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 2 clarify magnitude edited Jun 17 '13 at 22:03 David J. Harris 9,98211 gold badge2525 silver badges4848 bronze badges Let's start off with a simplified version where we're just interested in the probability of getting exactly 50 1's. In R, you can get this probability using the dbinom function. dbinom(50, size = 100, prob = 1/6) # about 10^-14  Multiplying by 6 gives the probability of 50 ones plus the probability of 50 twos, etc. As you pointed out, it's possible that you'd get 50 ones and 50 twos on the same run of 100 rolls, so this does a little bit of double-counting. How much double-counting? Well, there are n-choose-two pairs of numbers that could each get 50 hits. 6-choose-2 is 15 There are 30 ways to do this (see below), so we're double-counting 1530 instances out of the 6^100 possible sequences. 6 * dbinom(50, size = 100, prob = 1/6) - 1530 * (1/6)^100 # still about 10^-14  This gives me 8.229631e-14. According to Wikipedia, this is comparable to the probability of picking a given cell from the human body at random. As you expected, the double-counting issue is negligible. I initially said there were 15 ways because there are 15 unique ways to pick two of the six sides to be rolled 50 times. As @PeterFlom pointed out, this is off by a factor of 2, since you could have ones and then twos or twos and then ones. Let's start off with a simplified version where we're just interested in the probability of getting exactly 50 1's. In R, you can get this probability using the dbinom function. dbinom(50, size = 100, prob = 1/6) # about 10^-14  Multiplying by 6 gives the probability of 50 ones plus the probability of 50 twos, etc. As you pointed out, it's possible that you'd get 50 ones and 50 twos on the same run of 100 rolls, so this does a little bit of double-counting. How much double-counting? Well, there are n-choose-two pairs of numbers that could each get 50 hits. 6-choose-2 is 15, so we're double-counting 15 instances out of the 6^100 possible sequences. dbinom(50, size = 100, prob = 1/6) - 15 * (1/6)^100 # still about 10^-14  Let's start off with a simplified version where we're just interested in the probability of getting exactly 50 1's. In R, you can get this probability using the dbinom function. dbinom(50, size = 100, prob = 1/6) # about 10^-14  Multiplying by 6 gives the probability of 50 ones plus the probability of 50 twos, etc. As you pointed out, it's possible that you'd get 50 ones and 50 twos on the same run of 100 rolls, so this does a little bit of double-counting. How much double-counting? Well, there are n-choose-two pairs of numbers that could each get 50 hits. There are 30 ways to do this (see below), so we're double-counting 30 instances out of the 6^100 possible sequences. 6 * dbinom(50, size = 100, prob = 1/6) - 30 * (1/6)^100  This gives me 8.229631e-14. According to Wikipedia, this is comparable to the probability of picking a given cell from the human body at random. As you expected, the double-counting issue is negligible. I initially said there were 15 ways because there are 15 unique ways to pick two of the six sides to be rolled 50 times. As @PeterFlom pointed out, this is off by a factor of 2, since you could have ones and then twos or twos and then ones. 1 answered Jun 17 '13 at 21:35 David J. Harris 9,98211 gold badge2525 silver badges4848 bronze badges Let's start off with a simplified version where we're just interested in the probability of getting exactly 50 1's. In R, you can get this probability using the dbinom function. dbinom(50, size = 100, prob = 1/6) # about 10^-14  Multiplying by 6 gives the probability of 50 ones plus the probability of 50 twos, etc. As you pointed out, it's possible that you'd get 50 ones and 50 twos on the same run of 100 rolls, so this does a little bit of double-counting. How much double-counting? Well, there are n-choose-two pairs of numbers that could each get 50 hits. 6-choose-2 is 15, so we're double-counting 15 instances out of the 6^100 possible sequences. dbinom(50, size = 100, prob = 1/6) - 15 * (1/6)^100 # still about 10^-14