Lokasenna
Posts: 2077
Joined: 3/3/2012 From: Iowan in MD/DC Status: online

quote:
ORIGINAL: erstad quote:
ORIGINAL: Lokasenna quote:
ORIGINAL: erstad quote:
He should be doing one of them 38.5875% of the time. Being that I spend a large fraction of my life playing with numbers... Canoe said I roll one twentysided dice three times, take the largest number divisible by three, square it, and divide by itself. Assuming the antecedent for "itself" is the "largest number divisible by three" (as opposed to the squared number), the squaring and dividing is an identity function. Thus, he is basically taking the largest number divisible by three. To restrain Mandrake from discussing lusty women, he must roll at least one nine, and no multiples of three larger than 9. For exactly one 9, the probability is P(1 nines, no large x3) = 3[combinations] * 1/20 [p of a 9] * (16/20)^2 [p of the other two rolls being not a 9, 12, 15, or 18] = 0.096 Likewise, P(2 nines, no larger x3)= 3 * (1/20)^2 * 16/20 = 0.006 P(3 nines, no larger x3) = 1 * (1/20)^3 = 0.000 (rounding) So he plays Mandrake 10.2% of the time. Without going through the rest of the math, one can see by inspection that the probability of invading Hokkaido (option 6) or seeing miller in his BVDs (option 3) has to be smaller than this probability. The first two terms stay the same, but the 16/20 factor for "a different number which is a (nonmultiple of 3) (inclusiveor) (is smaller)) decreases. Therefore, the probability of (any of these options) is going to be less than 3*10.2%, i.e., less than 30.6%. So I believe Wirraway_Ace is correct in saying that Canoerebel does one of these options less than a third of the time, with all the usual caveats (Canoerebel is telling the truth, the die is fair, etc.). Although I acknowledge some possibility of error. I showed you mine, now you show me yours I didn't take into account any of the other results as only a roll of 3, 6, or 9 will give a valid result on his list. The probability that he rolls a 3, 6, or 9 on a 20sided die is 3/20 or 15%. Therefore, he will not play any games 85% of the time on any given roll. However, since he rolls 3 times we need to evaluate .85^3, which is 61.4125% of the time. 1  .614125 = .385875. I don't want to work out the individual probabilities, but shouldn't they all add up to that 38.5875%? You would be correct if he used any 3, 6, or 9. However, at least as I understood the description, he uses the largest multiple of 3 he rolled. So if he rolled a 9, but then later rolled a 15, the 15 would be used and as there is no entry 15 he would apply not be a variant. So not all 3, 6, and 9 rolls will result in a variant, again to my understanding of the method that I'm quite certain Canoerebel rigorously applies to all his games. Ah ha. There's the difference. Your numbers must be right! Although I could approximate by doing .7^3 instead of .85^3 (it's obviously not accurate, but it's more accurate than my previous).
