Ask the Wizard #42 roulette correction

removed
silly

This was addressed in another thread.

DorothyGale started that thread this way:
"While I was milking the pigs today, I thought of a little problem..." LOL

The thread you are looking for is
HERE

Thanks.
removed
silly

Quote: DorothyGale

For example, with N = 2, the answer is 2*(1/2) + 3*(1/4) + 4*(1/8) + 5*(1/16) + ...
--Ms. D.

MustangSally

To be really honest you haven't derived the formula, you just plugged in.
You should express the answer as an infinite sum first, since that is the natural outcome. Then you should derive the answer as a closed form expression.

For example with a coin the solution is:
2*( Probability you get both head and tales in 2 ) +
3*( Probability you get both head and tales in 3 ) +
4*( Probability you get both head and tales in 4 ) +
...
i*( Probability you get both head and tales in i ) +
...
infinity

But
i*( Probability you get both head and tales in i ) = i*p^(i-1) where p = 1/2

S== infinite sum from i=2 to infinity of [ i*p^(i-1) ]

S = 2*p + 3* p^2 + 4* p^3 + ... [equation 1]
S/p = 2 +(3* p^1 + 4* p^2 + ...) [equation 2]

S/p -S = 2 +p + p^2 + p^3+ ... [equation 3]=[equation 2]- [equation 1]

Define GS as a standard Geometric Sum
GS = p + p^2 + p^3+...
GS/p -1= p +p^2+...=GS
GS - p = p*GS
GS= p /(1-p) [equation 4]

S/p -S = 2 +p/(1-p) [equation 4] into [equation 3]
S(1/p-1) = (2-p)/ (1-p)
S(1-p)/p = (2-p)/ (1-p)
S = p*(2-p)/(1-p)^2 [algebraic simplification]
when p= 1/2 then S=3