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Today my professor said that if I roll two identical dice at the same time then there will be $21$ outcomes and the probability of getting sum of $7$ is $1/7$: \begin{array}{c c c c c c} \{1,1\}, & \{1,2\}, & \{1,3\}, & \{1,4\}, & \{1,5\}, & \{1,6\},\\ & \{2,2\}, & \{2,3\}, & \{2,4\}, & \{2,5\}, & \{2,6\},\\ & & \{3,3\}, & \{3,4\}, & \{3,5\}, & \{3,6\},\\ & & & \{4,4\}, & \{4,5\}, & \{4,6\},\\ & & & & \{5,5\}, & \{5,6\},\\ & & & & & \{6,6\} \end{array}

but when I roll two different dice, then the probability of getting sum of 7 is $1/6$ \begin{array}{c c c c c c} (1,1) & (1,2) & (1,3) & (1,4) & (1,5) & (1,6)\\ (2,1) & (2,2) & (2,3) & (2,4) & (2,5) & (2,6)\\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & (3,6)\\ (4,1) & (4,2) & (4,3) & (4,4) & (4,5) & (4,6)\\ (5,1) & (5,2) & (5,3) & (5,4) & (5,5) & (5,6)\\ (6,1) & (6,2) & (6,3) & (6,4) & (6,5) & (6,6) \end{array}

How this is possible?

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    $\begingroup$ The probability would be 1/7 for the first case if and only if all the outcomes are equally likely, which isn't the case. Take example as (1,1) and (1,2). $\endgroup$ – Eagle May 24 '19 at 22:13
  • $\begingroup$ Welcome to MathSE. This tutorial, which includes a discussion of arrays, explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig May 24 '19 at 23:06
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If you roll two identical dies, there are 21 outcomes, but they have different probabilities: probability to get $(1, 1)$ (or any outcome with the same numbers on both dies) is $\frac{1}{36}$, but probability to get $(1, 3)$ (or any outcome with different numbers) is $\frac{2}{36}$. So correct answer for two identical dies is $P((1, 6)) + P((2, 5)) + P((3, 4)) = \frac{6}{36} = \frac{1}{6}$ too.

Just having set of $n$ outcomes doesn't mean any outcome has probability of $\frac{1}{n}$.

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  • $\begingroup$ Writing $(1, 3)$ implies you have an ordered pair, which is not what you mean. You should write $\{1, 3\}$ instead. $\endgroup$ – N. F. Taussig May 24 '19 at 23:09
  • $\begingroup$ It's question of notation. We can agree to write the smaller number first, and don't consider $(3, 1)$ as an outcome at all. $\endgroup$ – mihaild May 24 '19 at 23:13
  • $\begingroup$ Since the probability is (situation)/(possible outcomes) how can you consider 21 outcomes as 36 ? $\endgroup$ – Adnan Kayıkçı May 24 '19 at 23:43
  • $\begingroup$ No, probability isn't always (number of outcomes we are interested in) / (number possible outcomes), because different outcomes can have different probabilities (and also grouping outcomes is arbitrary). Classic joke is there are just two outcomes: sun either will explode tomorrow, or will not. But clearly probabilities of this outcomes are different. $\endgroup$ – mihaild May 24 '19 at 23:50
  • $\begingroup$ Thank you I guess I got it $\endgroup$ – Adnan Kayıkçı May 24 '19 at 23:55
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If you roll two dice then there are $6*6=36$ possible outcomes.

6 of those outcomes add up to 7.

So yes , the probability of rolling the sum of 7 is $\frac{6}{36}=\frac{1}{6}$

But in your first statement you considered $(2,1)$ and $(1,2)$ as one outcome. These are Different possible outcomes. If you consider those as one outcome then there are 21 possible outcomes. In other words, there are 21 different possible number combinations that the dice could show. But in reality you must treat these separately to calculate probabilities.

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The second situation is easiest, so let's start there. In the second situation, you throw two six-sided dice. There are 6*6=36 different outcomes (for example [1,6] and [6,1] are two different outcomes), and six of them have a sum of 7. So 6/36 = 1/6 outcomes have a sum of 7, as you said.

In the first situation, the situation is unusual. Instead of picking out of the 6*6=36 different outcomes, you're supposed to pick from the set of outcomes where you can't tell the difference between the two dice. That is, [1,6] and [6,1] are considered to be the same event.

There are 21 different pairings in this case (not 23?): {1,1} ... {1,6}, {2,2}, ..., {2,6}, {3,3},...{3,6}, ..., ..., {5,6}, {6,6}. In the physically realistic case, the probability of each pairing is not the same. For example, when rolling dice it's twice as likely to get {2,4} than to get {3,3} because there are two of the 36 outcomes yield {2,4} and only one yields {3,3}. In a physically unrealistic case, you can imagine a model where you choose one of these pairings uniformly at random. There are only three of those {1,6}, {2,5}, {3,4}, so the chances of getting a sum of 7 are 3/21 = 1/7 in this physically unrealistic case.

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