Here's a problem making the rounds about probability. I mention it not to get the "correct" answer, but as a target to invite conversations about what we should take probability to mean.
Sally rolls two dice (6-sided, assumed fair). She shows one is a six. What is the probability that the other one is a six?
1) Either zero or one.
2) 1/2
3) 1/6
4) 1/11
5) 1/12
6) 1/36
7) Make up another answer or even reject the premise.
This is a variant of classic probability problem (usually stated in terms of sons or daughters). The correct answer is 1/11, as you proved, by enumerating the sample space, here:
Once you see that one of the dice is a six, you know the set we are dealing with is: {1,6; 2,6; 3,6; 4,6; 5,6; 6,6; 6,1; 6,2; 6,3; 6,4; 6,5}
Don't believe your own results? Let's do a simulation. The following code draws 1 million random samples of a simulated pair of fair 6-sided dice, and computes the proportion of samples where both die came up 6 among those where at least one die came up 6:
Code:
d1 <- sample.int(6, 1e6L, replace=TRUE)
d2 <- sample.int(6, 1e6L, replace=TRUE)
sum(d1 == 6 & d2 == 6) / sum(d1 == 6 | d2 == 6)
# [1] 0.09087517
The result, shown in the last line, is 1/11 to within a small sampling error.
However, there is actually an ambiguity in the wording of the problem. The probability that the second die is a 6, depends on how Sally came to learn that one die was a 6. The 1/11 probability is assumes that she looked at both dice, saw that at least one of the was a 6 and then asked you to compute the probability that the other one was also a 6. In this case, the sample space if the set of 11 outcomes in which at least one of the dice is a 6—that is, the sample space you enumerated. Those outcomes are equally likely, and so the answer is 1/11.
On the other hand, what if Sally only looked at one of the dice, saw that it was a six, and that no matter what outcome she saw, she would not look at the other die. Given those conditions, she then asked you to compute the probability that the other die is a six. Now, the sample space is simply {1, 2, 3, 4, 5, 6}. And since all 6 possible outcomes are equally likely, the answer is 1/6.
1) The outcome is already determined and in fact, was determined long ago as each cause led to a subsequent effect until a combination of material events plus the laws of nature resulted in the pre-determined outcome. The results are fixed and must be either zero or one, depending on those prior causes. The fact that we don't know the details is irrelevant - only adding an unnecessary subjective element. In the actual world the dice have already been rolled and their state is a matter of certain, historical fact.
That logic is unproductive. It doesn't help you win at poker, for example, filter the spam from your email, or find a lost hiker. So, it's pretty useless to think of probabilities of events that have already happened, but whose outcome you don't know, as having to be 0 or 1.
2) True, the results were determined, but the question is about my own estimation and state of knowledge. Since I cannot determine between zero or one and have no insight into the prior determinants, I must average the two and state "1/2".
This answer is indefensible. You might not know what the outcome is, but you know what a 6-sided die is, and it isn't something that has a 50-50 chance of landing 6. You have equal ignorance about each of the 11 possible outcomes, and so your subjective probability of each outcome is 1/11.
6) (I picked this one before I found it lacking.) The relevant randomizing event is the original roll. Anything after is tainted by agency (Sally's choices). We know that rolling two die will generate 36 possibilities, only one of which is a pair of sixes. Therefore, keeping only the original roll "pristine" we get 1/36.
This is just nonsense. You were given some information about the outcome. This restricts the sample space to those outcomes of which at least one die showed a six. There are 11 of them, not 36.
