평균 대 도박꾼의 잘못에 대한 회귀

한편으로는 평균에 대한 회귀가 있고 다른 한편으로는 도박꾼의 오류가 있습니다.

도박꾼의 오류는 Miller와 Sanjurjo (2019)에 의해 "임의의 시퀀스가 ​​반전에 대한 체계적인 경향을 가지고 있다는 잘못된 생각, 즉 유사한 결과의 줄무늬가 계속되는 것보다 더 끝날 가능성이 높다"는 것으로 정의됩니다. 다음 시험에서는 연속으로 시간이 불균형 적으로 떨어질 것으로 생각됩니다.

나는 마지막 경기에서 좋은 성적을 보였으며, 평균에 대한 회귀에 따르면 아마도 다음 경기에서 더 나쁜 성과를 낼 것입니다.

그러나 도박꾼의 오류에 따르면, 공정한 동전을 가정 할 때 다음 두 가지 확률을 고려하십시오.

1. 20 머리의 확률, 1 꼬리 = $0.5^{20} × 0.5 = 0.5^{21}$
2. 20 헤드의 확률, 1 헤드 = $0.5^{20} × 0.5 = 0.5^{21}$

그때...

간단한 예를 생각해 보자. 한 반의 학생들이 과목에 대해 100 가지 항목의 참 / 거짓 시험을 치른다. 모든 학생들이 모든 질문에서 무작위로 선택한다고 가정하십시오. 그런 다음 각 학생의 점수는 예상 평균이 50 인 독립적이고 동일하게 분포 된 무작위 변수 세트 중 하나를 실현합니다.

당연히 일부 학생들은 우연히 50 점 이상, 50 점 이하의 점수를 받게됩니다. 학생의 최고 점수 10 % 만 취하고 모든 항목에서 무작위로 다시 선택하는 두 번째 시험을 제공하면 평균 점수는 다시 50에 가까울 것으로 예상됩니다.

따라서이 학생들의 평균은 원래의 시험을 치른 모든 학생들의 평균으로 되돌아 가게됩니다. 학생이 원래 시험에서 채점 한 점수에 관계없이 두 번째 시험에서 채점을 가장 잘 예측하는 점수는 50입니다.

특별하게 만약 학생의 최고 점수 10 %만을 취하여 모든 항목에서 무작위로 다시 선택하는 두 번째 시험을한다면, 평균 점수는 다시 50에 가까울 것으로 예상됩니다.

도박꾼의 오류는 점수에 대해 동일한 확률을 기 대해서는 안되며 반드시 50에 가까울 필요는 없습니까?

Miller, JB, & Sanjurjo, A. (2019). 샘플 크기가 무시 될 때 경험이 도박꾼의 오류를 확인하는 방법

나는 "평균에 대한 회귀"라는 개념이 실제로 과거와 아무 관련이 없다는 것을 고려함으로써 혼란을 해결할 수 있다고 생각합니다. 실험의 각 반복에서 우리는 평균 결과를 기대한다는 것은 단지 tautological 관찰입니다. 따라서 이전에 평균 이상의 결과를 얻은 경우 더 나쁜 결과를 기대하거나 평균 이하의 결과를 얻는다면 더 나은 결과를 기대합니다. 핵심은 기대 자체 가 도박꾼의 오류와 마찬가지로 이전의 역사에 의존하지 않는다는 것입니다.

합리적 인 사람 (그리고 공정한 동전을 가정)과 같은 위치에 자신을 찾으려면 가장 좋은 방법은 추측하는 것입니다. 당신이 미신 도박꾼과 같은 위치에 자신을 발견한다면, 당신의 최선의 방법은 이전의 사건을보고 과거에 대해 추론을 정당화하려고하는 것입니다 - "와우, 머리는 예를 들어 핫 !, 분담금 시간" 또는 " 우리가 다른 머리를 볼 수있는 방법 은 없습니다. 이런 종류의 행진 확률은 엄청나게 낮습니다!".

도박꾼의 잘못은 20 동전의 모든 특정 문자열이 우리를 미치게 던질 가능성을 깨닫지 못합니다. HHTHHTTTHT를 뒤집을 가능성은 매우 낮습니다. 어떤 문자열이든 많은 다른 결과에서 발생하는 유일한 방법 이 있기 때문입니다 . 따라서, 이들 중 어느 것을 "가능성이있다"또는 "가능하지 않은 것"으로 부 풀리는 것은 모두 동등 할 수 있기 때문에 오류이다.

평균에 대한 회귀는 장기적으로 관측치가 유한 한 예상 값으로 수렴되어야한다는 올바른 근거입니다. 예를 들어-동전 던지기 20 개 중 10 개는 좋은 방법이라고 생각합니다. 최종 카운트를 달성하는 문자열이 훨씬 적기 때문에 15/20에 베팅 할 가능성은 상당히 낮습니다. 동전을 충분히 길고 뒤집어 놓으면 결국 50/50 정도의 물건으로 끝날 것입니다. 그러나 "줄무늬"또는 다른 불가능한 것으로 끝나지 않을 것입니다. 그것의 사건. 이것이이 두 개념의 차이점의 핵심입니다.

TL; DR : 평균에 대한 회귀는 시간이 지남에 따라 실험에서 예상되는 것과 유사한 분포를 얻게 될 것이라고 말합니다. 도박꾼의 잘못은 (잘못) 동전의 각 개인 플립이 이전 결과에 대한 기억 을 가지고 있으며 , 이는 다음 독립 결과에 영향을 줄 것이라고 말합니다 .

I always try to remember that regression toward the mean isn't a compensatory mechanism for observing outliers.

There's no cause-and-effect relationship between having an outstanding gambling run, then going 50-50 after that. It's just a helpful way to remember that, when you're sampling from a distribution, you're most likely to see values close to the mean (think of what Chebyshev's inequality has to say here).

Here's a simple example: you've decided to toss a total of 200 coins. So far you've tossed 100 of them and you've gotten extremely lucky: 100% came up heads (incredible, I know, but let's just keep things simple).

Conditional on 100 heads in the 100 first tosses, you expect to have 150 heads total at the end of the game. An extreme example of the gambler's fallacy would be to think that you still only expect 100 heads total (i.e. the expected value before starting the game), even after getting 100 in the first 100 tosses. The gambler fallaciously thinks the next 100 tosses must be tails. An example of regression to the mean (in this context) is that your head-rate of 100% is expected to fall to 150/200 = 75% (i.e. toward the mean of 50%) as you finish the game.

I could be wrong but I have always thought the difference to be in the assumption of independence.

In the Gambler's fallacy the issue is the misunderstanding of independence. Sure over some large N number of coin tosses you will be around a 50-50 split, but if by chance you are not then the thought that your next T tosses will help even out the odds is wrong because there each coin toss is independent of the previous.

Regression towards the mean is, where I see it used, some idea that draws are dependent on previous draws or a previous calculated average/values. For example let use NBA shooting percentage. If player A has made on average 40% of his shots during his career and starts off a new year by shooting 70% in his first 5 games its reasonable to think that he will regress to the mean of his career average. There are dependent factors that can and will influence his play: hot/cold streaks, teammate play, confidence, and the simple fact that if he were to maintain 70% shooting for the year he would absolutely annihilate multiple records that are simply impossible physical feats (under the current performance abilities of professional basket ball players). As you play more games your shooting percentage will likely drop closer to your career average.

The key is that we don't have any information that will help us with the next event (gambler's fallacy), because the next event isn't dependent on the previous event. We can make a reasonable guess about how a series of trials will go. This reasonable guess is the average aka our expected mean result. So when we watch a deviation in the mean trend back toward the mean, over time/trials, then we witnessing a regression to the mean.

As you can see regression to the mean is an observed series of actions, it isn't a predictor. As more trials are conducted things will more closely approximate a normal/Gaussian distribution. This means that I'm not making any assumptions or guess on what the next result will be. Using the law of large numbers I can theorize that even though things might be trending one way currently, over time things will balance themselves out. When they do balance themselves out the result set has regressed to the mean. It is important to note here that we aren't saying that future trials are dependent on past results. I'm merely observing a change in the balance of the data.

The gambler's fallacy as I understand it is more immediate in it's goals and focuses on prediction of future events. This tracks with what a gambler desires. Typically games of chance are tilted against the gambler over the long term, so a gambler wants to know what the next trial will be because they want to capitalize on this knowledge. This leads the gambler to falsely assume that the next trial is dependent on the previous trial. This can lead to neutral choices like:

The last five times the roulette wheel landed on black, so therefore next time I'm betting big on red.

Or the choice can be self-serving:

I've gotten a full house the last 5 hands, so I'm going to bet big because I'm on a winning streak and can't lose.

So as you can see there are few key differences:

1. Regression to the mean doesn't assume that independent trials are dependent like the gambler's fallacy.

2. Regression to the mean is applied over a large amount of data/trials, where the gambler's fallacy is concerned with the next trial.

3. Regression to the mean describes what has already taken place. Gambler's fallacy attempts to predict the future based on an expected average, and past results.

Are students with higher grades who score worse on retest cheaters?

The question received a substantial edit since the last of six answers.

The edited question contains an example of regression to the mean in the context of student scores on a $100$ question true-false test and an retest for the top performers on an equivalent test. The retest shows substantially more average scores for the group of top performers on the first test. What's going on? Were the students cheating the first time? No, it is important to control for regression to the mean. Test performance for multiple choice tests is a combination of luck in guessing and ability/knowledge. Some portion of the top performers' scores was due to good luck, which was not necessarily repeatable the second time.

Or should they just stay away from the roulette wheel?

Let's first assume that no skill at all was involved, that the student's were just flipping (fair) coins to determine their answers. What's the expected score? Well, each answer has independently a $50\%$ chance of being the correct one, so we expect $50\%$ of $100$ or a score of $50$.

But, that's an expected value. Some will do better merely by chance. The probability of scoring at least $60\%$ correctly according to the binomial distribution is approximately $2.8\%$. So, in a group of $3000$ students, the expected number of students to get a grade of $60%$ or better is $85$.

Now let's assume indeed there were $85$ students with a score of $60\%$ or better and retest them. What's the expected score on retest under the same coin-flipping method? Its still $50\%$ of $100$! What's the probability that a student being retested in this manner will score above $60\%$? It's still $2.8\%$! So we should expect only $2$ of the $85$ ($2.8\% \cdot 85$) to score at least $60\%$ on retest.

Under this setup it is a fallacy to assume an expected score on retest different from the expected score on the first test -- they are both $50\%$ of $100$. The gambler's fallacy would be to assume that the good luck of the high scoring students is more likely to be balanced out by bad luck on retest. Under this fallacy, you'd bet on the expected retest scores to be below $50$. The hot-handed fallacy (here) would be to assume that the good luck of the high scoring students is more likely to continue and bet on the expected retest scores to be above $50$.

Lucky coins and lucky flips

Reality is a bit more complicated. Let's update our model. First, it doesn't matter what the actual answers are if we are just flipping coins, so let's just score by number of heads. So far, the model is equivalent. Now let's assume $1000$ coins are biased to be heads with probability of $55\%$ (good coins $G$), $1000$ coins are biased to be heads with probability of $45\%$ (bad coins $B$), and $1000$ have equal probability of being heads or tails (fair coins $F$) and randomly distribute these. This is analogous to assuming higher and lower ability/knowledge under the test taking example, but it is easier to reason correctly about inanimate objects.

The expected score is $(55 \cdot 1000 + 45 \cdot 1000 + 50 \cdot 1000)/3000 = 50$ for any student given the random distribution. So, the expected score for the first test has not changed. Now, the probability of scoring at least $60\%$ correctly, again using the binomial distribution is $18.3\%$ for good coins, $0.2\%$ for bad coins, and of course $2.8\%$ still for the fair coins. The probability of scoring at least $60\%$ is, since an equal number of each type of coin was randomly distributed, the average of these, or $7.1\%$. The expected number of students scoring at least $60\%$ correctly is $21$.

Now, if we do indeed have $21$ scoring at least $60\%$ correctly under this setup of biased coins, what's the expected score on retest? Not $50\%$ of $100$ anymore! Now you can work it out with Bayes theorem, but since we used equal size groups the probability of having a type of coin given a outcome is (here) proportional to the probability of the outcome given the type of coin. In other words, there is a $86\% = 18.3\%/(18.3\% + 0.2\% + 2.8\%)$ chance that those scoring at least 60% had a good coin, $1\% = 0.2\%/(18.3\% + 0.2\% + 2.8\%)$ had a bad coin, and $13\%$ had a fair coin. The expected value of scores on retest is therefore $86\% \cdot 55 + 1\% \cdot 45 + 13\% \cdot 50 = 54.25$ out of $100$. This is lower than actual scores of the first round, at least $60$, but higher than the expected value of scores before the first round, $50$.

So even when some coins are better than others, randomness in the coin flips means that selecting the top performers from a test will still exhibit some regression to the mean in a retest. In this modified model, hot-handedness is no longer an outright fallacy -- scoring better in the first round does mean a higher probability of having a good coin! However, gambler's fallacy is still a fallacy -- those who experienced good luck cannot be expected to be compensated with bad luck on retest.

They are saying the same thing. You were mostly confused because no single experiment in the coin flip example has extreme result (H/T 50/50). Change it to "flipping ten fair coins at the same time in every experiment", and gamblers want to get all of them right. Then an extreme measurement would be that you happen to see all of them are heads.

Gambler fallacy: Treat each gamble outcome (coin flipping result) as IID. If you already know the distribution those IID shares, then the next prediction should come directly from the known distribution and has nothing to do with historical (or future) results (aka other IID).

Regression to the mean: Treat each test outcome as IID (since the student is assumed to be guessing randomly and have no real skill). If you already know the distribution those IID shares, then the next prediction comes directly from the known distribution and has nothing to do with historical (or future) results (aka other IID) (exactly as before up to here). But, by CLT, if you observed extreme values in one measurement (e.g by chance you were only sampling the top 10% students from the first test), you should know the result from your next observation/measurement will still be generated from the known distribution (and thus more likely to be closer to the mean than staying at the extreme).

So fundamentally, they both say the next measurement will come from the distribution instead of past results.

Let X and Y be two i.i.d. uniform random variables on [0,1]. Suppose we observe them one after another.

Gambler's Fallacy: P( Y | X ) != P( Y ) This is, of course, nonsense because X and Y are independent.

Regression to the mean: P( Y < X | X = 1) != P( Y < X ) This is true: LHS is 1, LHS < 1

Thanks your answers I think I could understand the difference between the Regression to the mean and Gambler's fallacy. Even more, I built a database to help me illustrate in the "real" case.

I built this situation: I collected 1000 students and I put them to do a test randomly answering questions .

The test score ranges from 01 to 05. As they are randomly answering questions, so each score has a 20% chance of being achieved. So for the first test the number of students with a score 05 should be something close to 200

(1.1) $1000*0,20$

(1.2) $200$

I Had 196 students with score 05 which is very close to the expected 200 students.

So I put those 196 students repeat the test is exepected 39 students with score 05.

(2.1) $196*0,20$

(2.2) $39$

Well, according to the result I got 42 students which is within the expected.

For those who got score 05 I put them to repeat the test and so and forth...

Therefore, the expected numbers were:

Expected RETEST 03

(3.1) $42*0,20$

(3.2) $8$

(3.3) Outcomes (8)

Expected RETEST 04

(4.1) $8*0,20$

(4.2) $1,2$

(4.3) Outcomes (2)

Expected RETEST 05

(4.1) $2*0,20$

(4.2) $0,1$

(4.3) Outcomes (0)

If I'm expecting for a student who gets score 05 four times I shall to face the probability of $0,20^4$, i.e, 1,2 student per 1000. However If I expect for a student who gets score 05 five times I should have at least 3.500 samples in order to get 1,12 student with score 05 in all tests

(5.1.) $0,20^5 = 0,00032$

(5.2.) $0,00032 * 3500 = 1.2$

Therefore the probability of the one student gets score 05 in the all 05 tests has nothing to do with his last score, I mean, I must not calculate the probability on the each test singly. I must look for those 05 tests like one event and calculate the probability for that event.