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Well, once again my intuition completely fails in a probability problem. My thinking was the more rolls you make the more uniform your distribution of the numb

Well, once again my intuition completely fails in a probability problem. My thinking was the more rolls you make the more uniform your distribution of the number showing becomes. In the limiting case as number of rolls -> infinity you are guaranteed to have 1/6 of the rolls be a 6. So the probability of at least 1/6 is less than 1 for a small number of rolls but approaches 1 the more rolls you do. Intuition is like a good friend who likes to mess with me at weird times.

mturmon 10y ago
You're on the right track with this idea about the flattening of the distribution as N grows. As in the article, let X = # sixes in N rolls and note that the expected value of X is N*p = N/6 in this case (fair dice). The event of interest is: {X >= N/6} We want the probability of this event as a function of N. It turns out this is a decreasing function of N, and that fact answers the question ("A is most likely"). But why does it decrease in N? As you say, the distribution of X flattens out as N increases. The probability mass is spread out over the numbers 0...N, with a peak at exactly N/6. (Not N/6 - 1, or N/6 + 1, but N/6.) About half the probability mass is between 0...N/6, and half is between N/6...N. So in general, the probability we care about (A, B, or C) is rather close to 1/2 (because it is just the right half of the distribution). Since the distribution is flattening out, as N increases, mass moves out of the exact center (X = N/6) and toward the left or right. Since the event of interest is {X >= N/6} and not {X > N/6}, the fact that the precise center is steadily losing mass (some moving left, some moving right) as N increases is critical. In short: Because the event of interest contains the very center, as more rolls are done, the overall distribution flattens, and the interval between [N/6...N] benefits less and less by containing the (ever-more-diluted) center point. This is another way to restate Stigler's explanation at the top-right corner of page 401 in the OP. ("The ranking ... reﬂects the fact that ... P(X = Np) decreases as N increases and the distribution spreads out.")
ykler 10y ago
In the limiting case the number of sixes will be larger than 1/6 half the time, smaller than 1/6 half the time, and exactly 1/6 only an infinitesimal amount of the time. Roughly speaking, with a smaller number of rolls you are more likely to get at least 1/6 sixes because the "exactly 1/6" case becomes more significant. However, this is a bit too rough because just because the expected number of sixes is 1/6 it doesn't follow (and isn't true) that you are equally likely to get more than 1/6 and less than 1/6 sixes. You are actually more likely to get less than 1/6, just not by enough to offset the chance of getting exactly 1/6. You can get intuition by just looking at a particular small case, like the six throws case, or to make the numbers even simpler you could look at three rolls of a three-sided die.

mturmon 10y ago

You're on the right track with this idea about the flattening of the distribution as N grows. As in the article, let X = # sixes in N rolls and note that the expected value of X is N*p = N/6 in this case (fair dice). The event of interest is: {X >= N/6} We want the probability of this event as a function of N. It turns out this is a decreasing function of N, and that fact answers the question ("A is most likely"). But why does it decrease in N? As you say, the distribution of X flattens out as N increases. The probability mass is spread out over the numbers 0...N, with a peak at exactly N/6. (Not N/6 - 1, or N/6 + 1, but N/6.) About half the probability mass is between 0...N/6, and half is between N/6...N. So in general, the probability we care about (A, B, or C) is rather close to 1/2 (because it is just the right half of the distribution). Since the distribution is flattening out, as N increases, mass moves out of the exact center (X = N/6) and toward the left or right. Since the event of interest is {X >= N/6} and not {X > N/6}, the fact that the precise center is steadily losing mass (some moving left, some moving right) as N increases is critical. In short: Because the event of interest contains the very center, as more rolls are done, the overall distribution flattens, and the interval between [N/6...N] benefits less and less by containing the (ever-more-diluted) center point. This is another way to restate Stigler's explanation at the top-right corner of page 401 in the OP. ("The ranking ... reﬂects the fact that ... P(X = Np) decreases as N increases and the distribution spreads out.")

ykler 10y ago

In the limiting case the number of sixes will be larger than 1/6 half the time, smaller than 1/6 half the time, and exactly 1/6 only an infinitesimal amount of the time. Roughly speaking, with a smaller number of rolls you are more likely to get at least 1/6 sixes because the "exactly 1/6" case becomes more significant. However, this is a bit too rough because just because the expected number of sixes is 1/6 it doesn't follow (and isn't true) that you are equally likely to get more than 1/6 and less than 1/6 sixes. You are actually more likely to get less than 1/6, just not by enough to offset the chance of getting exactly 1/6. You can get intuition by just looking at a particular small case, like the six throws case, or to make the numbers even simpler you could look at three rolls of a three-sided die.