What is the probability of getting doubles when 2 dice are rolled?

2 dices:

number of possible outcomes: $6^2$
number of possible pairs: $1$
probability: $6 / 6^2 = 1/6$.

3 dices:

number of possible outcomes: $6^3$
number of possible doubles: $\{(a,a,b),(a,b,a),(b,a,a)\}$ with $a,b\in\{1,2,3,4,5,6\}$ ($b \neq a$). For each possibe $b$. In total, you have $3*6*5$ possible pairs. probability = $3*6*5/6^3 = 90/216$.

4 dices:

number of possible outcomes: $6^4$
number of possible pairs:
number of possible $N$-tuples with $a,b,c$ outcomes $\{(a,a,b,c), (a,b,a,c), (a,b,c,a), \ldots \}$, with $a,b,c \in \{1, \ldots, 6\}$ times the possible values for $a,b,c=6*5*4$ ($b \neq c\neq a$).
The number of possible permutations of 4 elements with 2 repetitions is $4!/2!=12$, that must be multiplied by the number of possible different faces $6*5*4$.
probability: $12*6*5*4/6^4$

In general:

the number of possible permutations of $k$ with $2$ replicates is given by $k!/2!$, the number of possible distinct $k-1$ values is $6*5*\ldots*(6-(k-1)+1)$, and given that all possible permutations are $6^k$:

$$ P(k)=\frac{k!/2! * 6 * 5 * \ldots * (6-(k-1)+1)}{6^k}=\frac{\frac{k!}{2!} \frac{6!}{(6-(k-1))!}}{6^k} $$

From the formula, it's clear that with $k = 8$ the probability becomes 0.

One popular way to study probability is to roll dice. A standard die has six sides printed with little dots numbering 1, 2, 3, 4, 5, and 6. If the die is fair (and we will assume that all of them are), then each of these outcomes is equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on. But what happens if we add another die? What are the probabilities for rolling two dice?

To correctly determine the probability of a dice roll, we need to know two things:

• The size of the sample space or the set of total possible outcomes
• How often an event occurs

In probability, an event is a certain subset of the sample space. For example, when only one die is rolled, as in the example above, the sample space is equal to all of the values on the die, or the set (1, 2, 3, 4, 5, 6). Since the die is fair, each number in the set occurs only once. In other words, the frequency of each number is 1. To determine the probability of rolling any one of the numbers on the die, we divide the event frequency (1) by the size of the sample space (6), resulting in a probability of 1/6.

Rolling two fair dice more than doubles the difficulty of calculating probabilities. This is because rolling one die is independent of rolling a second one. One roll has no effect on the other. When dealing with independent events we use the multiplication rule. The use of a tree diagram demonstrates that there are 6 x 6 = 36 possible outcomes from rolling two dice.

Suppose that the first die we roll comes up as a 1. The other die roll could be a 1, 2, 3, 4, 5, or 6. Now suppose that the first die is a 2. The other die roll again could be a 1, 2, 3, 4, 5, or 6. We have already found 12 potential outcomes, and have yet to exhaust all of the possibilities of the first die.

The possible outcomes of rolling two dice are represented in the table below. Note that the number of total possible outcomes is equal to the sample space of the first die (6) multiplied by the sample space of the second die (6), which is 36.

The same principle applies if we are working on problems involving three dice. We multiply and see that there are 6 x 6 x 6 = 216 possible outcomes. As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify work. For two dice, there are 62 possible outcomes. For three dice, there are 63 possible outcomes. In general, if we roll n dice, then there are a total of 6n possible outcomes.

With this knowledge, we can solve all sorts of probability problems:

1. Two six-sided dice are rolled. What is the probability that the sum of the two dice is seven?

The easiest way to solve this problem is to consult the table above. You will notice that in each row there is one dice roll where the sum of the two dice is equal to seven. Since there are six rows, there are six possible outcomes where the sum of the two dice is equal to seven. The number of total possible outcomes remains 36. Again, we find the probability by dividing the event frequency (6) by the size of the sample space (36), resulting in a probability of 1/6.

2. Two six-sided dice are rolled. What is the probability that the sum of the two dice is three?

In the previous problem, you may have noticed that the cells where the sum of the two dice is equal to seven form a diagonal. The same is true here, except in this case there are only two cells where the sum of the dice is three. That is because there are only two ways to get this outcome. You must roll a 1 and a 2 or you must roll a 2 and a 1. The combinations for rolling a sum of seven are much greater (1 and 6, 2 and 5, 3 and 4, and so on). To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18.

3. Two six-sided dice are rolled. What is the probability that the numbers on the dice are different?

Again, we can easily solve this problem by consulting the table above. You will notice that the cells where the numbers on the dice are the same form a diagonal. There are only six of them, and once we cross them out we have the remaining cells in which the numbers on the dice are different. We can take the number of combinations (30) and divide it by the size of the sample space (36), resulting in a probability of 5/6.

The following is from Joseph Mazur’s new book, What’s Luck Got to Do with It?:

…there is an authentically verified story that sometime in the 1950s a [roulette] wheel in Monte Carlo came up even twenty-eight times in straight succession.  The odds of that happening are close to 268,435,456 to 1.  Based on the number of coups per day at Monte Carlo, such an event is likely to happen only once in five hundred years.

Assuming the math is right (we’ll check it later), can you find the flaw in his argument?  The following example will help.

The Probability of Rolling Doubles

Imagine you hand a pair of dice to someone who has never rolled dice in her life.  She rolls them, and gets double fives in her first roll.  Someone says, “Hey, beginner’s luck!  What are the odds of that on her first roll?”

Well, what are they?

There are two answers I’d take here, one much better than the other.

The first one goes like this.  The odds of rolling a five with one die are 1 in 6; the dice are independent so the odds of rolling another five are 1 in 6; therefore the odds of rolling double fives are

$$(1/6)*(1/6) = 1/36$$.

1 in 36.

By this logic, our new player just did something pretty unlikely on her first roll.

But wait a minute.  Wouldn’t ANY pair of doubles been just as “impressive” on the first roll?  What we really should be calculating are the odds of rolling doubles, not necessarily fives.  What’s the probability of that?

Since there are six possible pairs of doubles, not just one, we can just multiply by six to get 1/6. Another easy way to compute it: The first die can be anything at all.  What’s the probability the second die matches it?  Simple: 1 in 6.  (The fact that the dice are rolled simultaneously is of no consequence for the calculation.)

Not quite so remarkable, is it?

For some reason, a lot of people have trouble grasping that concept.  The chances of rolling doubles with a single toss of a pair of dice is 1 in 6.  People want to believe it’s 1 in 36, but that’s only if you specify which pair of doubles must be thrown.

Now let’s reexamine the roulette “anomaly”

This same mistake is what causes Joseph Mazur to incorrectly conclude that because a roulette wheel came up even 28 straight times in 1950, it was very likely an unfair wheel.  Let’s see where he went wrong.

There are 37 slots on a European roulette wheel.  18 are even, 18 are odd, and one is the 0, which I’m assuming does not count as either even or odd here.

So, with a fair wheel, the chances of an even number coming up are 18/37.  If spins are independent, we can multiply probabilities of single spins to get joint probabilities, so the probability of two straight evens is then (18/37)*(18/37).  Continuing in this manner, we compute the chances of getting 28 consecutive even numbers to be $$(18/37)^{28}$$.

Turns out, this gives us a number that is roughly twice as large (meaning an event twice as rare) as Mazur’s calculation would indicate.  Why the difference?

Here’s where Mazur got it right: He’s conceding that a run of 28 consecutive odd numbers would be just as interesting (and is just as likely) as a run of evens.  If 28 odds would have come up, that would have made it into his book too, because it would be just as extraordinary to the reader.

Thus, he doubles the probability we calculated, and reports that 28 evens in a row or 28 odds in a row should happen only once every 500 years.  Fine.

But what about 28 reds in a row?  Or 28 blacks?

Here’s the problem: He fails to account for several more events that would be just as interesting. Two obvious ones that come to mind are 28 reds in a row and 28 blacks in a row.

There are 18 blacks and 18 reds on the wheel (0 is green).  So the probabilities are identical to the ones above, and we now have two more events that would have been remarkable enough to make us wonder if the wheel was biased.

So now, instead of two events (28 odds or 28 evens), we now have four such events.  So it’s almost twice as likely that one would occur.  Therefore, one of these events should happen about every 250 years, not 500.  Slightly less remarkable.

What about other unlikely events?

What about a run of 28 numbers that exactly alternated the entire time, like even-odd-even-odd, or red-black-red-black?  I think if one of these had occurred, Mazur would have been just as excited to include it in his book.

These events are just as unlikely as the others.  We’ve now almost doubled our number of remarkable events that would make us point to a broken wheel as the culprit.  Only now, there are so many of them, we’d expect that one should happen every 125 years.

Finally, consider that Mazur is looking back over many years when he points out this one seemingly extraordinary event that occurred.  Had it happened anytime between 1900 and the present, I’m guessing Mazur would have considered that recent enough to include as evidence of his point that roulette wheels were biased not too long ago.

That’s a 110-year window.  Is it so surprising, then, that something that should happen once every 125 years or so happened during that large window?  Not really.

Slightly unlikely perhaps, but nothing that would convince anyone that a wheel was unfair.

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