mkendall
August 20th, 2010, 04:06 AM
Actually, confirmation of the correctness of my calculations.
I know that the Expectation (weighted mean average, expected value, first moment) of a fair X-sided die (dX) is E(dX) = (1+2+3+...+X)/X (which collapses to (1+X)/2) and that the Expectation is additive: E(2dX) = 2E(dX). Note that dX does not mean the infinitesimal.
Now, roll a die and subtract a fixed number with a minimum of 0. For example, roll a 6-sided die, subtract 3, any negative results become 0. So all we look at are when the die roll is greater than the number.
Abusing the notation, I find:
E(d6>1) = (0+1+2+3+4+5)/6 = 15/6
E(d6>2) = (0+0+1+2+3+4)/6 = 10/6
E(d6>3) = (0+0+0+1+2+3)/6 = 6/6
E(d6>4) = (0+0+0+0+1+2)/6 = 3/6
E(d6>5) = (0+0+0+0+0+1)/6 = 1/6
E(d6>6) = (0+0+0+0+0+0)/6 = 0
Generalizing this, E(dX>Y) = Summation i=1 to X-Y of i, all over X (looks much nicer in mathematical notation)
Taking this further, roll a die dX and subtract the result of a second die dY. The table above represents d6>d6. In this scenario,
E(dX>dY) = Summation j=1 to Y of Summation i=1 to X-j of i, all over X*Y.
Lastly, dX>dY+Z, roll of dX subtracting a roll of dY and a constant Z gives
E(dX>dY+Z) = Summation j=Z to Y+Z of Summation i=1 to X-j of i, all over X*Y.
So, any Statistics people out there, did I get it correct? If not, where have I screwed up?
I know that the Expectation (weighted mean average, expected value, first moment) of a fair X-sided die (dX) is E(dX) = (1+2+3+...+X)/X (which collapses to (1+X)/2) and that the Expectation is additive: E(2dX) = 2E(dX). Note that dX does not mean the infinitesimal.
Now, roll a die and subtract a fixed number with a minimum of 0. For example, roll a 6-sided die, subtract 3, any negative results become 0. So all we look at are when the die roll is greater than the number.
Abusing the notation, I find:
E(d6>1) = (0+1+2+3+4+5)/6 = 15/6
E(d6>2) = (0+0+1+2+3+4)/6 = 10/6
E(d6>3) = (0+0+0+1+2+3)/6 = 6/6
E(d6>4) = (0+0+0+0+1+2)/6 = 3/6
E(d6>5) = (0+0+0+0+0+1)/6 = 1/6
E(d6>6) = (0+0+0+0+0+0)/6 = 0
Generalizing this, E(dX>Y) = Summation i=1 to X-Y of i, all over X (looks much nicer in mathematical notation)
Taking this further, roll a die dX and subtract the result of a second die dY. The table above represents d6>d6. In this scenario,
E(dX>dY) = Summation j=1 to Y of Summation i=1 to X-j of i, all over X*Y.
Lastly, dX>dY+Z, roll of dX subtracting a roll of dY and a constant Z gives
E(dX>dY+Z) = Summation j=Z to Y+Z of Summation i=1 to X-j of i, all over X*Y.
So, any Statistics people out there, did I get it correct? If not, where have I screwed up?