5 ms·

>By switching, you have a 2/3 chance. I guess I need to read up on the Monty Hall problem again, because I always thought it was 1/2 chance if you switch (vs 1

by 22c 5y ago

>By switching, you have a 2/3 chance. I guess I need to read up on the Monty Hall problem again, because I always thought it was 1/2 chance if you switch (vs 1/3 chance if you don't).

albedoa 5y ago
That would mean you have a 5/6 chance of finding the car if you were allowed to pick both doors. You can see the problem with that. In reality, it's 2/3 and 1/3 respectively.
- 22c 5y ago
  If you started with 4 doors, where 3 of them had goats, wouldn't you have a 1/4 chance of picking the car the first time and a 1/3 chance of picking the car the second time?
  - lmm 5y ago
    1/4 chance if you keep your original door, 3/4 chance if you switch (assuming the host opens two doors to show goats). If the host opened one door (so you have your original door and two other doors to pick from) then it's a 3/8 chance if you switch, if that's what you're asking.
    - 22c 5y ago
      Yes, I was asking about the latter, thanks!
anyfoo 5y ago
It's been a while since I actively thought about the Monty Hall problem, I thought I understood it intuitively back then. I thought it became one of those "oh yeah, that's unintuitive, but once it clicks it's fine" things for me. I thought through it again, and I'm angry now. At least I'm not alone (got this link from the article): https://web.archive.org/web/20140413131827/http://www.decisionsciences.org/DecisionLine/Vol30/30_1/vazs30_1.pdf https://web.archive.org/web/20140413131827/http://www.decisi...
aonsager 5y ago
It's a 1/2 chance at the point where you're considering the switch. If you start the game with the intention of switching, you have a 2/3 chance of being successful because the winning strategy is to miss the car on your first pick.
- cjbest 5y ago
  No, I don’t think this is right. Regardless of your intended plan, you only had a 1/3 chance of picking correctly the first time, so switching gets you a 2/3 chance.
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- Nursie 5y ago
  No, it's a 2/3 chance at that point. When you choose initially, you have a 1/3 probability of getting the right one, leaving a 2/3 probability that the car is on one of the other two. The host reveals one of the other two. So that 2/3 probability applies to the remaining door. Here is a short C implementation that made it very clear to me... #include <stdio.h> #include <stdlib.h> int doround () { int car = rand() % 3; int firstchoice = rand() % 3; // host reveals one of the goat doors if (car == firstchoice) { // you changing to the other door after the reveal is a loss return 0; } if(car != firstchoice) { // you changing to the other door after the reveal is a win return 1; } } int main(int argc, char** argv) { int wins=0; for (int round=0; round < 1000; round++){ wins+=doround(); } printf("Worked in %d of %d rounds\n", wins, rounds); }
HWR_14 5y ago
It's a 1/3 chance if you pick before the second door is opened and do not switch. It's a 1/2 chance if you pick after the second door is opened, regardless of switching. It's a 2/3 chance if you pick before the second door is open and switch after the second door is opened. I recommend reading again, because I don't want to write up the logic. Of a bigger concern is that this was not true on "Let's Make a Deal". The problem assumes that Monty always offered the option to switch. In reality, he did not, and the offers were not random. That corrected for the issue to a large degree in practice.
mlboss 5y ago
Door A | Door B | Door C | —————————————————————————- Car. | Goat. |Goat.| Goat. | Car. | Goat.|. Goat. | Goat. | Car.|. The above tables shows all possible distribution of car and goats. If the initial choice was Door A, then you only have 1/3 probability. But after B/C door is opened the probability increases to 2/3 if switched. The clue is that the host will always open the door with a goat.
SailingSperm 5y ago
I find the Monty Hall problem becomes intuitive when scaled. 100 doors, 1 car, 99 goats. Pick a door. All doors open (with goats except 1) ... Do you think you picked the 1/100 door with the car, or that it's the other only one left standing.
- rho4 5y ago
  Beautiful short explanation. I always struggle getting this point across in discussions.
  - rho4 5y ago
    My hopes for this explanation were too high. I just tested it on my cooworkers. No luck. I reverted to suggesting to write a short simulation script to convince themselves by experimental data.
    - v-erne 5y ago
      I personally find the Generalized Monty Hall problem as unintuitive as the original one. But I belive that I found a batter way to make it intuitive (at least for people that have heard about classic and conditional probability). And it event worked on one of my friends :) If Your first choice was lucky and two goats remained in other gates than Monty does not change anything by showing You the goat in one of them (it't the same as if Monty selected random gate). Things becomes intresting if You was unlucky and selected Goat. Monty MUST show You the other goat (because there is only one available for him to select). And thus he introduces information to otherwise random selection (it stops beeing random). And by doing this he eliminates for You conditional possiblity of selecting second goat when You selected one in first round (and in this conditional scenariu swiching is sure bet which changes the overall odds to 2/3)
- Mordisquitos 5y ago
  Exactly. It also helps to emphasise that the eponymous host knows which door contains the prize and will never open the prize door before asking you whether you want to switch.
rocqua 5y ago
That doesn't add up. Literally, 1/3 + 1/2 does not add up to 1. What third option exist that has a 1/6 chance of happening? The basic argument is. If you don't switch, you have 1/3 chance of winning. So if you do switch you have 1/3 chance of losing. Hence if you switch you have 2/3 chance of winning. How the door you switched too went from 1/3 chance to double is the hard part to explain. But it must be the case.
aliceryhl 5y ago
When using the switch strategy, you win if you choose a goat initially. Since there are two goats and one car, this is a 2/3 chance.

by 22c 5y ago

>By switching, you have a 2/3 chance. I guess I need to read up on the Monty Hall problem again, because I always thought it was 1/2 chance if you switch (vs 1/3 chance if you don't).

albedoa 5y ago

That would mean you have a 5/6 chance of finding the car if you were allowed to pick both doors. You can see the problem with that. In reality, it's 2/3 and 1/3 respectively.

22c 5y ago
If you started with 4 doors, where 3 of them had goats, wouldn't you have a 1/4 chance of picking the car the first time and a 1/3 chance of picking the car the second time?
- lmm 5y ago
  1/4 chance if you keep your original door, 3/4 chance if you switch (assuming the host opens two doors to show goats). If the host opened one door (so you have your original door and two other doors to pick from) then it's a 3/8 chance if you switch, if that's what you're asking.
  - 22c 5y ago
    Yes, I was asking about the latter, thanks!

anyfoo 5y ago

It's been a while since I actively thought about the Monty Hall problem, I thought I understood it intuitively back then. I thought it became one of those "oh yeah, that's unintuitive, but once it clicks it's fine" things for me. I thought through it again, and I'm angry now. At least I'm not alone (got this link from the article): https://web.archive.org/web/20140413131827/http://www.decisionsciences.org/DecisionLine/Vol30/30_1/vazs30_1.pdf https://web.archive.org/web/20140413131827/http://www.decisi...

aonsager 5y ago

It's a 1/2 chance at the point where you're considering the switch. If you start the game with the intention of switching, you have a 2/3 chance of being successful because the winning strategy is to miss the car on your first pick.

cjbest 5y ago
No, I don’t think this is right. Regardless of your intended plan, you only had a 1/3 chance of picking correctly the first time, so switching gets you a 2/3 chance.
- deleted 5y ago
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Nursie 5y ago
No, it's a 2/3 chance at that point. When you choose initially, you have a 1/3 probability of getting the right one, leaving a 2/3 probability that the car is on one of the other two. The host reveals one of the other two. So that 2/3 probability applies to the remaining door. Here is a short C implementation that made it very clear to me... #include <stdio.h> #include <stdlib.h> int doround () { int car = rand() % 3; int firstchoice = rand() % 3; // host reveals one of the goat doors if (car == firstchoice) { // you changing to the other door after the reveal is a loss return 0; } if(car != firstchoice) { // you changing to the other door after the reveal is a win return 1; } } int main(int argc, char** argv) { int wins=0; for (int round=0; round < 1000; round++){ wins+=doround(); } printf("Worked in %d of %d rounds\n", wins, rounds); }

HWR_14 5y ago

It's a 1/3 chance if you pick before the second door is opened and do not switch. It's a 1/2 chance if you pick after the second door is opened, regardless of switching. It's a 2/3 chance if you pick before the second door is open and switch after the second door is opened. I recommend reading again, because I don't want to write up the logic. Of a bigger concern is that this was not true on "Let's Make a Deal". The problem assumes that Monty always offered the option to switch. In reality, he did not, and the offers were not random. That corrected for the issue to a large degree in practice.

mlboss 5y ago

SailingSperm 5y ago

I find the Monty Hall problem becomes intuitive when scaled. 100 doors, 1 car, 99 goats. Pick a door. All doors open (with goats except 1) ... Do you think you picked the 1/100 door with the car, or that it's the other only one left standing.

rho4 5y ago
Beautiful short explanation. I always struggle getting this point across in discussions.
- rho4 5y ago
  My hopes for this explanation were too high. I just tested it on my cooworkers. No luck. I reverted to suggesting to write a short simulation script to convince themselves by experimental data.
  - v-erne 5y ago
    I personally find the Generalized Monty Hall problem as unintuitive as the original one. But I belive that I found a batter way to make it intuitive (at least for people that have heard about classic and conditional probability). And it event worked on one of my friends :) If Your first choice was lucky and two goats remained in other gates than Monty does not change anything by showing You the goat in one of them (it't the same as if Monty selected random gate). Things becomes intresting if You was unlucky and selected Goat. Monty MUST show You the other goat (because there is only one available for him to select). And thus he introduces information to otherwise random selection (it stops beeing random). And by doing this he eliminates for You conditional possiblity of selecting second goat when You selected one in first round (and in this conditional scenariu swiching is sure bet which changes the overall odds to 2/3)
Mordisquitos 5y ago
Exactly. It also helps to emphasise that the eponymous host knows which door contains the prize and will never open the prize door before asking you whether you want to switch.

rocqua 5y ago

That doesn't add up. Literally, 1/3 + 1/2 does not add up to 1. What third option exist that has a 1/6 chance of happening? The basic argument is. If you don't switch, you have 1/3 chance of winning. So if you do switch you have 1/3 chance of losing. Hence if you switch you have 2/3 chance of winning. How the door you switched too went from 1/3 chance to double is the hard part to explain. But it must be the case.

aliceryhl 5y ago

When using the switch strategy, you win if you choose a goat initially. Since there are two goats and one car, this is a 2/3 chance.