Re: The Box Stops Here
- From: russotto@xxxxxxxxxxxxxxxxxxx (Matthew Russotto)
- Date: Thu, 13 Oct 2005 22:26:14 -0500
In article <XME3f.175$fc7.118@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
Dogstar <dogstarZeroZeroZero@xxxxxxxxxxxxxxx> wrote:
>The Game Show Host showed Bert the four boxes and said, "In one of these
>boxes in the Grand Prize. The other three boxes are empty. Please pick a
>box...".
>The boxes were marked with the letters A, B, C, and D. Bert picked box B.
>The Host (who knew which box contained the Grand Prize) opened box A, which
>was one of the empty boxes, and said, "If you wish, I will let you switch to
>a different box.".
>"Ah ha!", thought Bert, "I've seen something like this in the rec.puzzles
>newsgroup! I can improve my odds of winning if I switch boxes!".
>Bert switched to box C. The Host then opened box D, which was of course
>empty, and said, "I will let you switch one more time. Do you want to
>switch back to the box you picked originally?".
>Bert thought, "Hmmm. I switched away from box B because I knew it would
>improve my chances. Would switching back improve my chances, make them
>worse, or keep them the same?".
>
>If Bert wants the best odds for winning the Grand Prize, should he switch
>back to box B or keep box C? What are the odds?
The way the host behaves is critical, and not enough information is
provided to determine that.
On the assumption that the host always opens two boxes, and picks at
random among the available ones without regard to which one is picked
originally:
His original view of the probabilities
A B C D
1/4 1/4 1/4 1/4
His view after Box A was opened (by the Monty Hall reasoning)
A B C D
0 1/4 3/8 3/8
To figure out the probabilities after the host opens D, use Bayes Theorem
By Bayes
P(B|openD) = P(openD|B) * P(B)/P(openD)
P(B) = 1/4
P(openD|B) = 1
P(openD) = P(openD|B)*P(B) + P(openD|C)*P(C) + P(openD|D)*P(D)= 1/4 +
1/2 * 3/8 + 0 * 3/8 = 7/16
P(B|openD) = (1 * 1/4) / 7/16 = 4/7
P(C|openD) = 3/7
And he should switch back
However, if the host only opens the second box if he can open a box
you haven't ever picked (that is, if D is a winner he doesn't give the player
another choice), things are different. The fact that he ever got into this
situation means he picked a winner at least once, and by our previous
analysis we know he was more likely to have picked it the second time,
than the first. So he shouldn't switch
Quantitatively, using Bayes Theorem again
P(B|openD) = P(openD|B) * P(B)/P(openD)
P(B) = 1/4
P(openD|B) = 1
P(openD) = 5/8 (he'll open D whenever D isn't the winner)
P(B|openD) = 1 * (1/4)/(5/8) = 2/5
P(C|openD) = 3/5
So he'd better not switch back
.
- References:
- The Box Stops Here
- From: Dogstar
- The Box Stops Here
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