Probability of die rolls

[kkneafsey]

I am trying to find out what the formula is for rolling at least one "1" when I roll ten die simultaneously. I am not so much looking for the answer as I am the formula to figure out probabilities of other rolls. Please help!

[AlFas]

It's easier to work out the probability of rolling no '1's at all and then subtracting this value from 1 (all probabilities added together come to 1).

Given that the probabilty of not throwing a '1' with a single dice is 5/6, the probability of not throwing a '1' with 10 dice is (5/6)^10, [or 5/6 x 5/6 ten times].

So the probability of throwing at least one '1' from 10 throws is;

 1 - (5/6)^10    which is about 84%

[Tweed]

The probability is 1/6 * 10 which is 1 and 2/3. If you think about it like this. There's a 1 in 6 chance of rolling any number with a single die. So the chance of rolling a particular number with two dice is 1/6 + 1/6 =1/3 for three, you would add up 1/6 three times etc. So the chances of rolling at least one out of 10 rolls is a certainty.

[AlFas]

Tweed I think you need to re-think your theory. By your reckoning, if you flipped a coin twice you would always get 1 head and 1 tail!?

If you throw 10 dice you aren't necessarily going to get any 1's are you?

[Tweed]

Yes Allfas I agree. Ignore the answer. Don't know what I was thinking of.

[kkneafsey]

Thanks everyone...

[JohnPPotts]

The probability of exactly 1 "1" on ten dice is 10*(1/6)*((5/6) to the power 9). The general formula requires you to use Pascal's triangle (that's where the 10 at the start of the formula comes from) which takes a little explaining or a lot of Algebra!

[JohnPPotts]

...which is about 32%

[kkneafsey]

The probability of rolling "at least" one "one" is about 31% so I would have to think the probability of rolling exactly one "one" would be much much less than that.

[AlFas]

No no no.

The probability of rolling exactly one '1' is 32% and of rolling one or more '1s' is 84% as stated.

[kkneafsey]

That's right, my bad... Thanks for the help.

[Big Bear]

I think you'll find that the previous answers are all twaddle!!.

The correct answer for rolling a particular number on a die, in this case 1,  is 1 in 6, simply because there are a total of 6 numbers that can come up.

Taking this further, if you roll 2 dice simultaneously, the chances of rolling a 1 with the first die AND a 1 with the second die are 1 in 6 AND 1 in 6, which is 1 in 36 - you simply multiply the fractions together, thus: 1/6 x 1/6

For 3 dice: 1/6 x 1/6 x 1/6 = 1/216

For 4 dice: 1/6 x 1/6 x 1/6 x 1/6 = 1/1296

For x dice 1/6 raised to the xth power

So if you roll 10 dice simultaneously, the probability of all ten comin up '1' is 1/6 raised to the 10th power, which is 1/60466176 or, put another way, one chance in 60,466,176 - longer odds than winning the lottery!

You can get more info on probabilities at Dr Math at http://mathforum.org/dr.math/

 

[AlFas]

Big Bear, I agree with what you say, but you have answered a different question to the one asked (and correctly answered). You're working out the probabilty of getting 10 1s, we've been discussing getting at least one 1.