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I do the same thing by expanding it out to more doors, but when talking about it or explaining it to others (great cocktail party discussion btw), I use 100 doo

I do the same thing by expanding it out to more doors, but when talking about it or explaining it to others (great cocktail party discussion btw), I use 100 doors, to make the percentages easy to follow. Example - there are 100 doors, you pick 1 that has the grand prize. There's a 1% chance you picked the prize on your selection. Monty Hall eliminates 98 of the doors, meaning there are 2 left. There's STILL a 1% chance you picked the door with the prize, but that means there's a 99% chance the only remaining door has the prize. Obviously, you should switch.

Scriptor 13y ago
Ah, I used one million because it really hammers the scale of it in, but I like that with 100 you can easily track the percentages.
Jtsummers 13y ago
Another way is to take a step back and forget about revealing the goats. You have your initial selection (using your numbers) with a 1% chance of picking the prize. Now you're presented the option to leave it shut and open all other doors keeping the prize if it's in any one of them. So if the prize has equal odds of being in any given door, then opening 99 of them gives you a 99% chance of finding the prize versus the 1% chance of only opening one door. This has helped some people I've talked to realize that the odds the prize is in that remaining door (in the proper Monty Hall problem) is actually the sum of the odds it's in all the doors you didn't select. EDIT: I've also demo'd this using 3 cards (an ace and two others), but I like the idea of demonstrating with a larger scope. Call the ace of spades the prize, let the player select one card from the deck at random. Now they can keep that card or scour the remaining 51 cards for the ace of spades. A good tactile demonstration that they have a 1/52 chance (~2%) of having selected correctly the first time versus ~98% that they were wrong.
jwr 13y ago
What I find hard to undestand is why I should switch instead of randomly choosing again?
- Jtsummers 13y ago
  Consider the 3 door case, it's narrowed down to 2 doors. The one you selected has a 1/3 chance of being correct, the one you didn't has a 2/3 chance. If you randomly select between the two (50/50 odds) you come out with a 50% chance of picking the correct door. If you strictly switch, your odds are better, and if you stay the same the odds are worse.
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- adamio 13y ago
  Random choice would be optimal if the host didn't know the location of the car. If the host is blind & car wasn't revealed already, there's 1/2 chance. If the host knows & the car is hidden, then the remaining door has 2/3 chance

Scriptor 13y ago

Ah, I used one million because it really hammers the scale of it in, but I like that with 100 you can easily track the percentages.

Jtsummers 13y ago

Another way is to take a step back and forget about revealing the goats. You have your initial selection (using your numbers) with a 1% chance of picking the prize. Now you're presented the option to leave it shut and open all other doors keeping the prize if it's in any one of them. So if the prize has equal odds of being in any given door, then opening 99 of them gives you a 99% chance of finding the prize versus the 1% chance of only opening one door. This has helped some people I've talked to realize that the odds the prize is in that remaining door (in the proper Monty Hall problem) is actually the sum of the odds it's in all the doors you didn't select. EDIT: I've also demo'd this using 3 cards (an ace and two others), but I like the idea of demonstrating with a larger scope. Call the ace of spades the prize, let the player select one card from the deck at random. Now they can keep that card or scour the remaining 51 cards for the ace of spades. A good tactile demonstration that they have a 1/52 chance (~2%) of having selected correctly the first time versus ~98% that they were wrong.

jwr 13y ago

What I find hard to undestand is why I should switch instead of randomly choosing again?

Jtsummers 13y ago
Consider the 3 door case, it's narrowed down to 2 doors. The one you selected has a 1/3 chance of being correct, the one you didn't has a 2/3 chance. If you randomly select between the two (50/50 odds) you come out with a 50% chance of picking the correct door. If you strictly switch, your odds are better, and if you stay the same the odds are worse.
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adamio 13y ago
Random choice would be optimal if the host didn't know the location of the car. If the host is blind & car wasn't revealed already, there's 1/2 chance. If the host knows & the car is hidden, then the remaining door has 2/3 chance