Open Test..., A "mathematical" problem Options

[Gorath]

Hi,

Sometime ago we had a discussion about open rolls (like a stealth test) here on the dumpshock boards.

We developed a formula to calculate the expected value at an open roll.

I.e. for 1 die the expected value is E(1)=4.2
You get this from:
E(1)=1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 1/6*(E(1)+6) ==> E(1)=21/5=4.2

You can continue this, if E(2) is the expected value rolling 2 dice you get:

E(2)= 1/36*(E(2)+6)+ 10/36*(E(1)+6)+1/36*95 ==> E(2)=29/5=5.8

If you rewrite using Binomial factors you get:

E(2) = B(2,2) * (1/6)^2 * (5/6)^0 * (E(2)+6) + B(2,1) * (1/6)^1 * (5/6)^1 * (E(1)+6) + (1/6)^2 * R(2)

with R(x):= (Sum from k=1 to 5) (k^x-(k-1)^x)*k


==>

E(x):=B(x,x)*(1/6)^x * (5/6)^0 * (E(x)+6) + B(x, x-1) * (1/6)^(x-1) * (5/6)^1 * (E(x-1)+6) + B(x, x-2) * (1/6)^(x-2) * (5/6)^2 * (E(x-2)+6) + ... +B(x,2) * (1/6)^2 * (5/6)^(x-2) * (E(2)+6) + B(x,1) * (1/6)^1 * (5/6)^(x-1) * (E(1)+6) + (1/6)^x * R(x)

My question is:

(a) Can someone remember this discussion and has a link to the post?

(b) Is anyone good in math and can give a expicite formula (insteed of this recursive one) to calculate E(x)?