6 ms·

Isaac Newton as a Probabilist

70 points by micaeloliveira 10y ago

tome 10y ago
Here's my heuristic. I wonder if it can be made rigorous. Write P_n for the probability of throwing at least n sixes with 6n dice. 1. P_0 = 1 2. P_n -> 1/2 as n -> infinity, by the central limit theorem. 3. I can't see any reason that P_n shouldn't be monononic. Thus I expect that P_1 >= P_2 >= P_3.
- Chinjut 10y ago
  I had originally heuristically reasoned in a similar but opposite way, to the wrong answer. To wit: 1. P_{1/6} = 1/6 [when rolling one die, you have a 1/6 chance of getting more than 1/6 many sixes] 2. P_n -> 1/2 as n -> infinity, by the central limit theorem 3. I couldn't see any reason that P_n shouldn't be strictly monotonic. Thus I expected that P_1 < P_2 < P_3
  - thomasahle 10y ago
    If you calculated P_1 = 0.665 I suppose you would guess P_2 would be between that and P_infty = 1/2, thus smaller than P_1.
dschiptsov 10y ago
Probabilities could be correctly applied only for fully observable and well defined cases like of an unbiased dice or a fair coin. Everything else is a misapplication and should be regarded as a type error. Frequency based approach is the right tool, but it could explain nothing about the partially observable phenomena, except partial statistics of the outcomes observed so far. Attempts to derive any causal relationships from statistics of a partially observed phenomena which is not fully understood and well-defined is unscientific and is the source of so many misleading "studies" and overconfident decisions and policies. The observations that the sun is raising each day on the east does not tell anything about whether or not and why it will raise there tomorrow. Only physics offer a causal explanation.
mrcactu5 10y ago
let me give a ridiculous argument that all 3 events A,B,C are equally likely: * event A involves getting one 6 from six dice * event B involves getting two 6's from twelve dice * event C involves getting three 6's from eightteen dice since all three events are in proportion, 1:6=2:12=3:18 the events themselves must all be equally likely. -------------------------------------------------------------------------------------------------------------------------------- these faulty arguments must have been thrown out all over the place in Newton's time... and I doubt many discussions today are more careful. I can imagine a room full of politicians arguing over the likelihood of a certain event and three camps emerging as to whether they feel A, B, C is more likely --- throwing away the correct mathematical argument as too complicated. -------------------------------------------- what is the correct argument anyway? For the third event C(18,3)*(1/6)^3(5/6)^(18-3) I believe as we put larger and larger numbers we do get a limit, so these numbers are tending to zero: C(6*k,k)*(1/6)^k(5/6)^(5k) ---> 1/sqrt(2*pi*(5/6)*k) ---> 0 so that event A could be the most likely. This does not sound like Isaac Newton at all.. instead more like Jacob Bernoulli
- maverick_iceman 10y ago
  1- {(5/6)^18 +(18(5/6)^17)/6 + 1817*(5/6)^16/72}
- JadeNB 10y ago
  For what it's worth, the article formalises "one 6 from 6 dice" as "at least one 6", and similarly for the other events, which throws off your probability (which works with "exactly one 6") and hence your asymptotics.
- unclesaamm 10y ago
  You're right about the prevalence of guesswork about probability along laypeople (and everyone until Pascal was essentially a layperson). The Emergence of Probability is a very interesting book on this topic that sketches how statistica emerged from an odd mix of intuitive advances, from places like medicine and alchemy where before statistics you had to guess whether your medicine worked.
dpflan 10y ago
Here is the Newton-Pepys problem explained by Professor Joe Blitzstein in the Harvard class Stats110: https://www.youtube.com/watch?v=P7NE4WF8j-Q&feature=youtu.be&t=17m47s https://www.youtube.com/watch?v=P7NE4WF8j-Q&feature=youtu.be... We see that probability is counterintuitive and complete understanding can elude the strongest of minds.
ykler 10y ago
There are simple ways to intuit the answer to this problem (simpler than the "heuristic" argument the author gives), but Newton's argument is amazingly bad. Surprising he could have thought that way.
deleted 10y ago
[deleted]
te_platt 10y ago
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.

tome 10y ago

Here's my heuristic. I wonder if it can be made rigorous. Write P_n for the probability of throwing at least n sixes with 6n dice. 1. P_0 = 1 2. P_n -> 1/2 as n -> infinity, by the central limit theorem. 3. I can't see any reason that P_n shouldn't be monononic. Thus I expect that P_1 >= P_2 >= P_3.

Chinjut 10y ago
I had originally heuristically reasoned in a similar but opposite way, to the wrong answer. To wit: 1. P_{1/6} = 1/6 [when rolling one die, you have a 1/6 chance of getting more than 1/6 many sixes] 2. P_n -> 1/2 as n -> infinity, by the central limit theorem 3. I couldn't see any reason that P_n shouldn't be strictly monotonic. Thus I expected that P_1 < P_2 < P_3
- thomasahle 10y ago
  If you calculated P_1 = 0.665 I suppose you would guess P_2 would be between that and P_infty = 1/2, thus smaller than P_1.

dschiptsov 10y ago

Probabilities could be correctly applied only for fully observable and well defined cases like of an unbiased dice or a fair coin. Everything else is a misapplication and should be regarded as a type error. Frequency based approach is the right tool, but it could explain nothing about the partially observable phenomena, except partial statistics of the outcomes observed so far. Attempts to derive any causal relationships from statistics of a partially observed phenomena which is not fully understood and well-defined is unscientific and is the source of so many misleading "studies" and overconfident decisions and policies. The observations that the sun is raising each day on the east does not tell anything about whether or not and why it will raise there tomorrow. Only physics offer a causal explanation.

mrcactu5 10y ago

let me give a ridiculous argument that all 3 events A,B,C are equally likely: * event A involves getting one 6 from six dice * event B involves getting two 6's from twelve dice * event C involves getting three 6's from eightteen dice since all three events are in proportion, 1:6=2:12=3:18 the events themselves must all be equally likely. -------------------------------------------------------------------------------------------------------------------------------- these faulty arguments must have been thrown out all over the place in Newton's time... and I doubt many discussions today are more careful. I can imagine a room full of politicians arguing over the likelihood of a certain event and three camps emerging as to whether they feel A, B, C is more likely --- throwing away the correct mathematical argument as too complicated. -------------------------------------------- what is the correct argument anyway? For the third event C(18,3)*(1/6)^3(5/6)^(18-3) I believe as we put larger and larger numbers we do get a limit, so these numbers are tending to zero: C(6*k,k)*(1/6)^k(5/6)^(5k) ---> 1/sqrt(2*pi*(5/6)*k) ---> 0 so that event A could be the most likely. This does not sound like Isaac Newton at all.. instead more like Jacob Bernoulli

maverick_iceman 10y ago
1- {(5/6)^18 +(18(5/6)^17)/6 + 1817*(5/6)^16/72}
JadeNB 10y ago
For what it's worth, the article formalises "one 6 from 6 dice" as "at least one 6", and similarly for the other events, which throws off your probability (which works with "exactly one 6") and hence your asymptotics.
unclesaamm 10y ago
You're right about the prevalence of guesswork about probability along laypeople (and everyone until Pascal was essentially a layperson). The Emergence of Probability is a very interesting book on this topic that sketches how statistica emerged from an odd mix of intuitive advances, from places like medicine and alchemy where before statistics you had to guess whether your medicine worked.

dpflan 10y ago

Here is the Newton-Pepys problem explained by Professor Joe Blitzstein in the Harvard class Stats110: https://www.youtube.com/watch?v=P7NE4WF8j-Q&feature=youtu.be&t=17m47s https://www.youtube.com/watch?v=P7NE4WF8j-Q&feature=youtu.be... We see that probability is counterintuitive and complete understanding can elude the strongest of minds.

ykler 10y ago

There are simple ways to intuit the answer to this problem (simpler than the "heuristic" argument the author gives), but Newton's argument is amazingly bad. Surprising he could have thought that way.

deleted 10y ago

[deleted]

te_platt 10y ago

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.