Pure probability question

[gaming_mouse]

[ QUOTE ]

I presume you mean the statement above.

The part of the n-dimensional hypercube below the hyperplane x1+x2+...+xn=1 is a pyramid over the n-1 dimensional case. This has an n-dimensional volume of 1/n times the (n-1)-volume of the base, so the volume is 1/n! by induction.

There is also a volume-preserving linear transformation that takes the part with sum less than or equal to 1 to the part of the unit n-cube with x1<x2<x3<...<xn:

x1' = x1
x2' = x1+x2
x3' = x1+x2+x3
...
xn' = z1+x2+x3+...+xn.

This part of the unit n-cube has volume 1/n! by symmetry, as there are n! possible orderings of the coordinates.

[/ QUOTE ]

Thanks pzhon. Both of those solutions are nice, especially the second.

But why did you write out the expansion for e, as though the statement below it followed directly from that expansion (in your original answer in white)?

[pzhon]

[ QUOTE ]

But why did you write out the expansion for e, as though the statement below it followed directly from that expansion (in your original answer in white)?

[/ QUOTE ]
# draws necessary = Z1 + Z2 + Z3 + ...

where

Zi = 0 if the ith draw was not necessary
Zi = 1 if the ith draw was necessary (which happens with probability 1/(i-1)! )

The expected number of draws necessary is the sum of the probabilities of requiring the first draw (1/0!), the second draw (1/1!), the third draw (1/2!), etc.

[gaming_mouse]

ahhh... i got it. i didn't realize that was the expectation. i thought the statement below it was supposted to follow from it, rather than the other way around.

thanks,
gm

[bobman0330]

it is not immediately (or, to me, eventually) apparent why P(requiring draw N) = 1/(n-1)!

[gaming_mouse]

[ QUOTE ]
it is not immediately (or, to me, eventually) apparent why P(requiring draw N) = 1/(n-1)!

[/ QUOTE ]

check out pzhon's responses to my questions.... he gives two different proofs.