General Zod

(Model_Atheist) Zod: Two fair dice are rolled. Assuming at least one of them is a 6, what are the odds that both of them are a six?

He says the odds are 1/11.

Autonemesis

1 in 6, same as for a single die. Unless I misunderstood, we aren't considering rolls that do not have at least one die showing six. That ends up being the same as rolling a single die.

Friar Bellows

Sorry, Kind Bud, but Sartre's friend on IRC is correct. Note, he doesn't specify which die is a 6, just that at least one of them is a 6. And so we have the following possibilities:

6 1
6 2
6 3
6 4
6 5
6 6
1 6
2 6
3 6
4 6
5 6

1 in 11.

If he had said the left die must be a 6, then the answer would be 1 in 6. But he didn't specify which die, left or right. And so it's 1 in 11.

parkdalian

Wrong. We know that one die is a six, so we're only concerned with the roll of the other die.
Left or right doesn't matter.

6 1
6 2
6 3
6 4
6 5
6 6

1 6
2 6
3 6
4 6
5 6
6 6


One in six.

beausoleil

It depends on how you do the experiment.

Case 1. My friend rolls two dice. Without showing me the result, he shows me that one is a 6. Chance that the other is a six? 1 in 11. (There are 11 combinations of dice that have one 6, only one of them has a second 6.)

Case 2. I roll a dice. If it comes up 6 I roll a second dice. Chance that the other is a 6? 1 in 6.

From the way the problem is framed, it sounds like case 1 to me.

parkdalian

Are we kidding here? Take the dice he showed you and put it in your proverbial pocket. Forget that one, whether it was the left or right die, the top or bottom die, the die that was on Earth or the one that was rolled on Mars. Now take a peek at the other die. What are the chances of the unknown one being a six?

beausoleil

1 in 11 - try it!

There are 11 combinations of two dice that have at least one 6. Take that 6 away from each. How many 6's are there in the 11 other dice? One.

The two events ceased to be indepent when the 'at least one six' condition was imposed. You would be right if the second dice was rolled again after the first 6 had been removed.

[ July 03, 2002: Message edited by: beausoleil ]</p>

parkdalian

Okay I'll bite. There are actually 6 combinations, or 6 to be considered in each of two main contingencies.

Suppose we have a red die and a white die. In the first main contingency we have the red die as 6

That leaves:

red white
6 1
6 2
6 3
6 4
6 5
6 6

In the second contingency, the white die is the known six:

red white
1 6
2 6
3 6
4 6
5 6
6 6

Because of the two main contingencies within the information given (one die is a six, either red or white), the combinations are split into 2 groups with only one group considered separately, throwing the other combination group out, depending on the contingency.

In either contingency, of which one or the other must occur with 100 percent probability, the odds are 1 in 6 for 2 sixes.

Perhaps someone who is a real mathematician can explain this better?

Principia

This problem is similar to a classic, in which you are asked to give the probability to the following question: <a href="http://mathforum.org/dr.math/faq/faq.boy.girl.html" target="_blank">In a two-child family, one child is a boy. What is the probability that the other child is a girl?</a>?

The answer to the above question is 2/3. Note that the solution depends heavily on the wording of the question. In particular, the question above is in essence asking, Of all the two-child families in the world with one boy, what fraction of them also have a girl? So, by conditioning the problem, all families with two girls are never considered.

Now, let's consider Sartre's question: Two fair dice are rolled. Assuming at least one of them is a 6, what are the odds that both of them are a six?

The question explicitly asks what is the probability that two sixes were rolled, given that one of them is a 6. In other words, out of a sufficiently large number of fair two-dice throws with one 6 showing (regardless of which of the dice is showing), what fraction of them might be two 6s?

The sample space is thus:

1 6
2 6
3 6
4 6
5 6
6 6
6 5
6 4
6 3
6 2
6 1

[NB: Some of the posters mistakingly double-count the (6 6) case twice.]

So the answer is 1 out of 11.

Now, here's the more interesting question. Boys and girls, like real-world dice are not fair. Suppose one occurs with probability p and the other with probability q, what are the probabilities of the above problems?

liquid

Both your maths are 'correct', it's your interpretations that are different, and that is down to the wording given - it's somewhat ambiguous. So ponder over the semantics...