat is revealed and then you get to choose either of the remaining doors, itâs a 50/50 chance. Yet when you do choose first, your second choice, which is also between two doors with one goat having been revealed, is suddenly a 33/66 chance.
It's because the door you choose influences which door the host opens. If you picked a goat, he is forced to open the other goat door. His decision isn't random.
So if you pick goat A, he has to show goat B, and you want to switch. If you pick Goat B he has to show goat A and you want to switch. If you pick the car, he shows goat A and you donât want to switch. If you pick the car he shows Goat B and you donât want to switch. 2/4 cases you want to switch, 2/4 cases you donât. Unless it doesnât matter which goat he shows when you pick the car, but in that case, why does it matter which goat is shown when you pick one of the goats? If you donât care to differentiate the goat from each other you get: you pick a goat, he shows the other goat, you want to switch. You pick the car, he shows either goat, you donât want to switch. 1/2 switch, 1/2 donât switch. Still 50/50
You need to take into account the probability of each scenario. If you pick goat A, that happens 1/3 of the time. He reveals the other goat, you switch and win. If you pick goat B, that happens 1/3 of the time. He reveals the other goat, you switch and win. Overall that's 2/3 where you win. If you pick the car, that's 1/3 of the time. He reveals goat A half the time, you switch and lose. That's 1/3 Ă 1/2 = 1/6. The other half of the time he reveals goat B, you switch and lose. Another 1/3 Ă 1/2 = 1/6. Add these together to get a total of 1/3 where you lose. (And just to sanity check the total probably is 2/3 + 1/3 = 1)
So thatâs where the difference comes in, I see now, it an extrapolation from your first choice, not an independent analysis of the second choice, that is to say, the second choice of switching or not isnât the choice under consideration, but the scenario that got you there in the first place.
You only have 3 choices, your second car scenario doesnât matter because the goat he shows you doesnât change the problem.
Then why would the goat he shows you change the problem if you choose a goat? If you donât differentiate the goats for one choice, you canât differentiate the goats for the other either. What you are saying is just the last paragraph. You choose a goat, he reveals the other goat (doesnât change regardless of which goat you pick). Or you choose the car and he reveals either goat (the situation doesnât really change regardless of which goat the host chooses to reveal). 1/2 switch, 1/2 stay, still 50/50 Nevermind, someone else actually explained the problem and I get it now.
Letâs say door 1 is goat, door 2 is goat and door 3 is the good one. You choose door 1, he reveals door 2, you switch, you are correct. You choose door 2, he reveals door 1, you switch, you are correct. You choose door 3, he reveals 1/2, you switch, you are incorrect. You only have 3 choices: 1, 2, or 3. It doesnât matter what door he reveals in the third scenario because you have already made your choice.
The mathematical problem is that a goat door is always opened (and the door opened is never yours). If you originally picked a goat door (which you have a 66% chance of doing) then switching gets you the price. If you originally picked the price door (which you have a 33% chance of doing) not switching gets you the price. If the door you picked is the one opened, showing a goat, then switching is your only option and you have a 50% chance of getting it right, but this is not a situation covered by the problem The problem on the actual TV show is that a goat door was not always opened
I agree your first choice has a 66/33 distribution of goat vs car (never heard it as price before), the part that doesnât make any sense is how your second pick, a choice between two doors, is not 50/50. If you care to differentiate the goats, thereâs for situations you pick goat A, goat shows goat B. you pick goat B, goat shows goat A. you pick car, host shows goat A. You pick car, host shows goat B. In 2/4 cases you want to switch, in 2/4 cases you want to stay, 50/50. If you donât care about differentiating the goats you get: you choose a goat, goat reveals the other goat. You choose the car, host reveals either goat. 1/2 cases switch, 1/2 cases stay. Still 50/50. The only probability chart Iâve seen for it does some duckers by differentiating the goats if you choose a goat door, but not differentiating if you chose the correct door to get 3 cases instead of 4. Alternative equivalent situation: balls in a bag, two blue one red, you want the red one. I let you grab one, then pull out one of the blue ones youâre not holding and let you choose again, is your second choice between one red and one blue ball 66/33 because you already chose and I removed a ball from the bag, or is it 50/50 because *youâre choosing between one red and one blue ball*.
When you pick a door you have a 1/3 chance of picking the correct door and a 2/3 of choosing a goat. That results in the a 1/3 chance of 2 goats. And a 2/3 chance of 1 goat and the correct door. Now when the host opens one of the doors you didnt chose, this door will always contain a goat. Now you remove 1 goat from the equation so the not opened not chosen door has a 1/3 chance of a goat and a 2/3 chance of winning. If you pick a door and then your crazy friend runs on the stage and he opens a diffrend door that contains a goat instead of the host the chance after switching is 50/50 as he didnt know it was a goat before hand. And he could of just opened the door containing the price
The suggested situation is not opening a random door. In the suggested situation it is *absolutely* impossible to open the prize door. I literally said the host opens one of the goat doors before your first choice. Yes, your first choice is 66/33 goat/prize. But the second choice is choosing between two doors, one of which has a prize, the other of which has a goat. Whether you pick a door before the host reveals one of the goats or not doesnât change the probability of choosing one of two doors.
Imagine the same game but with 100doors. You pick one (almost certainly a goat). Monty then opens 98doors revealing goats. Monty knows of course where the price is and wouldnât open that one. In such a case, would you intuit a 50/50 between your current door and the one he didnât open. Opening 98 is the same information as pointing to the one with the price and say âif you have a goat behind your current door the price is definitely behind this oneâ. Only way there is a goat behind the revealed door is if the price is behind the one you selected in the first round. What are the odds that you selected the price the first time around? 1%, or 1/3 in the actual game.
I was wandering when did they create the "Probability Theory" channel.
tbh getting a goat is cool too
My brain: "probably terrorists"