Probablility question.

[Energize]

How do you work out probability for dice rolling?

Say you roll 3 normal dice what are the chances of rolling 2 ones?

Something like

1/6 x 1/6 x 1/6?

I did loads of these diagrams but forgotten how to do them now.

 

[SiriusB]

or = +

and = x

So, odds of rolling a 1 and a 1 and a 1 is 1/6 x 1/6 x 1/6!

SiriusB

 

[daz]

OK hang on, i think it's

3C2 * [(1/6)^2] * [(5/6)^1] = 7% chance of rolling two 1's if you roll 3 dice.

 

[Docaroo]

yep you mulitply the odds each time you "roll" the dice...

So 2 1's in a row is 1/6 x 1/6 = 1/36 !!!!

This works because when you roll the dice there are 36 possible outcomes to the roll...

1 + 1
1 + 2
1 + 3
....
2 + 1.... etc

 

[daz]

yep you mulitply the odds each time you "roll" the dice...

So 2 1's in a row is 1/6 x 1/6 = 1/36 !!!!

Yes, but rolling two 1's from the three dice is slightly different. The first one can be a 5, the second and third could be a 1, et cetera, et cetera, so all of these combinations have to be taken into account.

 

[Haircut]

OK hang on, i think it's

3C2 * [(1/6)^2] * [(5/6)^1] = 7% chance of rolling two 1's if you roll 3 dice.

Yep, I got that as well.

Also should point out that this is the figure for exactly two 1s, if he wants at least two 1s then I make that 8.3%

 

[Energize]

Ah forgot to mention that, it is at least 2 ones. Im sure I can't remember the forumula having been that complicated, I guess we didn't learn that at gcse after all.

 

[daz]

Google can calculate binomial coefficients!

I typed in 3 choose 2.... and it works!

 

[Haircut]

Ah forgot to mention that, it is at least 2 ones. Im sure I can't remember the forumula having been that complicated, I guess we didn't learn that at gcse after all.

It's not that complicated really - try thinking of it this way:

For at least two 1s, you need two of the dice to be 1 and the other can be anything.

two 1s in two dice is 1/6 * 1/6 = 1/36
Now the probability of the other die being anything is 1/6 * 6 = 1

As it doesn't matter which two dice have the two 1s we need to multiply this by the number of ways we can pick two dice from 3 possible choices. There are three ways to do this, so you get 1/36 * 1 * 3 = 1/12 = 8.3%

 

[Energize]

Ah thanks that makes it simple.

 

[daz]

Now the probability of the other die being anything is 1/6 * 6 = 1

This should read the probability of anything other than a one is 5/6 * 1

 

[Haircut]

This should read the probability of anything other than a one is 5/6 * 1

Only if you want to specify exactly two 1s and no more. He's already said he wants at least two 1s.

 

[daz]

Sorry dude, I haven't done probability in a long while.

 

[Reeve]

wouldnt it be
1/6*1/6*1/6 for 1,1,1
and add on 1/6*1/6*5/6 since there is 3 different ways to get two ones on three dice you would multiply that by 3

so 1/216 + 15/216 = 16/216 = 4/54