Nice bit of maths on the BBC

[Sean Hewitt]

Greg - It's all about 'conditional' probability. In the Monty Hall problem, you are right that the chance of the car being behind either of the remaining doors is 50/50 if we know nothing else. That's basic probability. The trick is that we do actually have a little bit more information than that. We know that the host knows where the car is, and that changes things.

If you picked the car originally (a 1 in 3 chance remember) then the host showing you a goat is irrelevant as both of the doors he could have opened have a goat behind them. Switching in the scenario is a bad choice.

BUT, if you picked a goat originally (which you will have done 2 times out of every 3) then only one of the remaining doors hides a goat, and the other the car. The host knows where the car is and shows you the goat. This is not random, this is extra information. This means that the door he does not show you must have the car behind it. So in this scenario you should switch.

That is why switching is the right choice. 2 out of 3 times you will have picked a goat at the start and if you did, the host will effectively tell you which door hides the car by eliminating the other goat.

[John McKenna]

Hi Sean (and Greg), in The Man Who Loved Numbers, Paul Hoffman says Marilyn vos Savant explained the upshot of the Monty Hall problem like this - when you switch you (potentially) win twice out of three and lose once but when you don't you (potentially) win only once and lose twice.

My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home. (Monty Hall)

[Greg Breed]

That's a bit more helpful Sean thanks. I think I'm getting there

[Paul Dargan]

I've had to explain the 'restricted choice' version of this to a few bridge players over the years and as others have said adding doors/card helps.

Imagine being asked to pick 1 card from a deck and not look at it. You're hoping to pick the Ace of spades to win a prize (car). All of the other cards/doors are goats. Now you have one card with a 1/52 chance of winning and the host has 51 cards with 51/52 chance of winning.

Now the host looks at his cards one by one, checking each one to make sure it's not the Ace of spades(car) and then showing you "look I've got another goat". Eventually the host has only one card remaining, the same as you. Do you want to swap?

Well the chances of originally picking the ace of spades for you card haven't changed - that's still 1/52. And the host's chances haven't changed - that's still 51/52 but now rather than that being spread across 51 cards it's concentrated in a single card - the one you are being given the chance to swap for!

I hope that helps some who might not be quite over the line with the problem yet, it certainly helped me to get my head round it.

Paul

[Steven Gardner]

The easiest way I find of thinking about it is using a flow chart to map out the six options. Say the car is behind door A

Pick A > Swap > Goat
Pick A > Stick > Car
Pick B > Swap > Car
Pick B > Stick > Goat
Pick C > Swap > Car
Pick C > Stick > Goat

So by always swapping you win 2/3 times. Simple, right?

[Greg Breed]

Thanks Paul and Steven, that makes even more clear. Hopefully it will sink in now

[MartinCarpenter]

The problem with that sort of very simple thing is that it does break down for sundry possible behaviours of the host You really do have just the two cases left, its just that the chance of reaching one of them is reduced under the normal assumptions.