B Dice calculation of p(7), two different values?

[PeroK]

Summary:: Six sided fair dice (two) labeled standard 1-6; two methods of calculating p(7) get different p?

Scenario 1]
The dice are rolled out of your sight and you are then asked the p(7), before you are allowed to look at the results.

Calculate p(7) by looking at table of 36 possibilities and counting the number of sum cells totals that equal 7, there are six of them, so

p(7) = 6/36 = 1/6

Scenario 2]
Same as in 1 except before you're requested for your calculation of p(7) you are informed that one of the faces shows a six.

You calculate p(7) by looking at the table of 36 possibilities and counting the number of possibilities that include six (the cells comprised of row 6 and column 6) so you count 11 of these cells whose row or column value is 6, and notice that 2 of them indicate 7 as their cell sums so

p(7) = 2/11 > 1/6

Sorry to come in late. The question is ambiguous and has no well-defined answer unless you describe how the information about the faces has been obtained. This is the same as the "second child" problem.

To show how the question has no unique answer, consider these three example cases:

1) The dice are thrown. Someone looks at the first die and tells you what it reads. In this case, you are told "one of the faces shows an $n$", where $n$ is any number between one and six with equal probability.

In this scenario, the probability that the total is $7$ remains $1/6$.

2) The dice are thrown. Someone looks at both dice and if there is a six they tell you. In this case, you are told "one of the faces shows a $6$" with probability $11/36$. And, you are presumably told (or can infer) that neither die shows a $6$ with probability $25/36$.

In this scenario, the probability that the total is $7$ is slightly changed (as described and simulated above).

3) The dice are thrown. A passer-by is brought in off the street and asked to look at the dice and say something. They say: "one of the faces shows a $6$".

In this scenario, you cannot compute a well-defined probability without making further assumptions about human behaviour and, in particular: if the dice show $2$ and $6$, say, how likely it is that the person will choose the $2$ or the $6$? The natural assumption is perhaps scenario 1: that the choice is equally likely. But, who knows, people in general may have a strong bias towards seeing the number six.

 

[PeroK]

Here is perhaps a cleaner way to look at this. Imagine that the data comes from a computer program. It simulates the roll of two dice then tells you something.

The first program checks (potentially both dice) and reports either: (at least) one face shows a six; or, not. This is the implied/assumed scenario.

The second program checks the first die and reports: (at least) one face shows an $n$, where $n = 1-6$. Note the equal probability for the different $n$ here. This is the second scenario.

The third program checks first for a one, then for a two etc. and reports the first number it finds: (at least) one face shows an $n$, where $n = 1-6$. Note now that there is not an equal probability of each $n$. And, in fact, if this program reports a $6$, then we know for certain that both dice are $6$. Otherwise, it would have found and reported the lower number.

The probabilities depend not just on the data that comes out, but the process the computer is following to generate the data. Without knowing the process, the question is ambiguous.

 

[sysprog]

If we interpret the situation as that the reporter is confronted by 2 randomly rolled dice, and always truthfully reports "at least one of the dice shows $n$", where $n$ = 1, 2, 3, 4, 5, or 6, and of the 36 possible cases, in the 30 of them that are such that the 2 dice show different values, the reporter decides at random which number to report, whereas in the 6 cases in which the 2 dice show the same value, the reporter has only that value to report, so he reports that value, then in the in the 30/36 cases in which the numbers are different, there is a 1/2 chance for each of the 2 numbers that the reporter will report that number, whereas in the 6 cases in which both dice show the same value, there is a 1/1 chance of the reporter reporting the double number, which means that compared to the 30 different-number cases, the 6 double-number cases are double-weighted, which results in the same 1/6 chance of the dice summing to 7 as for 2 independent dice about which nothing is reported.

Breaking up that sentence in hope of improved perusability:

If we interpret the situation as that the reporter is confronted by 2 randomly rolled dice, and always truthfully reports "at least one of the dice shows $n$", where $n$ = 1, 2, 3, 4, 5, or 6,

and of the 36 possible cases, in the 30 of them that are such that the 2 dice show different values, the reporter decides at random which number to report,

whereas in the 6 cases in which the 2 dice show the same value, the reporter has only that value to report, so he reports that value,

then in the in the 30/36 cases in which the numbers are different, there is a 1/2 chance for each of the 2 numbers that the reporter will report that number,

whereas in the 6 cases in which both dice show the same value, there is a 1/1 chance of the reporter reporting the double number,

which means that compared to the 30 different-number cases, the 6 double-number cases are double-weighted,

which results in the same 1/6 chance of the dice summing to 7 as for 2 independent dice about which nothing is reported.