# How do we know that the probability of rolling 1 and 2 is 1/18?

Since my first probability class I have been wondering about the following.

Calculating probabilities is usually introduced via the ratio of the "favored events" to the total possible events. In the case of rolling two 6-sided dice, the amount of possible events is $36$, as displayed in the table below.

\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}

If we therefore were interested in calculating the probability of the event A "rolling a $1$ and a $2$", we would see that there are two "favored events" and calculate the probability of the event as $\frac{2}{36}=\frac{1}{18}$.

Now, what always made me wonder is: Let's say it would be impossible to distinguish between the two dice and we would only observe them after they were rolled, so for example we would observe "Somebody gives me a box. I open the box. There is a $1$ and a $2$". In this hypothetical scenario we would not be able to distinguish between the two dice, so we would not know that there are two possible events leading to this observation. Then our possible events would like that:

\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}

and we would calculate the probability of event A as $\frac{1}{21}$.

Again, I am fully aware of the fact that the first approach will lead us to the correct answer. The question I am asking myself is:

How do we know that $\frac{1}{18}$ is correct?

The two answers I have come up with are:

• We can empirically check it. As much as I am interested in this, I need to admit that I haven't done this myself. But I believe it would be the case.
• In reality we can distinguish between the dice, like one is black and the other one blue, or throw one before the other or simply know about the $36$ possible events and then all the standard theory works.

My questions to you are:

• What other reasons are there for us to know that $\frac{1}{18}$ is correct? (I am pretty sure there must be a few (at least technical) reasons and this is why I posted this question)
• Is there some basic argument against assuming that we cannot distinguish between the dice at all?
• If we assume that we cannot distinguish between the dice and have no way to check the probability empirically, is $P(A) = \frac{1}{21}$ even correct or did I overlook something?

Thank you for taking your time to read my question and I hope it is specific enough.

• The simple answer: because this is probability of distinguishable events. There are probabilistic models in physics of indistinguishable events (e.g. Einstein-Bose statistic). – Tim Jul 25 '16 at 17:08
• This is one reason there are axioms of probability: you can know that $1/18$ is correct when you can deduce it using solely the axioms and the rules of logic. – whuber Jul 25 '16 at 17:41
• Use a pair of dice where one is red and the other green. You can tell them apart, but someone with red-green color-blindness can't. Should the probabilities be based on what you see or what he sees? – Monty Harder Jul 25 '16 at 19:02
• While all the posted answers were very informative (thank you to everybody who contributed!) and mostly made me realise that in fact - no matter how one puts it - dice are distinguishable, I think @Tim 's answer was exactely what I was looking for (dziękuję bardzo)! I did some further research on this topic and really liked this article and this video. – elmo Jul 25 '16 at 19:47
• @ELM it's nice to hear it :) For completeness I added my own answer. – Tim Jul 25 '16 at 20:23

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.

• Why did I choose this as the best answer? As I stated above, all the answers were very informative (thank you again to everybody who invested time, I really appriciate it!) and also showed me that it is not necessary for me to be able to distinguish between the dice myself as long as the dice can objectively be distinguished. But as soon as they can be objectively distinguished it was clear to me that the events in the second scenario are not equally probable, so for me the Bose-Einstein-model was what I was looking for. – elmo Jul 25 '16 at 21:30

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).

• this. being able to distinguish the dice does not change the result. The observer cannot act on the result. (unless magic?). The dice don't care if you can make the difference between red and blue. – njzk2 Jul 26 '16 at 19:36
• "you incorrectly assumed all of these outcomes are equally likely" I think this is the key part and probably the most direct answer to the original question. – Gediminas Jul 27 '16 at 8:41

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.

• Welcome to our site! You can use Latex formatting for the math here by putting dollar signs around it, e.g. $a^x$ produces $a^x$ – Silverfish Jul 25 '16 at 18:59

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.