Woher wissen wir, dass die Wahrscheinlichkeit, 1 und 2 zu würfeln, 1/18 beträgt?

Seit meiner ersten Wahrscheinlichkeitsklasse habe ich mich über Folgendes gewundert.

Die Berechnung der Wahrscheinlichkeiten erfolgt in der Regel über das Verhältnis der "favorisierten Ereignisse" zu den insgesamt möglichen Ereignissen. Beim Würfeln von zwei 6-seitigen Würfeln beträgt die Anzahl der möglichen Ereignisse $36$ , wie in der folgenden Tabelle dargestellt.

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

Wenn wir daher daran interessiert wären, die Wahrscheinlichkeit des Ereignisses A zu berechnen, "eine $1$ und eine $2$ ", würden wir sehen, dass es zwei "bevorzugte Ereignisse" gibt, und die Wahrscheinlichkeit des Ereignisses als 2 berechnen$\frac{2}{36}=\frac{1}{18}$ .

Was mich immer gefragt hat, ist: Sagen wir, es wäre unmöglich, zwischen den beiden Würfeln zu unterscheiden, und wir würden sie erst beobachten, nachdem sie gewürfelt wurden, also würden wir zum Beispiel beobachten, dass "jemand mir eine Schachtel gibt. Ich öffne die Schachtel. Es gibt eine und eine 2 ". In diesem hypothetischen Szenario wären wir nicht in der Lage, zwischen den beiden Würfeln zu unterscheiden, sodass wir nicht wissen würden, dass es zwei mögliche Ereignisse gibt, die zu dieser Beobachtung führen. Dann würden unsere möglichen Ereignisse so aussehen:$1$$2$

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

und wir würden die Wahrscheinlichkeit von Ereignis A als 1 berechnen .$\frac{1}{21}$

Auch hier bin ich mir der Tatsache voll bewusst, dass der erste Ansatz uns zur richtigen Antwort führen wird. Die Frage, die ich mir stelle, ist:

Woher wissen wir, dass ist richtig?$\frac{1}{18}$

Die zwei Antworten, die ich mir ausgedacht habe, sind:

• Wir können es empirisch überprüfen. So sehr ich daran interessiert bin, muss ich zugeben, dass ich das nicht selbst getan habe. Aber ich glaube, das wäre der Fall.
• In Wirklichkeit können wir zwischen den Würfeln unterscheiden, als wäre einer schwarz und der andere blau, oder wirf einen vor den anderen oder kenne einfach die möglichen Ereignisse und dann funktionieren alle Standardtheorien.$36$

Meine Fragen an Sie sind:

• Welche anderen Gründe gibt es für uns zu wissen, dass ist richtig? (Ich bin mir ziemlich sicher, dass es ein paar (zumindest technische) Gründe gibt und deshalb habe ich diese Frage gestellt.)$\frac{1}{18}$
• Gibt es ein grundlegendes Argument gegen die Annahme, dass wir überhaupt nicht zwischen den Würfeln unterscheiden können?
• Wenn wir annehmen, dass wir nicht zwischen den Würfeln unterscheiden können und keine Möglichkeit haben, die Wahrscheinlichkeit empirisch zu überprüfen, ist sogar richtig oder habe ich etwas übersehen?$P(A) = \frac{1}{21}$

Vielen Dank, dass Sie sich die Zeit genommen haben, meine Frage zu lesen, und ich hoffe, dass sie spezifisch genug ist.

Stellen Sie sich vor, Sie haben Ihren fairen sechsseitigen Würfel geworfen und haben ⚀. Das Ergebnis war so faszinierend, dass Sie Ihren Freund Dave angerufen und ihm davon erzählt haben. Da er neugierig war, was er bekommen würde, wenn er seinen schönen sechsseitigen Würfel warf, warf er ihn und bekam ⚁.

Ein Standardwürfel hat sechs Seiten. Wenn Sie nicht schummeln, landen Sie mit gleicher Wahrscheinlichkeit auf jeder Seite, dh zu 6 Mal. Die Wahrscheinlichkeit, dass du ⚀ wirfst, ist wie bei den anderen Seiten $1$$6$ . Die Wahrscheinlichkeit, dass du ⚀ wirfstunddein Freund ⚁ wirft, ist$\tfrac{1}{6}$ da die beiden Ereignisseunabhängig sindund wir unabhängige Wahrscheinlichkeiten multiplizieren. Anders ausgedrückt, es gibt36Anordnungen solcher Paare, die leicht aufgelistet werden können (wie Sie es bereits getan haben). Die Wahrscheinlichkeit des gegenteiligen Ereignisses (du wirfst ⚁ und dein Freund wirft ⚀) ist ebenfalls$\tfrac{1}{6} \times \tfrac{1}{6} = \tfrac{1}{36}$$36$ . Die Wahrscheinlichkeiten, die du wirfstunddein Freund wirft,oderdie du wirfstunddein Freund wirft, sindexklusiv, also addieren wir sie$\tfrac{1}{36}$ . Unter allen möglichen Vereinbarungen gibt es zwei, die diese Bedingung erfüllen.$\tfrac{1}{36} + \tfrac{1}{36} = \tfrac{2}{36}$

Woher wissen wir das alles? Nun, aus Gründen der Wahrscheinlichkeit , der Kombinatorik und der Logik, aber diese drei brauchen etwas Faktenwissen, auf das sie sich verlassen können. Wir wissen aufgrund der Erfahrung von Tausenden von Spielern und einigen Physikern, dass es keinen Grund zu der Annahme gibt, dass ein fairer sechsseitiger Würfel eine andere als die gleichwahrscheinliche Chance hat, auf jeder Seite zu landen. Ebenso haben wir keinen Grund zu der Annahme, dass zwei unabhängige Würfe irgendwie zusammenhängen und sich gegenseitig beeinflussen.

$2$$1$$6$$21$

Alles was man braucht , was gesagt, kommentieren , dass solche Modelle sind möglich, aber nicht für Dinge wie Würfel. In der auf empirischen Beobachtungen basierenden Teilchenphysik zeigte sich beispielsweise, dass die Bose-Einstein-Statistik nicht unterscheidbarer Teilchen (siehe auch das Problem der Sterne und Balken ) geeigneter ist als das Modell unterscheidbarer Teilchen. Einige Anmerkungen zu diesen Modellen finden Sie in Probability oder Probability via Expectation von Peter Whittle oder in Band 1 von Eine Einführung in die Wahrscheinlichkeitstheorie und ihre Anwendungen von 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.

Wenn Sie also im Szenario mit geschlossener Box die Box öffnen und eine 1 und eine 2 sehen, wissen Sie nicht, ob dies der Fall ist $(1,2)$ oder $(2,1)$, 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.