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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 $7$