Thread: a probability question

THANK YOU IN ADVANCE

Originally Posted by Narek

THANK YOU IN ADVANCE

You just have to calculate the probabilities of both cases: rolling 8 with 2 dice, and rolling 8 with 3 dice.

Starting with the 2 dice: which combinations of dice rolls yield 8?
- 2 with dice1, and 6 with dice2
- 6 with dice1, and 2 with dice1
(these are two different events!)
...
Count them, and count how many different combinations of dice rolls (with two dice) are possible overall.

Before you start with the 3 dice, it helps to look at the probability distribution of 2-dice-rolls, as there's a simple pattern to it.

we consider P1 for rolling 2 dices:
( we roll dices together )

so we have 5 pairs of numbers which if summed up, makes 8
so the probability is:

P1 = 5 / 6^2 = 5/36

for 3 dices we have 18 pairs of numbers which if summed up, makes 8
so the probability is:

P2 = 18 / 6^3 = 18/216

from here it is obvious that P1 > P2
__________________________________________________ ___________

Am I true? any mistake?

Originally Posted by Narek

we consider P1 for rolling 2 dices:
( we roll dices together )

so we have 5 pairs of numbers which if summed up, makes 8
so the probability is:

P1 = 5 / 6^2 = 5/36

Correct.

Originally Posted by Narek

for 3 dices we have 18 pairs of numbers which if summed up, makes 8
so the probability is:

P2 = 18 / 6^3 = 18/216

I think you made a counting mistake.
D1 = 1: then D2 + D3 = 7, therefore: 6 triples
D1 = 2: then D2 + D3 = 6, therefore: 5 triples
..
D1 = 6: then D2 + D3 = 2, therefore: 1 triple
Total: 21 triples

Originally Posted by Narek

from here it is obvious that P1 > P2
__________________________________________________ ___________

Am I true? any mistake?

That conclusion still stands