# Math Help - couple of dice probability questions

1. ## couple of dice probability questions

i am not sure if this is in the right place, however if it is not, could a mod or admin kindly move it, thankyou.

ok, i am stumped trying to work out the formula for a spreadsheet to work out a couple of things. i have googled but nothing seemed to cover my exact topic

firstly, how would i work out the probability of rolling over a certain number on a six sided die which if not over that number could be re rolled once

secondly, how would i work out the probability of two dice being over a certain number?

2. Originally Posted by Magnus
how would i work out the probability of two dice being over a certain number?
Here are all the possible outcomes from tossing a pair of dice.
There are thirty-six of them.
Count the pairs that give the required result.
Its probability is that number divided by 36.
(1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5), (6,6)

3. Hello, Magnus!

How would i work out the probability of rolling over a certain number
on a six-sided die, which, if not over that number, could be re-rolled once?

Of course, it depends on the chosen number.
. . We must consider the five possible choices.

[1] Roll a number greater than 1
. . (a) It can be done on the first roll: . $P(x > 1,\text{ 1st roll}) = \frac{5}{6}$
. . (b) If the first roll is not > 1, $P(x \not{>} 1) = \frac{1}{6}$
. . . . the probability that the second is > 1 is: $\frac{5}{6}$
. . . . $P(x > 1,\text{ 2nd roll}) \:=\:\frac{1}{6}\cdot\frac{5}{6}\:=\:\frac{5}{36}$

. . Hence: . $P(x > 1)\:=\:\frac{1}{6}+\frac{5}{6} \:=\:\frac{35}{36}$

[2] Roll a number greater than 2.
. . (a) It can be done on the first roll: . $P(x > 2,\text{ 1st roll}) = \frac{4}{6}$
. . (b) If the first roll is not > 2, $P(x \not{>} 2) = \frac{2}{6}$
. . . . the probability that the second is > 2 is: $\frac{4}{6}$
. . . . $P(x > 2,\text{ 2nd roll}) \:=\:\frac{2}{6}\cdot\frac{4}{6} \:=\:\frac{8}{36}$

. . Hence: . $P(x > 2)\:=\:\frac{4}{6} + \frac{8}{36}\:=\:\frac{32}{36}\:=\:\frac{8}{9}$

Get the idea?