OT: Statistical question

[whisky-dave]

yeah, but you won't get your goat ;-(

[Man at B&Q]

Three doors A (with the prize), B and C with no prize

You have three initial choices: (1a) You pick A, host discards B. If you stick you win, switch and you lose. (1b) You pick A, host discards C. If you stick you win, switch and you lose.

(2) You pick B, host discards C. If you stick you lose, switch and you win.

(3) You pick C, host discards B. If you stick you lose, switch and you win.

Out of your three initial choices, A, B or C, subsequently sticking has one chance of wining (after choosing A) but two of losing (after choosing B or C). Switching has two chances of winning (if you chose B or C) but one of losing (if you chose A).

It is not 50:50 as the game includes knowledge of what happened before it got to the choice between stick or switch.

I can't make it any simpler for you.

MBQ

[Geoff Clare]

Two were primed to fizz; one was safe.

That's what James said, and he pointed out that the ratio between

50% and 33.3% agreed with the result they got (60 wins, 40 losses) but unfortunately it's wrong. By switching, your odds go up to 2/3, not 1/2 (as several others in the thread have explained). Their result was still reasonably close to the theoretical 66.7 wins out of 100.

It's a shame the Man Lab producer didn't get hold of an actual mathematician for James to ask to explain it.

[whisky-dave]

yes but what is the proabliity of a person picking 2 goats rather than 1 ca r and one goat. 2/3 X 1/2 compared with 1/3 X 1/2 so by changing doors but this is overc two choices where in reality it's 50/50 and it doesn;t ma tter how many doors are opened, but we only know this because in the end on ly 2 doors will be left.

[John Rumm]

I think this is the bit that is misdirecting you...

The significant point is that the game has not changed. The odds that your door is right are still 1 in 3. The odds that one of the other two is right are still 2 in 3.

Even when he shows you one that is wrong, the odds of one of the doors you did not choose being right, is still 2 in 3.

So as soon as he opens one of them, you now have a choice of stay with your 1 in 3 chance door, or switch to the only remaining of the 2 in 3 chance doors.

Its easier to think it through intuitively if you rerun with more doors. Say 1000. In that case you chose a door randomly, and get a 1 in 1000 chance of being right, and a 999/1000 of being wrong. You in effect have two sets of doors. Yours with the 1/1000 chance, and the hosts with a

999/1000 chance. So you know with a very strong probability that one of the host's set of doors contains the prise rather than yours.

He then eliminates all but one of the doors in his set. Remember that the odds situation remains the same - your set has a 1/1000, and his set has a 999/1000. So the choice then becomes stay with your 1/1000 chance, or switch to the set with a 999/1000 chance. One assumes that your would not choose any of the 997 he has opened and shown to have no prise.

[Dave Liquorice]

But you are not picking again, you have already picked once and have an option of sticking or switching. Also the host has thrown out a losing choice.

[Tim Streater]

Indeed, and for it to be a theory it has to have been backed up by experiment (otherwise it's just a hypothesis). It only takes one contrary result to demote a theory.

[Steve Firth]

Roger Mills wrote: [snip]

You can play the game yourself and see the outcome.

[Andy Burns]

Several posters (who get it) seem to think that it's more obvious with a larger number of doors, but I think (to those who haven't got it yet) it's no more obvious, or even confirms their incorrect logic ...

Certainly nobody has had a Bisto moment in response to being told to think about more doors.

[Man at B&Q]

Try teling that to the Warmistas.

MBQ

[jgharston]

The important point - which is often not phased very clearly - is that the host *ALWAYS* opens another door.

JGH

[Tim+]

yet... ;-)

They might have but are now just keeping quiet.

Tim

[John Rumm]

I was hoping that putting it into the context of "sets" may help - the bigger numbers make the probability difference between the two sets far more stark.

Think of it in terms of a lottery where you buy one ticket, and someone else buys *all* of the others. So its now absolutely certain that one of you has the winning ticket. Its almost certain that the one with the large set of tickets will be that one. If they then check *all* of their tickets for a winner, and then show you all but one of their tickets, and they only show you losing tickets, their set is still the one with a n-1 in n chance of winning, and yours is still the 1 in n set.

I guess we will have to see ;-)

[John Rumm]

your system now ?.

Except you don't know your guess so far has been correct. All you know is that your first guess had a 1/1000 chance of being correct, and the doors you did not choose had a 999/1000 chance of being correct.

[harry]

It makes no difference whether you change your choice or not.

[Tim+]

is the wrong answer....

Tim

[harry]

Exactly how I see it.

[Clive George]

You, Roger and NT. You're all wrong. Try it and see.

[Steve Firth]

Is the answer that an idiot would give.

And an idiot did.

[John Rumm]

Indeed, and it touches on information theory discussions we have had in the past about baud rates etc.