Statistics question about rolling dice...?

Question

Since I screwed up when I typed this the first time, let me try again.

How would you calculate the odd of rolling at least one of two specific numbers with multiple dice (for example 1 and 5). I know the odds for a single die would be 1/3.

For what it's worth, this is not a homework question. This is for personal interest and it has been so long since I took statistics that I don't remember how this works.

Answer 1

Let's take the example of getting at least one 1 or 5 in n dice. It is easier to work out the probability of getting zero 1s or 5s in n dice.

P(none in 1 die) = 2/3
P(none in n dice) = (2/3)^n

So required prob = 1 - (2/3)^n
It will be the same answer for whatever two numbers you pick. Note that as n gets larger, the prob of getting at least one of them approaches 1.

Answer 2

So, you know the probability of rolling at least one of two specific numbers (say 1 and 5) is 1/3.

Lets suppose you roll a die n times (note: This is exactly the same thing as rolling n dice once.)

Each roll of the die is independent of every other roll and the probability of "success", which we define as getting at least one of two specific numbers is 1/3 from roll to roll. So you are dealing with a bernoulli/binomial experiment.

Then, let X be a binomial random variability with p = 1/3. X represents the number of times (out of n) that we achieved a "success" when we rolled the die. What we want to know is the probability of getting at least one of two specific numbers in n rolls of the die. This is expressed as

P(X >= 1) = 1 - P(X < 1) = 1 - P(X = 0) = 1 - (1 - p)^n = 1 - (2/3)^n

n..........P(X >= 1)
1..........1/3 = 0.3333
2..........5/9 = 0.5556
3..........19/27 = 0.7037
4..........65/81 = 0.8025
5..........211/243 = 0.8683
10........58025/59049 = 0.9827

Math (and Stats) Rule!

Answer 3

You have to specify how many dice you have... if you are picking 2 numbers here (1, 5), I'm assuming you're using 2 dice.

With 2 dice, there are 36 possible outcomes.
The probability of rolling at least one of two specific numbers is P(rolling 1) + P (rolling 5) + P (rolling 1 and 5)

= (6 + 6 - 1)/36
= 11/36

Odd is 11: 25

Answer 4

For the answer, you would multiply the probability of rolling a number greater than 3 which would be 3/6 or 1/2 simplified by the probability of rolling an odd number which is also 3/6 or 1/2 simplified. Multiply 1/2 x 1/2 which equals 1/4.

Answer 5

First, it sounds like you are confusing probabilities with odds. Odds are ratios of probabilities.
The probability of rolling a 1 and a 5 with two dice is (1/6)*(1/6) = 1/36
That would be true with any 2 numbers.

Answer 6

One minus the odds of not getting either one on either die. The odds of not getting them on one is 2/3. The odds of not getting them on both is 2/3 times 2/3, or 4/9. So the odds of getting at least one is 5/9. For a third die, it's then 8/27 for not getting any, and therefore 19/27 that at at least one will appear.

Answer 7

well, the first roll is obviously 2/6 like you said.

the second roll...

there are 36 different combinations (6*6)

1,1-1,6; 2,1-2,6; so on and so forth

the 1-? and 5-? obviously already satisfy having a 1 or a 5.

of the rest, there are 2-1,3-1,4-1,6-1,2-5,3-5,4-5,6-5; or 8 more... so 2 dice = 20/36=5/9

The probability of getting something is 1-(the probability of NOT getting it) - I know, what a pessimistic view.

The probability each round of NOT getting it is 2/3

and as you keep going, that number goes down.

(2/3)^n ;where n is the number of rolls

1-(2/3)^n will be your answer

CHECK:

1/3 if n=1
5/9 if n=2

Answer 8

If you are particularly looking for 2 dice, you should search the web for craps odds. You should be able to find what you need there.

Answer 9

(chances of getting 1)*(chances of getting 5)*combinations(2 combinations, 1,5 and 5,1)
1/6*1/6*2=2/36