# Re: MathBard's bags

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Posted by Joel on November 01, 2002 at 10:41:51:

In Reply to: Re: MathBard's bags posted by Denis Borris on October 31, 2002 at 23:41:08:

: : Sorry - no unnerstand problem. He expects to count 500 less than what?

: With his 4 bags in the total, he'd EXPECT to count
: X-1 bags before his 4th bag appears, it being the
: Xth bag...or his 4th bag would be the Xth bag
: coming down the carousel (or whatever you call
: that rolling thing!).

: But he's only got 3 bags; so the 3rd bag is now
: EXPECTed to be the Yth bag; X - Y = 500
: Btw, there's a simple formula for this:
: m(n+1) / m+1
: n = total number of bags

OK, well that kind of gives the whole thing away (9984).
That's a very interesting little formula. Taking position number in a queue to be the variable X, that gives the expected value of X in one easy step. The long way: if you had 2 bags out of a total of 5, they could be arranged 5C2 ways: 5!/2!*3! = 10 ways.
(1,2) (1,3) (1,4) (1,5) (2,3) (2,4) (2,5) (3,4) (3,5) (4,5). So X=2 in 1 of 10 arrangements, X=3 in 2 out of the 10, X=4 in 3 out of 10 and X=4 in 4 out of 10. Assuming each position is equally likely, the expected value of X is 2*.1 + 3*.2 +4*.3 + 5*.4 = 4. And your formula rolls all of that into one just: 2*6/3 = 4. Neat.

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