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 Pick a box! (Posted on 2002-03-28)
You are shown three boxes, and told that one of them contains a prize. You are then asked to pick one box, and if that box is the one with the prize, you will win it. After picking a box, you are shown that one of the other two boxes is empty, and offered a chance to change your selection.

Should you do this? Would changing your choice to the other remaining box affect your odds of winning? Why or why not?

 See The Solution Submitted by levik Rating: 4.2857 (14 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(4): YES! ~33/66! All possibilites shown! | Comment 30 of 42 |
(In reply to re(3): YES! ~33/66! All possibilites shown! by jduval)

I apolgize for yelling, but that's what it seemed like you were doing.  Like Tristan said, the ones listed twice were representative of their likelihood.

Imagine a pie chart with nine pieces of the same size.

1-3, you pick box one.  4-6, you pick box two.  7-9, you pick box three.

1, 4, 7, prize in box one.  2, 5, 8, prize in box two.  3, 6, 9, prize in box three.

Clearly, these have equal probablilty.

Now, divide slices 1, 5, and 9 each into two equal halves.  This represents the possibilities of which box the host reveals.  The other 6 slices are not diveded because the host only has one reveal choice.

In the 6 small slices (equal in total size and probability to three regular slices), changing would be bad.  In the 6 other (regular size) slices, changing would be good.  In 3/9 of the pie, change is bad.  In 6/9 of the pie, change is good.

Again, I'm sorry for yelling, but "NO! 50/50! All possibilities shown!"  seemed to me like yelling.

 Posted by Dustin on 2005-03-03 00:15:47

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