조건부 확률에 대한 직관을 개발하는 방법?

iTunes와 YouTube에서 찾을 수있는 Harvard의 Statistics 110 : Probability 과정 의 비디오 강의 에서이 문제 가 발생 했습니다. 여기에 요약하려고했습니다.

표준 데크에서 무작위로 2 장의 카드를받습니다.

1. 우리가 하나 이상의 에이스를 가졌을 때 두 카드가 에이스 일 확률은 얼마입니까?

$$P(both\ aces | have\ ace) = \frac{P(both\ aces, have\ ace)}{P(have\ ace)}$$

적어도 하나의 ACE는 모두 에이스있는 경우, 교차 단지 감소 될 수 암시 갖는 사람 $P(both\ aces)$

$$P(both\ aces | have\ ace) = \frac{P(both\ aces)}{P(have\ ace)}$$

이것은 단지

$$P(both\ aces | have\ ace) = \frac{4C2\ /\ 52C2}{1-48C2\ /\ 52C2}=\frac{1}{33}$$

2. 스페이드 에이스를 가졌을 때 두 카드가 에이스 일 확률은 얼마입니까?

$$P(both\ aces | have\ ace\ of\ spades)=\frac{P(both\ aces, have\ ace\ of\ spades)}{P(have\ ace\ of\ spades)}$$

$$P(both\ aces | have\ ace\ of\ spades)=\frac{(3C1*1C1)\ /\ 52C2}{2!*\frac{51}{52}*\frac{1}{51}}=\frac{1}{17}$$

자,이 예제를 따라 어딘가에서 잃어 버렸습니다 ...

후자는 분명히 3 과 동일합니다. 이 해답이 될 것 (나에게) 많은 이해된다. 스페이드의 에이스가 있다고 들었다면3개의 에이스와51개의 카드가더있다는 것을 알 수 있습니다.$\frac{3}{51}$$3$$51$

그러나 이전 예에서 수학은 괜찮아 보입니다 (그리고 강사가 틀린 경우이 예를주지 않을 것이라고 생각합니다 ...). 그러나 나는 이것을 머리로 감쌀 수는 없습니다.

이 문제에 대한 직관을 어떻게 얻습니까?

직관을 돕기 위해 두 가지 이벤트 (결과 집합)를 시각화 해보십시오.

1. 주어진 정보 인 컨디셔닝 이벤트.

2. 조건부 이벤트로, 확률을 찾고자합니다.

조건부 확률은 두 번째 확률을 첫 번째 확률로 나누어서 구합니다.

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두 장의 카드를 무작위로 처리하는 방법 은 입니다. 이러한 거래를 시각화하는 편리한 방법은 거래 된 첫 번째 카드를 지정하는 행 (예 : 행)과 거래에서 두 번째 카드를 열이있는 테이블에 배치하는 것입니다. 누락 된 부분을 지정하는 타원 ( ⋯ ) 과 함께이 표의 일부가 있습니다. 두 카드가 같을 수 없으므로 테이블의 주 대각선을 따라 항목이 없습니다. 행과 열은 에이스부터 왕까지 주문됩니다.$52 \times 51$$\cdots$

질문은 에이스에 중점을 둡니다. "우리는 하나 이상의 에이스를 가지고있다"는 정보는 처음 네 개의 행 또는 첫 네 개의 열 내에서 쌍을 찾습니다. 우리는 이러한 행과 열을 채색하여이를 개략적으로 시각화 할 수 있습니다. 나는 그들에게 붉은 색을 칠 했지만 두 에이스가 나타나는 곳에 검은 색으로 채색했습니다.

$2\times 6 = 12$$2\times (4\times 48) = 384$$12+384=396$ pairs on which you are conditioning, as represented by both the red and black areas. Because all such pairs are equally likely, the chance of the former is

$$\frac{12}{396} = \frac{1}{33}.$$

It is the black fraction of the red+black region.

The second question asserts "we have the ace of spades." This corresponds only to the very first row and column:

Now there are just $2\times 3 = 6$ such pairs with two aces and $2\times 48=96$ other pairs with the ace of spades, for a total of $96+6 = 102$ such pairs. Reasoning exactly as before, the chance of two aces is

$$\frac{6}{102} = \frac{1}{17}.$$

Again it is the black fraction of the red+black region.

For reference, the last figure includes the previous one shown in pink and gray. Comparing these regions reveals what happened: in going from the first question to the second, the number of pairs in the conditioning event (pink) dropped to about one-quarter of its original count (red), while the number of pairs in question dropped by only one-half (from gray to black, $12$ to $6$).

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I have found such schematic figures to be helpful even--perhaps especially--when trying to understand more complicated concepts of probability, such as filtrations of sigma algebras.

A different way to set up a problem that leads to the second calculation is the following:

You draw two cards from the deck. What is the probability of two aces given that the first card you drew was an ace?

This phrasing makes it easier to contrast with the first calculation. The underlying chance of having picked two aces does not change, but the condition to have the first card as an ace is more restrictive than the condition if either being an ace. This means that in the conditional probability calculation the desired combination has to occur among fewer options, so it has a larger probability.

The two different phrasings (ace of spades versus first card as ace) are similar, because they break the symmetry / exchangeability between the aces: the suit or the order cannot be arbitrarily swapped.

At the beginning it was hard to me to have some intuition about.

One idea is to take the problem to the limit. In this case as Steve noted one identical problem is: My neighbor has two kids--you know that one of them is a boy. What's the probability that she has two boys.

Thje first idea is, ok, I have one boy, the other child has 1/2 chance to be a girl and 1/2 to be a boy, but in this case you are not taking all the information that gives you the fact (at least you have a boy) because it is implicit that this boy can be the youngest child being the oldest a girl or viceversa or both are boys which means that only one of the three possible outcomes is favorable.

As I said this is easier taking the problem to the limit...

Case1: Abstract case identical to "we have one ace"-> In this case imagines My neighbour has not 2 children but 27, and you know 26 are boys, the probability of this is almost zero. In this case it is clear that this information gives you a lot of information that the probabilistically speaking the remain child is a girl. To be precise, you will have one case with 27 boys, let's say a tuple (b,b,b,b,b,b...,b) and 27 cases with 1 girl and 26 boys (g,b,b,b...), (b,g,b,b,b...), so the probability of all boys is 1/27, in general it will be 1/(N+1)

case2: Concrete information. This would be identical to "We have the ace of spades" or "we have the first card being an ace". In this case imagine our neighbour have 26 children all boys and is pregnant with the 27th. What is the probability that the 27th will be a boy?

With case2 I am pretty sure we all can have a grasp of the intuition needed for this kind of not-so-obvious conditional probabilities problems.

If you want to become rich, you have to bet on the first case with 26 boys and a 27th because the lack of concrete information means a lot of probabilistic energy on the remain child while in the second case, the entropy is huge, we have not information to know where to bet.

I hope this is useful

How one can tell the answer is 3/51 without calculating?

If you took the ace of spades in the first place. I know which cards ar in the package. So there is stil 3 aces on 51 cards. so for the second one, you have 3/51 chances to have two aces.

And how to understand the difference between the two scenarios intuitively?

It's because "Have one ace" is included in "Have two aces". But "Have the ace of spades" is not included in "Have two aces". This is the difference.

In fact, if you have two ace, you have one but maybe not the ace of spades So it's not the same probability.

This answered was for a other post which has been moved on this one ..