Làm thế nào để chúng ta biết rằng xác suất cán 1 và 2 là 1/18?

Kể từ lớp xác suất đầu tiên của tôi, tôi đã tự hỏi về những điều sau đây.

Tính toán xác suất thường được giới thiệu thông qua tỷ lệ của "sự kiện ưa thích" trên tổng số sự kiện có thể. Trong trường hợp lăn hai con xúc xắc 6 mặt, số lượng sự kiện có thể xảy ra là $36$ , như được hiển thị trong bảng dưới đây.

$$\begin{array} {|c|c|c|c|c|c|c|} \hline &1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ \hline 2 & (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ \hline 3 & (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ \hline 4 & (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ \hline 5 & (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ \hline 6 & (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \\ \hline \end{array}$$

Do đó, nếu chúng ta quan tâm đến việc tính xác suất của sự kiện A "cán và 2 ", chúng ta sẽ thấy có hai "sự kiện được ưa thích" và tính xác suất của sự kiện là $1$$2$ .$\frac{2}{36}=\frac{1}{18}$

Bây giờ, điều luôn làm tôi tự hỏi là: giả sử không thể phân biệt giữa hai con xúc xắc và chúng ta sẽ chỉ quan sát chúng sau khi chúng được lăn, vì vậy, ví dụ chúng ta sẽ quan sát "Ai đó đưa cho tôi một hộp. Tôi mở hộp. Có và 2 ". Trong kịch bản giả thuyết này, chúng tôi sẽ không thể phân biệt giữa hai con xúc xắc, vì vậy chúng tôi sẽ không biết rằng có hai sự kiện có thể dẫn đến quan sát này. Sau đó, các sự kiện có thể của chúng tôi sẽ như thế:$1$$2$

$$\begin{array} {|c|c|c|c|c|c|} \hline (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6) \\ \hline & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ \hline & & (3,3) & (3,4) & (3,5) & (3,6) \\ \hline & & & (4,4) & (4,5) & (4,6) \\ \hline & & & & (5,5) & (5,6) \\ \hline & & & & & (6,6) \\ \hline \end{array}$$

và chúng tôi sẽ tính xác suất của sự kiện A là .$\frac{1}{21}$

Một lần nữa, tôi hoàn toàn nhận thức được thực tế rằng cách tiếp cận đầu tiên sẽ dẫn chúng ta đến câu trả lời đúng. Câu hỏi tôi đang tự hỏi mình là:

Làm thế nào để chúng ta biết rằng có đúng không?$\frac{1}{18}$

Hai câu trả lời tôi đã đưa ra là:

• Chúng ta có thể kiểm tra theo kinh nghiệm Nhiều như tôi quan tâm đến điều này, tôi cần phải thừa nhận rằng tôi đã không tự mình làm điều này. Nhưng tôi tin rằng nó sẽ là trường hợp.
• Trong thực tế, chúng ta có thể phân biệt giữa súc sắc, giống như một màu đen và một màu xanh khác, hoặc ném cái này trước cái kia hoặc chỉ đơn giản là biết về sự kiện có thể và sau đó tất cả các lý thuyết tiêu chuẩn hoạt động.$36$

Câu hỏi của tôi cho bạn là:

• Còn lý do nào khác để chúng ta biết rằng có đúng không? (Tôi khá chắc chắn rằng phải có một vài lý do (ít nhất là về mặt kỹ thuật) và đây là lý do tại sao tôi đăng câu hỏi này)$\frac{1}{18}$
• Có một số lập luận cơ bản chống lại giả định rằng chúng ta không thể phân biệt giữa súc sắc?
• Nếu chúng ta giả sử rằng chúng ta không thể phân biệt giữa súc sắc và không có cách nào để kiểm tra xác suất theo kinh nghiệm, thì thậm chí chính xác hoặc tôi đã bỏ qua một cái gì đó?$P(A) = \frac{1}{21}$

Cảm ơn bạn đã dành thời gian để đọc câu hỏi của tôi và tôi hy vọng nó đủ cụ thể.

Imagine that you threw your fair six-sided die and you got ⚀. The result was so fascinating that you called your friend Dave and told him about it. Since he was curious what he'd get when throwing his fair six-sided die, he threw it and got ⚁.

A standard die has six sides. If you are not cheating then it lands on each side with equal probability, i.e. $1$ in $6$ times. The probability that you throw ⚀, the same as with the other sides, is $\tfrac{1}{6}$. The probability that you throw ⚀, and your friend throws ⚁, is $\tfrac{1}{6} \times \tfrac{1}{6} = \tfrac{1}{36}$ since the two events are independent and we multiply independent probabilities. Saying it differently, there are $36$ arrangements of such pairs that can be easily listed (as you already did). The probability of the opposite event (you throw ⚁ and your friend throws ⚀) is also $\tfrac{1}{36}$. The probabilities that you throw ⚀, and your friend throws ⚁, or that you throw ⚁, and your friend throws ⚀, are exclusive, so we add them $\tfrac{1}{36} + \tfrac{1}{36} = \tfrac{2}{36}$. Among all the possible arrangements, there are two meeting this condition.

How do we know all of this? Well, on the grounds of probability, combinatorics and logic, but those three need some factual knowledge to rely on. We know on the basis of the experience of thousands of gamblers and some physics, that there is no reason to believe that a fair six-sided die has other than an equiprobable chance of landing on each side. Similarly, we have no reason to suspect that two independent throws are somehow related and influence each other.

You can imagine a box with tickets labeled using all the $2$-combinations (with repetition) of numbers from $1$ to $6$. That would limit the number of possible outcomes to $21$ and change the probabilities. However if you think of such a definition in term of dice, then you would have to imagine two dice that are somehow glued together. This is something very different than two dice that can function independently and can be thrown alone landing on each side with equal probability without affecting each other.

All that said, one needs to comment that such models are possible, but not for things like dice. For example, in particle physics based on empirical observations it appeared that Bose-Einstein statistic of non-distinguishable particles (see also the stars-and-bars problem) is more appropriate than the distinguishable-particles model. You can find some remarks about those models in Probability or Probability via Expectation by Peter Whittle, or in volume one of An introduction to probability theory and its applications by William Feller.

I think you are overlooking the fact that it does not matter whether "we" can distinguish the dice or not, but rather it matters that the dice are unique and distinct, and act on their own accord.

So if in the closed box scenario, you open the box and see a 1 and a 2, you don't know whether it is $(1,2)$ or $(2,1)$, because you cannot distinguish the dice. However, both $(1,2)$ and $(2,1)$ would lead to the same visual you see, that is, a 1 and a 2. So there are two outcomes favoring that visual. Similarly for every non-same pair, there are two outcomes favoring each visual, and thus there are 36 possible outcomes.

Mathematically, the formula for the probability of an event is

$$\dfrac{\text{Number of outcomes for the event}}{\text{Number of total possible outcomes}}.$$

However, this formula only holds for when each outcome is equally likely. In the first table, each of those pairs is equally likely, so the formula holds. In your second table, each outcome is not equally likely, so the formula does not work. The way you find the answer using your table is

Probability of 1 and 2 = Probability of $(1,2)$ + Probability of $(2,1)$ = $\dfrac{1}{36} + \dfrac{1}{36} = \dfrac{1}{18}$.

Another way to to think about this is that this experiment is the exact same as rolling each die separately, where you can spot Die 1 and Die 2. Thus the outcomes and their probabilities will match with the closed box experiment.

Lets imagine that the first scenario involves rolling one red die and one blue die, while the second involves you rolling a pair of white dice.

In the first case, can write down every possible outcome as (red die, blue die), which gives you this table (reproduced from your question):

$$\begin{array} {|c|c|c|c|c|c|c|} \hline \frac{\textrm{Blue}}{\textrm{Red}}&1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & (1,1) & \mathbf{(1,2)} & (1,3) & (1,4) & (1,5) & (1,6) \\ \hline 2 & \mathbf{(2,1)} & (2,2) & (2,3) & (2,4) & (2,5) & (2,6) \\ \hline 3 & (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6) \\ \hline 4 & (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6) \\ \hline 5 & (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6) \\ \hline 6 & (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \\ \hline \end{array}$$ Our idealized dice are fair (each outcome is equally likely) and you've listed every outcome. Based on this, you correctly conclude that a one and a two occurs with probability $\frac{2}{36}$, or $\frac{1}{18}.$ So far, so good.

Next, suppose you roll two identical dice instead. You've correctly listed all the possible outcomes, but you incorrectly assumed all of these outcomes are equally likely. In particular, the $(n,n)$ outcomes are half as likely as the other outcomes. Because of this, you cannot just calculate the probability by adding up the # of desired outcomes over the total number of outcomes. Instead, you need to weight each outcome by the probability of it occurring. If you run through the math, you'll find that it comes out the same--one doubly-likely event in the numerator out of 15 double-likely events and 6 singleton events.

The next question is "how could I know that the events aren't all equally likely?" One way to think about this is to imagine what would happen if you could distinguish the two dice. Perhaps you put a tiny mark on each die. This can't change the outcome, but it reduces the problem the previous one. Alternately, suppose you write the chart out so that instead of Blue/Red, it reads Left Die/Right Die.

As a further exercise, think about the difference between seeing an ordered outcome (red=1, blue=2) vs. an unordered one (one die showing 1, one die showing 2).

The key idea is that if you list the 36 possible outcomes of two distinguishable dice, you are listing equally probable outcomes. This is not obvious, or axiomatic; it's true only if your dice are fair and not somehow connected. If you list the outcomes of indistinguishable dice, they are not equally probable, because why should they be, any more than the outcomes "win the lottery" and "don't win the lottery" are equally probable.

To get to the conclusion, you need:

• We are working with fair dice, for which all six numbers are equally probable.
• The two dice are independent, so that the probability of die number two obtaining a particular number is always independent of what number die number one gave. (Imagine instead rolling the same die twice on a sticky surface of some kind that made the second roll come out different.)

Given those two facts about the situation, the rules of probability tell you that the probability of achieving any pair $(a,b)$ is the probability of achieving $a$ on the first die times that of achieving $b$ on the second. If you start lumping $(a,b)$ and $(b,a)$ together, then you don't have the simple independence of events to help you any more, so you can't just multiply probabilities. Instead, you have made a collection of mutually exclusive events (if $a \neq b$), so you can safely add the probabilities of getting $(a,b)$ and $(b,a)$ if they are different.

The idea that you can get probabilities by just counting possibilities relies on assumptions of equal probability and independence. These assumptions are rarely verified in reality but almost always in classroom problems.

If you translate this into terms of coins - say, flipping two indistinguishable pennies - it becomes a question of only three outcomes: 2 heads, 2 tails, 1 of each, and the problem is easier to spot. The same logic applies, and we see that it's more likely to get 1 of each than to get 2 heads or 2 tails.

That's the slipperiness of your second table - it represents all possible outcomes, even though they are not all equally weighted probabilities, as in the first table. It would be ill-defined to try to spell out what each row and column in the second table means - they're only meaningful in the combined table where each outcome has 1 box, regardless of likelihood, whereas the first table displays "all the equally likely outcomes of die 1, each having its own row," and similarly for columns and die 2.

Let's start by stating the assumption: indistinguishable dice only roll 21 possible outcomes, while distinguishable dice roll 36 possible outcomes.

To test the difference, get a pair of identical white dice. Coat one in a UV-absorbent material like sunscreen, which is invisible to the naked eye. The dice still appear indistinguishable until you look at them under a black light, when the coated die appears black while the clean die glows.

Conceal the pair of dice in a box and shake it. What are the odds you'll get a 2 and a 1 when you open the box? Intuitively you might think "rolling a 1 and a 2" is just 1 of 21 possible outcomes because you can't tell the dice apart. But if you open the box under a black light, you can tell them apart. When you can tell the dice apart, "rolling a 1 and a 2" is 2 of 36 possible combinations.

Does that mean a black light has the power to change the probability of obtaining a certain outcome, even if the dice are only exposed to the light and observed after they've been rolled? Of course not. Nothing changes the dice after you stop shaking the box. The probability of a given outcome can't change.

Since the original assumption depends on a change that doesn't exist, it's reasonable to conclude that the original assumption was incorrect. But what about the original assumption is incorrect - that indistinguishable dice only roll 21 possible outcomes, or that distinguishable dice roll 36 possible outcomes?

Clearly the black light experiment demonstrated that observation has no impact on probability (at least on this scale - quantum probability is a different matter) or the distinctness of objects. The term "indistinguishable" merely describes something which observation cannot differentiate from something else. In other words, the fact that the dice appear the same under some circumstances (i.e. that they aren't under a black light) and not others has no bear on the fact that they are truly two distinct objects. This would be true even if the circumstances under which you're able to distinguish between them are never discovered.

In short: your ability to distinguish between the dice being rolled is irrelevant when analyzing the probability of a particular outcome. Each die is inherently distinct. All outcomes are based on this fact, not on an observer's point of view.

We can deduce that your second table does not represent the scenario accurately.

You have eliminated all the cells below and left of the diagonal, on the supposed basis that (1, 2) and (2, 1) are congruent and therefore redundant outcomes.

Instead suppose that you roll one die twice in a row. Is it valid to count 1-then-2 as an identical outcome as 2-then-1? Clearly not. Even though the second roll outcome does not depend on the first, they are still distinct outcomes. You cannot eliminate rearrangements as duplicates. Now, rolling two dice at once is the same for this purpose as rolling one die twice in a row. You therefore cannot eliminate rearrangements.

(Still not convinced? Here is an analogy of sorts. You walk from your house to the top of the mountain. Tomorrow you walk back. Was there any point in time on both days when you were at the same place? Maybe? Now imagine you walk from your house to the top of the mountain, and on the same day another person walks from the top of the mountain to your house. Is there any time that day when you meet? Obviously yes. They are the same question. Transposition in time of untangled events does not change deductions that can be made from those events.)

If we just observe "Somebody gives me a box. I open the box. There is a $1$ and a $2$", without further information, we don't know anything about the probability.

If we know that the two dice are fair and that they have been rolled, then the probability is 1/18 as all other answer have explained. The fact we don't know if the die with 1 o the die with 2 was rolled first doesn't matter, because we must account for both ways - and therefore the probability is 1/18 instead of 1/36.

But if we don't know which process led to having the 1-2 combination, we can't know anything about the probability. Maybe the person who handed us the box just purposely chose this combination and stuck the dice to the box (probability=1), or maybe he shacked the box rolling the dice (probability=1/18) or he might have chosen at random one combination from the 21 combinations in the table you gave us in the question, and therefore probability=1/21.

In summary, we know the probability because we know what process lead to the final situation, and we can compute probability for each stage (probability for each dice). The process matters, even if we haven't seen it taking place.

To end the answer, I'll give a couple of examples where the process matters a lot:

• We flip ten coins. What's the probability getting heads all of ten times? You can see that the probability (1/1024) is a lot smaller than the probability of getting a 10 if we just choose a random number between 0 and 10 (1/11).
• If you have enjoyed this problem, you can try with the Monty Hall problem. It's a similar problem where the process matters much more than what our intuition would expect.

The probability of event A and B is calculated by multiplying both probabilities.

The probability of rolling a 1 when there are six possible options is 1/6. The probability of rolling a 2 when there are six possible options is 1/6.

1/6 * 1/6 = 1/36.

However, the event is not contingent on time (in other words, it is not required that we roll a 1 before a 2; only that we roll both a 1 and 2 in two rolls).

Thus, I could roll a 1 and then 2 and satisfy the condition of rolling both 1 and 2, or I could roll a 2 and then 1 and satisfy the condition of rolling both 1 and 2.

The probability of rolling 2 and then 1 has the same calculation:

1/6 * 1/6 = 1/36.

The probability of either A or B is the sum of the probabilities. So let's say event A is rolling 1 then 2, and event B is rolling 2 then 1.

Probability of Event A: 1/36 Probability of Event B: 1/36

1/36 + 1/36 = 2/36 which reduces to 1/18.