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Mar
9
comment Winning a restricted game of Nim?
Would it remain $n ~mod ~3$ if I were allowed to remove, say, 1 or 3 sticks from each pile (but not 2)?
Mar
9
comment Winning a restricted game of Nim?
ah, ok. I think I get it now. So let me double check. I first perform $(n ~mod ~3)$ where $n$ is the number of sticks in each pile. THEN I take the digital sum on the result?
Mar
9
comment Winning a restricted game of Nim?
Ok, but suppose I start with a different set of piles. Say I start with three sticks in pile 2 instead of 4 (and everything else remains the same)? I can't have fractional Grundy numbers.
Mar
9
comment Winning a restricted game of Nim?
ok, so the first pile would become a pile of zero, second and third would become a pile of 1, and the last one would become a pile of 2?
Mar
9
comment Winning a restricted game of Nim?
So I would convert my piles to base 3 then? Or take $mod~3$ after taking the digital sum?
Mar
9
revised Winning a restricted game of Nim?
added 3 characters in body
Mar
9
comment Winning a restricted game of Nim?
Yea, my mistake; I typed too fast...I'm so ashamed at myself for being a computer scientist haha.
Mar
9
asked Winning a restricted game of Nim?
Mar
8
accepted Determining Grundy Numbers for an inverted takeaway game
Mar
8
asked Determining Grundy Numbers for an inverted takeaway game
Mar
8
awarded  Teacher
Mar
8
revised Using Poisson Distribution
added 216 characters in body
Mar
8
answered Using Poisson Distribution
Mar
4
comment Conditional expectation of number of dice rolls
Now that I'm looking at it again, I'm having a harder time understanding where $EX$ came from than where $\mu_k$ came from. The recursive nature if $EX$ is hard to grapple
Mar
4
comment Conditional expectation of number of dice rolls
Can you elaborate more on how you came up with that expression for $\mu_k$? It does work, but I don't really understand where you got it from
Mar
3
awarded  Critic
Mar
2
comment Conditional expectation of number of dice rolls
And I'm using Introduction to Probability Models by Sheldon Ross
Mar
2
revised Conditional expectation of number of dice rolls
added 18 characters in body
Mar
2
comment Conditional expectation of number of dice rolls
Whoops, I'm sorry! I typed too fast...X and Y are random variables, not events. I'll fix that right now.
Mar
2
comment Conditional expectation of number of dice rolls
Ok, so then I'm just dealing with $\frac{P(X=x)}{P(Y=1)}$? But then $\sum_{x=1}^\infty x\frac{(5/6)^{x-1}(1/6)}{(1/6)} = \sum_{x=1}^\infty x(5/6)^{x-1} = 36$, which is not $7$