Game Show

[HotStuff]

You are a contestant on a game show and you’ve won thru to the final stage. The host shows you 3 doors, behind one of them has been randomly placed tonight’s star prize. The host asks you to choose a door. Once you have done so he opens one of the other two doors and shows you that there is no prize there. He then offers you a choice, to stick with the door you have already chosen or change your choice to the other unopened door.

What should you do to maximise your chances of winning? Should you stick with your chosen door? Should you change to the other unopened door? Or doesn’t it make a difference?

[OllieMartin]

I think I know this one, but I don't want the first response to be the correct answer, it kills the thread and I like it

[zazen999]

I know as well

[OllieMartin]

Can I say, can I say?

[HotStuff]

It's an old chestnut that was popular probably 10 or more years ago, I remember it causing much heated discussion in the office where I was working at the time

[OllieMartin]

My answer:

When you first guess you've got a 1/3 chance of getting the right door, so odds are you will guess wrongly.
In that case, one of the two remaining doors is the correct answer, and the host will open the one remaining incorrect door. It therefore stands to reason that the one remaining door (not the one you chose, not the one the host chose) is the correct door.

Therefore, you are best to change your mind to the remaining door to maximise your chance of being correct. This only fails 1/3 of the time when you pick the correct door straight away rather than failing 2/3 of the time by sticking with your first choice.

Voila

[FionaH]

That was on Numbers the other night. Only way I would have heard of it

[binley100]

waaaay too much for my brain to cope with

[zazen999]

That was on Numbers the other night. Only way I would have heard of it

On Pumpkin Becki?

[HotStuff]

That was on Numbers the other night. Only way I would have heard of it

As in Numb3rs the cop show? I've never seen it, is it good?

[Caro]

I'm sure my old statistics teacher would have disagreed

Previous events do not necessarily influence the probability of the next outcome. For example, if you toss a coin 5 times and each times it comes out "heads", the probability of "heads" the next time is still only 50% (let's not get pedantic about wighted coins or expert coin throwers ).

Thus with the given example, if you have two doors remaining, and the prize was placed randomly behind one of them, it's a 50:50 chance which one it's behind.

Ot at least, that's my take on it

[HotStuff]

But surely when the prize was placed there was a one third chance of it being placed behind each door. Whichever door you first choose there will always be atleast one without a prize behind it for the host to open. If when you first choose there is a one third chance of you being right then there is a two thirds chance of you being wrong. That isn't changed by the host opening a door, so if you stick with the door you first chose there's a one third chance your right. If you move to the other unpoened door there's a two thirds chance your right.

[Caro]

Going back to my example then.

The chances of throwing 5 "heads" in a row befors you even toss the coin is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32.

The chances of throwing 6 "heads" in a row befors you even toss the coin is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/64.

However, after you've thrown the first 5 heads, the chances of throwing the 6th is still 1/2. The chance of it being a head just is not influenced by previous events.

Same with the doors

Have I explained or just confuddled ?

[OllieMartin]

Going back to my example then.

The chances of throwing 5 "heads" in a row befors you even toss the coin is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/32.

The chances of throwing 6 "heads" in a row befors you even toss the coin is 1/2 x 1/2 x 1/2 x 1/2 x 1/2 x 1/2 = 1/64.

However, after you've thrown the first 5 heads, the chances of throwing the 6th is still 1/2. The chance of it being a head just is not influenced by previous events.

Same with the doors

Have I explained or just confuddled ?

What you say makes sense, but what you're not factoring in is the fact that the host shows you what's behind one of the doors. This alters the odds dramatically.
Imagine the odds of tossing the last "heads" if someone could show you for certain that it wouldn't land on tails!

[Caro]

What you say makes sense, but what you're not factoring in is the fact that the host shows you what's behind one of the doors. This alters the odds dramatically.
Imagine the odds of tossing the last "heads" if someone could show you for certain that it wouldn't land on tails!

This is fun.

Yes, but it alters the odds from 1/3:1/3:1/3 to 1/2:1/2

I agree before a door is shown to you, there is a 1 in 3 chance you have picked the right door. But I still maintain that when one has been eliminated, that elimination does not make your choice anything other than 1 in 2 of the 2 remaining doors. You're left with 2 doors and it's an equal chance the prize is behind either, therefore that must be 50:50.