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If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. Thus switching, at this

If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. Thus switching, at this point, is a 50-50 proposition. Or 1/3 to 1/3 if that makes more sense. Switching is the advantageous move only when the host always reveals a zonk first, and that's why it does in fact matter "how the TV show operates." Another show, Deal or No Deal, often came down to 2 suitcases (from an initial 26), one of which was worth a six-figure amount and the other a tiny amount. Something like $200,000 vs $500. The logic of the Monty Hall problem would suggest that since the player had 1/26 chance of initially choosing $200,000, they should switch suitcases at the end. But because all 24 previous reveals were random, there is actually no advantage to doing so.

HarHarVeryFunny 3y ago
> If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. No... Remember one door has been opened and the car was not there, so we now know that it is guaranteed the car is behind one of the other two doors. If there was only 1/3 chance of it being behind either door, then that's the same as saying there's only a 2/3 (1/3 + 1/3) chance of there being a car anywhere, when we know for sure (3/3 chance) that it is somewhere! The reality is that the contestant's original pick has a 1/3 chance of being right, and the other other two doors have a combined 2/3 chance of the car being behind one of them. Once one of those doors has been opened and the car shown not to be there, then this now means this combined 2/3 chance lies with the other door (which you should therefore switch to). Note that the probabilities now add up as expected: 1/3 chance original pick + 2/3 remaining choice = 3/3.
- default-kramer 3y ago
  What do you think is incorrect about the code here which produces 1/2 if Monty randomly chooses? https://news.ycombinator.com/item?id=39515082 https://news.ycombinator.com/item?id=39515082
  - HarHarVeryFunny 3y ago
    Nothing .. I was wrong. Switching only wins when Monty avoids opening the car door. If Monty opens a random door, then switching makes no difference. The flaw in my logic above for the random case is that it's 1/3 chance orig pick was right, 1/3 chance unopened door, plus 1/3 chance that Monty opened the door with the car behind meaning you never even got a choice to switch.
- kurikuri 3y ago
  The host’s knowledge does contribute to the 2/3, because that knowledge prevents an outcome from occurring. Let us consider the switching strategy and instead say selecting the correct door awards 6 points. Our question: what is the expected value of this strategy when the host knows, vs when the host doesn’t know. Consider the scenario where the player’s first choice is the winner. This happens 1/3 of the time, and the host’s knowledge doesn’t change their behavior. With the switching strategy, the player always loses in this case, so the contribution of this scenario towards the expected value for either case(host knowing vs not) is 0. Consider the other scenario where the player’s first selection does not contain the car, and note that this situation happens 2/3 of the time. For convenience, label the player’s selected door 1, and the car containing door 2. If we assume the host has knowledge, then the host will select door 3. If the player decides to stay, they lose. In this scenario, the switch strategy always wins, so its contribution towards the expected value when the host knows is (2/3 * 6) = 4. If we assume the host doesn’t have knowledge, he will pick either door 2 or 3 with equal probability. In the event he selects door 2, the winning door, so half the time, the player loses because they can’t switch. This occurs half of the time within this scenario, so this sub -scenario contributes 0 points to our expected value. If the host selects door three, then the switch strategy wins, meaning this sub-scenario contributes (2/3 * 1/2 * 6) = 3 points to our overall expected value. When the host has knowledge, the scenario where the player loses because of the host selection will not occur, leading to the 2/3-probability. Without the host having this knowledge, our event space changes and the player switch strategy only has a 1/2 chance of winning.

HarHarVeryFunny 3y ago

> If the host's first reveal is randomly chosen, when you get down to two doors left each of them has a 1/3 chance of containing the car. No... Remember one door has been opened and the car was not there, so we now know that it is guaranteed the car is behind one of the other two doors. If there was only 1/3 chance of it being behind either door, then that's the same as saying there's only a 2/3 (1/3 + 1/3) chance of there being a car anywhere, when we know for sure (3/3 chance) that it is somewhere! The reality is that the contestant's original pick has a 1/3 chance of being right, and the other other two doors have a combined 2/3 chance of the car being behind one of them. Once one of those doors has been opened and the car shown not to be there, then this now means this combined 2/3 chance lies with the other door (which you should therefore switch to). Note that the probabilities now add up as expected: 1/3 chance original pick + 2/3 remaining choice = 3/3.

default-kramer 3y ago
What do you think is incorrect about the code here which produces 1/2 if Monty randomly chooses? https://news.ycombinator.com/item?id=39515082 https://news.ycombinator.com/item?id=39515082
- HarHarVeryFunny 3y ago
  Nothing .. I was wrong. Switching only wins when Monty avoids opening the car door. If Monty opens a random door, then switching makes no difference. The flaw in my logic above for the random case is that it's 1/3 chance orig pick was right, 1/3 chance unopened door, plus 1/3 chance that Monty opened the door with the car behind meaning you never even got a choice to switch.
kurikuri 3y ago
The host’s knowledge does contribute to the 2/3, because that knowledge prevents an outcome from occurring. Let us consider the switching strategy and instead say selecting the correct door awards 6 points. Our question: what is the expected value of this strategy when the host knows, vs when the host doesn’t know. Consider the scenario where the player’s first choice is the winner. This happens 1/3 of the time, and the host’s knowledge doesn’t change their behavior. With the switching strategy, the player always loses in this case, so the contribution of this scenario towards the expected value for either case(host knowing vs not) is 0. Consider the other scenario where the player’s first selection does not contain the car, and note that this situation happens 2/3 of the time. For convenience, label the player’s selected door 1, and the car containing door 2. If we assume the host has knowledge, then the host will select door 3. If the player decides to stay, they lose. In this scenario, the switch strategy always wins, so its contribution towards the expected value when the host knows is (2/3 * 6) = 4. If we assume the host doesn’t have knowledge, he will pick either door 2 or 3 with equal probability. In the event he selects door 2, the winning door, so half the time, the player loses because they can’t switch. This occurs half of the time within this scenario, so this sub -scenario contributes 0 points to our expected value. If the host selects door three, then the switch strategy wins, meaning this sub-scenario contributes (2/3 * 1/2 * 6) = 3 points to our overall expected value. When the host has knowledge, the scenario where the player loses because of the host selection will not occur, leading to the 2/3-probability. Without the host having this knowledge, our event space changes and the player switch strategy only has a 1/2 chance of winning.