- #1

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- 47

## Summary:

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

## Main Question or Discussion Point

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

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

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

A closer look reveals that if you had been informed that one of the faces showed a 5, 4, 3, 2, or 1, you would find the same number of cells containing that row-column number (eleven of them), and the same number of cells totaling 7, (two of them).

So it would not have mattered if you were informed of one of the faces or not; you could have just imagined that it must be some number and used this method in the first scenario... what is wrong with this method? If the total cells selected came to 12 instead of 11, the p(7) would be right... Some reason the row-column intersection cell needs to be counted twice?

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, sop(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 sop(7) = 2/11 > 1/6

A closer look reveals that if you had been informed that one of the faces showed a 5, 4, 3, 2, or 1, you would find the same number of cells containing that row-column number (eleven of them), and the same number of cells totaling 7, (two of them).

So it would not have mattered if you were informed of one of the faces or not; you could have just imagined that it must be some number and used this method in the first scenario... what is wrong with this method? If the total cells selected came to 12 instead of 11, the p(7) would be right... Some reason the row-column intersection cell needs to be counted twice?