Impeccable logic

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But the single attempt scenario is a hypothetical one, because I wouldn't cope with being in the company of Noel Edmonds without punching him and being dragged off him by several heavies.
So you've met the **** aswell ;)
 
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Apparently Noel wrote to the Cosmos for a woman by the end of 2005, and indeed one came. She didn't hang around for too long, though.
 
There is a random and equal chance that any two boxes will be left. One on the table and one in the room. Random chance doesn't favour any of the boxes so the two that are left will have an equal chance of being any two boxes. It may be £250K and £100 or any other combination. Therefore there is no greater chance by swapping the boxes.
 
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isnt it a game of choice rather than chance, therefore the choice of box may be swayed by subtle differences, maybe the box is a slightly lighter shade or the lighting is slightly different or a physchological influence of some sort could stack any odds one way or t'other.
Not suggesting for one minute that a devious producer would resort to such dirty tactics ;)
 
As much as it pains me to say this ... Joe-90 is absolutely correct, the chance is 50:50 and it matters not whether you swap ... Chance will be the same.

MW
 
MW i've been saying that for 2 days... probability can't be swayed.... it can only be 50:50, whatever Noel says
 
cant believe no-ones thought of the possibility of the coin landing on its edge!! surely thats gotta be taken into account here! :LOL:
 
imamartian said:
but 18 of those 19 times there won't be £250k in the box
There is in the scenario that Kes has defined, viz:

...by a combination of blind chance and good fortune, whittle the lot down to two boxes, yours and one out there. The boxes have in them £250,000 and £1.
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Thermo said:
why will 19 out of 20 times the prize be in the other box.
It's the conventional way of computing and expressing a probability of this type.

19/20 is the probability of the average number of times that a given prize will appear in the set of boxes not chosen by the contestant, if the sample size is large enough to eradicate anomolies.
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joe-90 said:
Random chance doesn't favour any of the boxes so the two that are left will have an equal chance of being any two boxes.
Not in the scenario that Kes has defined.
 
19 times out of 20 (of the times that only two boxes are left unopened), the desired prize will be in one of the boxes that isn't the 'owned' box.

Therefore if you don't swap you will win that desired prize only once in 20 attempts.

correct

Therefore if you swap you will win that desired prize 19 times in 20 attempts.

Err, no.

I agree that, at the outset of the show, the player has odds of 19-1 of holding the desired prize. However, the odds reduce sequentially to 18-1, 17-1, 16-1, etc, as the boxes are opened. Each time a box is opened it is a game in its own right. We are told that the desired prize is still in the show with only 2 boxes left. Therefore, the odds in the final game have reduced (as explained) to 1-1, which is evens.

Don't make the mistake of thinking its only one game. It isn't. It is one show with 19 games of reducing odds.
 
Kes you don't have a pet bird called moz do you?
I'm a bit too dim to get the joke here, unless it's a pun on Cosmos (Kesmoz?), but no.

I wonder why so many posts to this thread happen after the pubs have closed?

After some research, it's the knowledege that Monty Hall has that determines his actions, and modifies the probability. Noel has neither the knowledge nor the capability of chosing a box to open.

Rgds.
 
kevnurse said:
I know.

Therefore if you swap you will win that desired prize 19 times in 20 attempts.
Err, no.
Perhaps you misunderstand the word "attempts". Perhaps not. Either way, you're wrong.

I agree that, at the outset of the show, the player has odds of 19-1 of holding the desired prize. However, the odds reduce sequentially to 18-1, 17-1, 16-1, etc, as the boxes are opened. Each time a box is opened it is a game in its own right.
That theory is flawed.

We are told that the desired prize is still in the show with only 2 boxes left.
Indeed so.

Therefore, the odds in the final game have reduced (as explained) to 1-1, which is evens.
That's a non sequitur.

Don't make the mistake of thinking its only one game. It isn't. It is one show with 19 games of reducing odds.
That isn't the correct way to compute the probability of a given prize being in the singled-out box. But you've certainly cottoned on to the popular way of getting it wrong.
 
Softus. Random chance doesn't have memory or favour one box over another. Any box could be the box on the table. Substituting any other box won't change that basic principle. Random chance is - well - random. ;)
 
This is all getting a bit boring and pointless, don't you all think?
 
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