Home Forum Chat Forum The Monty Hall Problem – a simple maths/probability problem for you all……..

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• The Monty Hall Problem – a simple maths/probability problem for you all……..
• Rob Hilton
Member

Go from 2:40

miketually
Subscriber

Am I missing the point here?

Yes?

Surely there are now 2 doors, and there is no more likelihood that the car is behind one or the other? So, how do you increase your chances of winning by switching to the other door with equal probability of having a car behind?

There’s a 2/3 chance that there will be a car behind one of the two doors that you didn’t choose first. Effectively, you’re being given the chance to choose both those doors instead of your first chance.

So, you’re twice as likely to win the car if you switch.

GrahamS
Subscriber

Am I missing the point here? Surely there are now 2 doors, and there is no more likelihood that the car is behind one or the other?

Well if that messes with your head then try this old thread for a probability reality check ðŸ˜€

Cougar
Subscriber

Am I missing the point here? Surely there are now 2 doors, and there is no more likelihood that the car is behind one or the other?

If the premise was simply “one of these doors has a car behind it, which one?” then you’d be right. But that’s not what’s been set up.

You’re presented with two doors. However, the odds of winning have been stacked by previous events; ie, the two doors are not equal.

At the start of the game, you pick a door and have a 33% (rounded) chance of being a winner, yes? Think about that. You can now do what you like with the other doors, add more, remove one, paint them pink, set them on fire, write “here be goats” on them; it does not, can not change the likelyhood that your door contains a car or a goat – it will always be 33%, that was the choice you originally made.

In the second round, you’re given another option. You can keep your existing choice (33% chance of being right, remember) or you can trade that choice for another door which, now that we’ve removed an obvious wrong choice, represents all other odds other than the one you already hold. 100% – 33% = 67%.

TrekEX8
Member

Thanks Cougar, however…..at the beginning, the chance of winning was1/3. If the host opened two doors, to reveal goats, would the odds still remain at 1/3 that opening the remaining door would reveal a car?

Farticus
Subscriber

Simplest to illustrate by going through all the options.

Assume you pick the one in bold in each of the 3 options below. Whatever you choose, Monty opens a goat* door (the one crossed out). So, sticking means you win 1/3, switching means you win 2/3.

G G C becomes G G C. Stick = Win. Switch = Lose.

G G C becomes G G C. Stick = Lose. Switch = Win.

G G C becomes G G C. Stick = Lose. Switch = Win.

As far as i know this should work with sheep, cows or anything else. I’m surprised questions weren’t asked about Monty’s interest in goats.

I_did_dab
Subscriber

this was demonstrated on BBC Horizon with toy cars and toy goats. One player always changed the other always stuck with the original choice. After 100 goes the pile of cars for the always changes was much bigger than the always stays in line with the probability theory…

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