would you swop?

[robsan77]

I have 3 boxes and within 1 of them is something with value, which you are trying to find.

You can pick 1 box,

I then open and show you 1 of the remaining boxes I know doesnt contain the item

You can now either stick with your original choice or swop to the other box.


what do you do and why???

:hmm: :hmm: :hmm:

 

[Ceejay]

Swap. There is a higher probability of choosing the article of value. I won't give away my "answer" yet though

 

[Cussword]

Have to agree, there's a 50/50 chance with the swap :thumb:

 

[Tony]

Stick with my original choice.

I'm bloody minded.

And I've got a beard.

 

[robsan77]

Have to agree, there's a 50/50 chance with the swap :thumb:

Its not 50:50

 

[Ceejay]

Indeed. The chances are 2/3

 

[Springer]

Though a wife may be involved here, from the title :lol: I'm off this sounds like that game that my in-laws watch every day. :)

 

[richc]

Swap. There is a higher probability of choosing the article of value. I won't give away my "answer" yet though

Yup, it's a pretty well known probability example.

 

[31bb3]

Personally i have the attitude S#!t or bust so id stick with my initial choice

 

[Springer]

Personally i have the attitude S#!t or bust so id stick with my initial choice

Are we back to wife's again :lol:

 

[commsbiff]

Took me a minute or two to figure out why it wasn't 50/50 and the swap would be better. Nice question :thumb:

 

[Doh]

:rofl:

 

[Cussword]

Its not 50:50

Ok, I'm thick (& I've got a beard too), if youve opened one of them, there must only be two left & I have a choice of the two the one I have or the one I can swap. Is that not 50/50?

 

[commsbiff]

I think it's 1/3 chance of winning if you stick with what you've got, or it is 2/3 chance if you swap. My reasoning (right or wrong):

You choose A from a choice of A, B and C.
That means A has a 1/3 chance of winning.
B and C also have 1/3 chance each.

So A is 1/3, the total of evreything that's not A (ie B+C at this point) is 2/3.
The only thing that will change is that you will remove a zero chance (because you know it's empty) from the 'not A' side. So the remaining B or C (it doesn't matter which) is 2/3 chance.

Am I right? :geek:
Or horribly embarrassed :cry:

 

[Sean_Mc]

:wha: could you just not punch Noel Edmonds in the face and run off with the two boxes

 

[graysalchemy]

Though a wife may be involved here, from the title :lol: I'm off this sounds like that game that my in-laws watch every day. :)

:rofl: :rofl:

 

[Tony]

could you just not punch Noel Edmonds in the face

:shock:

:rofl:

 

[Ceejay]

I think it's 1/3 chance of winning if you stick with what you've got, or it is 2/3 chance if you swap. My reasoning (right or wrong):

You choose A from a choice of A, B and C.
That means A has a 1/3 chance of winning.
B and C also have 1/3 chance each.

So A is 1/3, the total of evreything that's not A (ie B+C at this point) is 2/3.
The only thing that will change is that you will remove a zero chance (because you know it's empty) from the 'not A' side. So the remaining B or C (it doesn't matter which) is 2/3 chance.

Am I right? :geek:
Or horribly embarrassed :cry:

100% correct mate :

I'd also be up for subjecting a certain gameshow host to a blow upside the chops. :lol:

 

[Beardy Wierdy]

Well Noel Edmunds is a beardy wierdy so i would hope he has some sympathy for me and gives me a subtle hint.

 

[Cussword]

Doh :?