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This is a great clarification of how this problem works. I STILL can't grasp why you'd switch. Since you have no idea which door it is, couldn't the 2/3 proba

by zodPod 11y ago

This is a great clarification of how this problem works. I STILL can't grasp why you'd switch. Since you have no idea which door it is, couldn't the 2/3 probability be applied to either his door or your door? For all you know, the one that he removed was just one random one of the goat doors. Your chance of picking the car was 1/3 before, if you could have the car already, why would it be better to switch now that he removed which ever would definitely be a goat? It seems like you still have a 50/50 chance of getting the car and your chance of the one that remains (unpicked/unremoved) is just as likely to be the other goat as it is to be the car.. That said, I made a simulation of it where it does the following: 1. Assign random locations behind doors, 2. Player randomly chooses door, 3. Host excludes door (random if both are goats, goat if one is a car), 4. Player either switches, or doesn't switch based on what's chosen for that simulation (the full set gets one or the other) After running 50,000,000 then dividing success/tries, my number is nearly exactly the same between the two. Is my simulation working incorrectly?

wrsh07 11y ago
I find it clearer if you have 100 doors. Pick one. The host eliminates 98 doors [with no prize behind them]. Do you switch? Or do you stick with your original guess? As the parent mentioned, switching gives you a 99/100 chance of being correct, staying gives you a mere 1/100 chance.
- ricardobeat 11y ago
  I don't see why the other door has 99x more probability of being correct. It's still 50/50, original door or remaining door, the other 98 don't change the odds.
  - pklausler 11y ago
    Your initial pick has a 1/100 chance of being right. If you switch, and you were right, you lose. On the other hand, your initial pick has a 99/100 chance of being wrong, and if you were wrong, and you switch, you win. So switch.
    - deleted 11y ago
      [deleted]
  - umanwizard 11y ago
    Let's play a game to make this more clear. I'm thinking of a number between 1 and 1000. First, guess what it is.
    - cormacrelf 11y ago
      2
      - umanwizard 11y ago
        Well, before you finalize your guess, I'll give you a hint: it's either 2, or 437. It's not any other number. Keep in mind I made my choice before we started, and it hasn't changed. Do you want to change your guess to 437, or stick with 2?
        mikecb 11y ago
        This is a nice reinterpretation.
        joemi 11y ago
        I've always flip-flopped on grasping the Monty Hall thing (sometimes I'd think I understand, other times not so much)... but this little guessing game really clarified it for me, once and for all. Thanks!
  - yaur 11y ago
    the behavior of the game host as he will never reveal the door with the car this is the key bit. If the door you picked is the correct one (1% chance) the host can open any door. If you picked incorrectly (99% chance) he can open any door except the winning door so the remaining door will be the correct one 99% of the time.
- mettamage 11y ago
  Thanks for clarifying a very counter intuitive problem. I always struggled with this one, not anymore :) Sometimes scaling things up is the best way to understand things. Who knew!
baddox 11y ago
> For all you know, the one that he removed was just one random one of the goat doors. The key observation is that the problem states that Monty Hall opens an empty door. It's not clear whether Monty knows which doors are empty, but it doesn't actually matter, since the problem states that he will always open an empty door. As others have said, your choice as a contestant is whether to take one door (the one you initially pointed at), or to take the other two doors (at least one of which is obviously empty). Stated that way, it's pretty obvious that taking 2 doors is better than taking 1 door.
pklausler 11y ago
Your simulation is not correct. Look at it this way: 1/3 of the time, you pick the right door up front. If you switch, you lose. 2/3 of the time, you didn't pick the right door. If you switch, you win. So switch.
jinfiesto 11y ago
From the initial setup you have a 1/3 chance of choosing the correct door and a 2/3 chance of choosing the wrong door. The key to this problem is the behavior of the game host as he will never reveal the door with the car. So, after he reveals the contents of one of the doors, if you choose not to switch, you still only have a 1/3 chance of having chosen the door with the car. On the other hand, if you choose to switch, 2/3 of the time you will have chosen incorrectly initially, and under the switching strategy, those 2/3ds of the time you are guaranteed to win the car. Hence, you have a 2/3ds chance of winning under the switching strategy, vs a 1/3d chance of winning under the "staying" strategy. Hope this clarifies things.
logfromblammo 11y ago
The important bit is that Monty will never choose to show you the winning door, and he will always show you a door after you pick a door and before asking you if you want to switch. He always opens a losing door. The location of the prize and the player choice of door are independent events. Monty's choice is dependent on whether the player choice and the prize location are the same. Monty will never open the door the player picked. Monty will never reveal the prize. Those rules generate exploitable information for the player. Here is the complete probability tree, and two possible strategies of play: Prize behind door A; P = 1/3 ALWAYS STAY strategy: Player picks door A; P = 1/3 * 1/3 = 1/9 P(WIN) = 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18 Monty shows Door B; P = 1/3 * 1/3 * 1/2 = 1/18 = 6/18 Player stays with A; WIN = 1/3 Player switches to C; LOSS P(LOSS) = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 Monty shows Door C; P = 1/3 * 1/3 * 1/2 = 1/18 = 6/9 Player stays with A; WIN = 2/3 Player switches to B; LOSS Player picks door B; P = 1/3 * 1/3 = 1/9 ALWAYS SWITCH strategy: Monty shows Door C; P = 1/3 * 1/3 * 1/1 = 1/9 P(WIN) = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 Player stays with B; LOSS = 6/9 Player switches to A; WIN = 2/3 Player picks door C; P = 1/3 * 1/3 = 1/9 P(LOSS) = 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18 Monty shows Door B; P = 1/3 * 1/3 * 1/1 = 1/9 = 6/18 Player stays with C; LOSS = 1/3 Player switches to A; WIN Prize behind door B; P = 1/3 Player picks door A; P = 1/3 * 1/3 = 1/9 Monty shows Door C; P = 1/3 * 1/3 * 1/1 = 1/9 Player stays with A; LOSS Player switches to B; WIN Player picks door B; P = 1/3 * 1/3 = 1/9 Monty shows Door A; P = 1/3 * 1/3 * 1/2 = 1/18 Player stays with B; WIN Player switches to C; LOSS Monty shows Door C; P = 1/3 * 1/3 * 1/2 = 1/18 Player stays with B; WIN Player switches to A; LOSS Player picks door C; P = 1/3 * 1/3 = 1/9 Monty shows Door A; P = 1/3 * 1/3 * 1/1 = 1/9 Player stays with C; LOSS Player switches to B; WIN Prize behind door C; P = 1/3 Player picks door A; P = 1/3 * 1/3 = 1/9 Monty shows Door B; P = 1/3 * 1/3 * 1/1 = 1/9 Player stays with A; LOSS Player switches to C; WIN Player picks door B; P = 1/3 * 1/3 = 1/9 Monty shows Door A; P = 1/3 * 1/3 * 1/1 = 1/9 Player stays with B; LOSS Player switches to C; WIN Player picks door C; P = 1/3 * 1/3 = 1/9 Monty shows Door A; P = 1/3 * 1/3 * 1/2 = 1/18 Player stays with C; WIN Player switches to B; LOSS Monty shows Door B; P = 1/3 * 1/3 * 1/2 = 1/18 Player stays with C; WIN Player switches to A; LOSS If your simulation does not conform with the mathematically derived result, your simulation is incorrect.
brobinson 11y ago
The table here [1] helped me understand it when I first heard of this problem. [1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solutions https://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solu...
grahamburger 11y ago
The most clear way that I've heard it described is by scaling it up to 100 doors or so, and then removing 98 doors instead of removing 1. (This is mentioned elsewhere in the thread.) It would be interesting to see the results of your test using this method. It seems intuitively clear with 100 doors that switching is the better method.
SeoxyS 11y ago
This is the one article that made me fully grasp it forever. I've known about this for ages, but it never really clicked until I read it put this way: http://waitbutwhy.com/2016/03/the-jellybean-problem.html http://waitbutwhy.com/2016/03/the-jellybean-problem.html

by zodPod 11y ago

wrsh07 11y ago

I find it clearer if you have 100 doors. Pick one. The host eliminates 98 doors [with no prize behind them]. Do you switch? Or do you stick with your original guess? As the parent mentioned, switching gives you a 99/100 chance of being correct, staying gives you a mere 1/100 chance.

ricardobeat 11y ago
I don't see why the other door has 99x more probability of being correct. It's still 50/50, original door or remaining door, the other 98 don't change the odds.
- pklausler 11y ago
  Your initial pick has a 1/100 chance of being right. If you switch, and you were right, you lose. On the other hand, your initial pick has a 99/100 chance of being wrong, and if you were wrong, and you switch, you win. So switch.
  - deleted 11y ago
    [deleted]
- umanwizard 11y ago
  Let's play a game to make this more clear. I'm thinking of a number between 1 and 1000. First, guess what it is.
  - cormacrelf 11y ago
    2
    - umanwizard 11y ago
      Well, before you finalize your guess, I'll give you a hint: it's either 2, or 437. It's not any other number. Keep in mind I made my choice before we started, and it hasn't changed. Do you want to change your guess to 437, or stick with 2?
      - mikecb 11y ago
        This is a nice reinterpretation.
      - joemi 11y ago
        I've always flip-flopped on grasping the Monty Hall thing (sometimes I'd think I understand, other times not so much)... but this little guessing game really clarified it for me, once and for all. Thanks!
- yaur 11y ago
  the behavior of the game host as he will never reveal the door with the car this is the key bit. If the door you picked is the correct one (1% chance) the host can open any door. If you picked incorrectly (99% chance) he can open any door except the winning door so the remaining door will be the correct one 99% of the time.
mettamage 11y ago
Thanks for clarifying a very counter intuitive problem. I always struggled with this one, not anymore :) Sometimes scaling things up is the best way to understand things. Who knew!

baddox 11y ago

> For all you know, the one that he removed was just one random one of the goat doors. The key observation is that the problem states that Monty Hall opens an empty door. It's not clear whether Monty knows which doors are empty, but it doesn't actually matter, since the problem states that he will always open an empty door. As others have said, your choice as a contestant is whether to take one door (the one you initially pointed at), or to take the other two doors (at least one of which is obviously empty). Stated that way, it's pretty obvious that taking 2 doors is better than taking 1 door.

pklausler 11y ago

Your simulation is not correct. Look at it this way: 1/3 of the time, you pick the right door up front. If you switch, you lose. 2/3 of the time, you didn't pick the right door. If you switch, you win. So switch.

jinfiesto 11y ago

From the initial setup you have a 1/3 chance of choosing the correct door and a 2/3 chance of choosing the wrong door. The key to this problem is the behavior of the game host as he will never reveal the door with the car. So, after he reveals the contents of one of the doors, if you choose not to switch, you still only have a 1/3 chance of having chosen the door with the car. On the other hand, if you choose to switch, 2/3 of the time you will have chosen incorrectly initially, and under the switching strategy, those 2/3ds of the time you are guaranteed to win the car. Hence, you have a 2/3ds chance of winning under the switching strategy, vs a 1/3d chance of winning under the "staying" strategy. Hope this clarifies things.

logfromblammo 11y ago

The important bit is that Monty will never choose to show you the winning door, and he will always show you a door after you pick a door and before asking you if you want to switch. He always opens a losing door. The location of the prize and the player choice of door are independent events. Monty's choice is dependent on whether the player choice and the prize location are the same. Monty will never open the door the player picked. Monty will never reveal the prize. Those rules generate exploitable information for the player. Here is the complete probability tree, and two possible strategies of play: Prize behind door A; P = 1/3 ALWAYS STAY strategy: Player picks door A; P = 1/3 * 1/3 = 1/9 P(WIN) = 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18 Monty shows Door B; P = 1/3 * 1/3 * 1/2 = 1/18 = 6/18 Player stays with A; WIN = 1/3 Player switches to C; LOSS P(LOSS) = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 Monty shows Door C; P = 1/3 * 1/3 * 1/2 = 1/18 = 6/9 Player stays with A; WIN = 2/3 Player switches to B; LOSS Player picks door B; P = 1/3 * 1/3 = 1/9 ALWAYS SWITCH strategy: Monty shows Door C; P = 1/3 * 1/3 * 1/1 = 1/9 P(WIN) = 1/9 + 1/9 + 1/9 + 1/9 + 1/9 + 1/9 Player stays with B; LOSS = 6/9 Player switches to A; WIN = 2/3 Player picks door C; P = 1/3 * 1/3 = 1/9 P(LOSS) = 1/18 + 1/18 + 1/18 + 1/18 + 1/18 + 1/18 Monty shows Door B; P = 1/3 * 1/3 * 1/1 = 1/9 = 6/18 Player stays with C; LOSS = 1/3 Player switches to A; WIN Prize behind door B; P = 1/3 Player picks door A; P = 1/3 * 1/3 = 1/9 Monty shows Door C; P = 1/3 * 1/3 * 1/1 = 1/9 Player stays with A; LOSS Player switches to B; WIN Player picks door B; P = 1/3 * 1/3 = 1/9 Monty shows Door A; P = 1/3 * 1/3 * 1/2 = 1/18 Player stays with B; WIN Player switches to C; LOSS Monty shows Door C; P = 1/3 * 1/3 * 1/2 = 1/18 Player stays with B; WIN Player switches to A; LOSS Player picks door C; P = 1/3 * 1/3 = 1/9 Monty shows Door A; P = 1/3 * 1/3 * 1/1 = 1/9 Player stays with C; LOSS Player switches to B; WIN Prize behind door C; P = 1/3 Player picks door A; P = 1/3 * 1/3 = 1/9 Monty shows Door B; P = 1/3 * 1/3 * 1/1 = 1/9 Player stays with A; LOSS Player switches to C; WIN Player picks door B; P = 1/3 * 1/3 = 1/9 Monty shows Door A; P = 1/3 * 1/3 * 1/1 = 1/9 Player stays with B; LOSS Player switches to C; WIN Player picks door C; P = 1/3 * 1/3 = 1/9 Monty shows Door A; P = 1/3 * 1/3 * 1/2 = 1/18 Player stays with C; WIN Player switches to B; LOSS Monty shows Door B; P = 1/3 * 1/3 * 1/2 = 1/18 Player stays with C; WIN Player switches to A; LOSS If your simulation does not conform with the mathematically derived result, your simulation is incorrect.

brobinson 11y ago

The table here [1] helped me understand it when I first heard of this problem. [1] https://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solutions https://en.wikipedia.org/wiki/Monty_Hall_problem#Simple_solu...

grahamburger 11y ago

The most clear way that I've heard it described is by scaling it up to 100 doors or so, and then removing 98 doors instead of removing 1. (This is mentioned elsewhere in the thread.) It would be interesting to see the results of your test using this method. It seems intuitively clear with 100 doors that switching is the better method.

SeoxyS 11y ago

This is the one article that made me fully grasp it forever. I've known about this for ages, but it never really clicked until I read it put this way: http://waitbutwhy.com/2016/03/the-jellybean-problem.html http://waitbutwhy.com/2016/03/the-jellybean-problem.html