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

poop example gonefishin…. there are millions of combinations on the lottery to simplfy it to win or lose isn’t really the same

Crispo your basing it on the fact of just looking at the two doors.
Think of it this way, the host has to show you 998 goats. Before he show’s you them either he has the car in the 999 doors he has or you do in your 1 door. What is more likely?

all the pies, same principal but even more important to switch

No need to tell me 🙂 I understand the need to swap, my extrapolation example was an attempt to convince the errr, unconvinced 😉

If anyone doesn’t believe you should switch, I’ve a tenner here and three cups. Bring your own tenner and we’ll play.

One tenner three cups, it’s the next Internet sensation!

gravity sucks – I don’t know, which is more likely??
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? So, how do you increase your chances of winning by switching to the other door with equal probability of having a car behind?

Go from 2:40

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.

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 😀

This too

http://www.singletrackworld.com/forum/topic/monday-evening-puzzle

😀

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%.

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?

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.

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…