The solution
Dec. 14th, 2004 02:01 pm![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
yendiWarning: This is the solution as presented by a liberal-arts person who hasn't taken a math class in over twelve years. So I may have left gaps unfilled.
Okay, let's go over the one fact that's not likely to be disputed:
After picking your initial box (which we'll call Box A), you have a 1/3 (or roughly 33.3%) chance of having picked the $5000.
That means, of course, that there is a 2/3 or ~67% chance of that money being in one of Box B or Box C.
Now, remember (as![[livejournal.com profile]](https://www.dreamwidth.org/img/external/lj-userinfo.gif)
trochee emphasized), the Prizemaster knows where the money is.
He removes a box (we'll call it Box B) that was known not to contain any money, and never did.
Box A still retains a 33% chance of being the correct box. That's because it was picked before the state of Box B was known.
But remember -- the chances of the money being in one of the other two boxes were 2/3. Box B is out of the way. That means that the odds for the money being in Box C (which is, of course, "one of Box B or Box C" remain 2/3.
Find this troubling?
The argument that most folks have with this is best articulated byforce_of_will in this comment and the follow up. And had the Prizemaster been choosing whether to remove a box, or if the PM was choosing randomly and happened to pick an empty one, this logic would be correct.
Let's pretend there are 100 boxes. You pick one. Then the Prizemaster, knowing where the money is, removes 98 of the remaining boxes which contain no money. He again offers you the chance to switch. Still think the box in your hand, at 1%, is the way to go?
(that last example is swiped from Raymond Smullyan, btw)
Okay, let's go over the one fact that's not likely to be disputed:
After picking your initial box (which we'll call Box A), you have a 1/3 (or roughly 33.3%) chance of having picked the $5000.
That means, of course, that there is a 2/3 or ~67% chance of that money being in one of Box B or Box C.
Now, remember (as
trochee emphasized), the Prizemaster knows where the money is. He removes a box (we'll call it Box B) that was known not to contain any money, and never did.
Box A still retains a 33% chance of being the correct box. That's because it was picked before the state of Box B was known.
But remember -- the chances of the money being in one of the other two boxes were 2/3. Box B is out of the way. That means that the odds for the money being in Box C (which is, of course, "one of Box B or Box C" remain 2/3.
Find this troubling?
The argument that most folks have with this is best articulated by
force_of_will in this comment and the follow up. And had the Prizemaster been choosing whether to remove a box, or if the PM was choosing randomly and happened to pick an empty one, this logic would be correct.Let's pretend there are 100 boxes. You pick one. Then the Prizemaster, knowing where the money is, removes 98 of the remaining boxes which contain no money. He again offers you the chance to switch. Still think the box in your hand, at 1%, is the way to go?
(that last example is swiped from Raymond Smullyan, btw)