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# Probability & Statistics/Probability

Question
QUESTION: Hi Clyde,

I have a very easy question for you, but I just want to confirm it.

If you have 2 dice, what is the probability if you roll the dice that you get an even number on one and an odd number on the other? Is it 6/12 (1/2) chance of getting it?

Thanks

ANSWER: Yes -- this is an instance where the "gut" answer is right. Whatever the first die is rolled, the second die has a 1/2 chance of matching the evenness/oddness (which is called "parity").

You can confirm this by doing a probability tree or table, but there's really no need. Half of all outcomes have matching parity, the other half do not:

(1,1),(1,3),(1,5),(2,2),(2,4),(2,6),(3,1),(3,3),(3,5),(4,2),(4,4),(4,6),(5,1),(5,3),(5,5),(6,2),(6,4),(6,6)

(1,2),(1,4),(1,6),(2,1),(2,3),(2,5),(3,2),(3,4),(3,6),(4,1),(4,3),(4,5),(5,2),(5,4),(5,6),(6,1),(6,3),(6,5)

I am assuming the premise here is that the dice are "distinguishable" -- which is, in my opinion, always the correct interpretation. That's why I've used ordered pairs above.

If the dice are "indistinguishable" (which can only be true "in theory" and not in real life), then the set of outcomes are NOT ordered pairs like I listed above. In this case, oddly enough, the probability is not 1/2. The outcomes are:

{1,1},{1,3},{1,5},{2,2},{2,4},{2,6},{3,3},{3,5},{4,4},{4,6},{5,5},{6,6}

{1,2},{1,4},{1,6},{2,3},{2,5},{3,4},{3,6},{4,5},{5,6}

There are 21 total outcomes, and 12 of them have the same parity, which gives probability 12/21 = 4/7, which is about 57%. Here, as I said, this is "non-real" in the sense that "indistinguishable dice" don't exist. The two objects are, in reality, physically distinct.

This may be too much depth, the answer you want is indeed 1/2, but this other case is interesting to deal with. The way to "rectify" this problem is to deal with the case that the dice match separately. In the real life setting (5,6) and (6,5) are separate outcomes, which makes a non-match of this type twice as likely as a match of the type (6,6). In the case that the dice are indistinguishable, {5,6} and {6,5} are the same unordered set, so the outcome is {5,6} in either case and this is equally likely as {6,6}.

So how do we break it down?

Well, in real life there is a 5/6 chance the dice are different, and a 1/6 chance they are the same. If they are the same, then they have the same parity. If they are different, there is a 2/5 chance they are the same parity (there are only two other options, because if you roll a 3, you can't pick 3 as the matching parity, only 1 or 5).

That gives you (5/6)(2/5) + (1/6)(1/1) = 1/3 + 1/6 = 1/2 probability.

Why did we break it down this way? Because we can break it down the same way for the other case, and the numbers change slightly and give you a different outcome.

In this hypothetical world where the dice are indistinguishable, the way you can think of it is that the chances of rolling non-doubles is cut in half -- or, alternatively, the chances of rolling doubles is doubled.

So, if you roll the first die and then look at the second, because of this bias, there is a 2/7 chance of rolling doubles, and a 5/7 chance of not doing so.

That gives you (5/7)(2/5) + (2/7)(1/1) = 2/7 + 2/7 = 4/7 probability.

And that's the in-depth answer to your question. But, as I said, yes indeed, in real life the "gut" answer of 1/2 can easily be checked as correct.

---------- FOLLOW-UP ----------

QUESTION: Thanks Clyde.

I thought I would let you know that the question I used states one die is a normal one, numbered 1 to 6. The second on the other hand is numbered 1,1,2,3,6,6. Seeing as the second has three odd numbers anyway, I would assume the outcome is the same.

Is this correct or not?

Yes, the answer is still 1/2 for two reasons:

1. In the "real life" case, the dice are distinguishable, the answer is 1/2, and everything is good.

2. In the "not real life case" the dice are indistinguishable, but because the two dice are not the same, we can assume that this is not a valid case at all. (One could still do the computation here, it would be somewhat different and very much "not real-life.")

Probability & Statistics

Volunteer

#### Clyde Oliver

##### Expertise

I can answer all questions up to, and including, graduate level mathematics. I do not have expertise in statistics (I can answer questions about the mathematical foundations of statistics). I am very much proficient in probability. I am not inclined to answer questions that appear to be homework, nor questions that are not meaningful or advanced in any way.

##### Experience

I am a PhD educated mathematician working in research at a major university.

Organizations
AMS

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Various research journals of mathematics. Various talks & presentations (some short, some long), about either interesting classical material or about research work.

Education/Credentials
BA mathematics & physics, PhD mathematics from a top 20 US school.

Awards and Honors
Various honors related to grades, various fellowships & scholarships, awards for contributions to mathematics and education at my schools, etc.

Past/Present Clients
In the past, and as my career progresses, I have worked and continue to work as an educator and mentor to students of varying age levels, skill levels, and educational levels.