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Simple probability question

D

duragezic

Lifer
Call me a retard, and I've even had AP Stats in high school, but I don't remember jack, and I got this question for a totally unrelated class:

A pair of dice is rolled 10 times, what is the probability that there is a 7 (sum of) at least once? I was thinking along the lines of there's a possibility of 6 out of 36 rolls is a 7, so 1/6, then times 10, but that's over 1... Is it just (1/6)^10 instead?
 
Each roll is independent so it's usually easier to calculate the probably that none of the rolls will be 7, and subtract that from 1.0.
 
Originally posted by: duragezic
Call me a retard, and I've even had AP Stats in high school, but I don't remember jack, and I got this question for a totally unrelated class:

A pair of dice is rolled 10 times, what is the probability that there is a 7 (sum of) at least once? I was thinking along the lines of there's a possibility of 6 out of 36 rolls is a 7, so 1/6, then times 10, but that's over 1... Is it just (1/6)^10 instead?
Click to expand...

I'm not positive, but I think what you want to do is add up all the possibilities. So:

You get one 7: 1/6*(5/6)^9
You get two 7's: (1/6)^2*(5/6)^8
.
.
.
Do that for all of them, and then add up all 10 fractions. I think that's how you do it.
 
If you rolled the dice once, the chances are 1/6.

If you roll the dice twice... the chances are ONE 7 + TWO 7s = 1/6 + 1/36.

If you roll the dice thrice... the chances are ONE 7 + TWO 7s + THREE 7s = 1/6 + 1/36 + 1/6^3....

So if you roll it N times, chances are SUM(1/(6^n), where n = 1 to N..

In your case, if you roll it 10 times, the chances of getting 7 AT LEAST one time is SUM(1/(6^n) n=1 to 10 is 0.19999999, or 20%.

BOW!
 
if you roll a pair of dice these are the probabilities that you get a certain sum
p(2) = 1/36 = (1+1)
p(3) = 2/36 = (1+2) (2+1)
p(4) = 3/36 = (1+3) (2+2) (3+1)
p(5) = 4/36 = (1+4) (2+3) (3+2) (4+1)
p(6) = 5/36 = (1+5) (2+4) (3+3) (4+2) (5+1)
p(7) = 6/36 = (1+6) (2+5) (3+4) (4+3) (5+2) (6+1)
p(8) = 5/36 = (2+6) (3+5) (4+4) (5+3) (6+2)
p(9) = 4/36 = (3+6) (4+5) (5+4) (6+3)
p(10) = 3/36 = (4+6) (5+5) (6+4)
p(11) = 2/36 = (5+6) (6+5)
p(12) = 1/36 = (6+6)

So on a single roll:
p(7) = 6/36 = 1/6
p(not 7) = 5/6
p(not 7 for 10 straight rolls) = p(not 7)^10 = (5/6)^10 = 9765625/60466176 ~= 16.15%

so the probability that you get one or more seven rolls somewhere in those 10 is (1-.1615) = 83.85%
 
Originally posted by: Kyteland
if you roll a pair of dice these are the probabilities that you get a certain sum
p(2) = 1/36 = (1+1)
p(3) = 2/36 = (1+2) (2+1)
p(4) = 3/36 = (1+3) (2+2) (3+1)
p(5) = 4/36 = (1+4) (2+3) (3+2) (4+1)
p(6) = 5/36 = (1+5) (2+4) (3+3) (4+2) (5+1)
p(7) = 6/36 = (1+6) (2+5) (3+4) (4+3) (5+2) (6+1)
p(8) = 5/36 = (2+6) (3+5) (4+4) (5+3) (6+2)
p(9) = 4/36 = (3+6) (4+5) (5+4) (6+3)
p(10) = 3/36 = (4+6) (5+5) (6+4)
p(11) = 2/36 = (5+6) (6+5)
p(12) = 1/36 = (6+6)

So on a single roll:
p(7) = 6/36 = 1/6
p(not 7) = 5/6
p(not 7 for 10 straight rolls) = p(not 7)^10 = (5/6)^10 = 9765625/60466176 ~= 16.15%

so the probability that you get one or more seven rolls somewhere in those 10 is (1-.1615) = 83.85%
Click to expand...

WINNAR! This is the absolutely positively correct answer.

 
Originally posted by: Kyteland
if you roll a pair of dice these are the probabilities that you get a certain sum
p(2) = 1/36 = (1+1)
p(3) = 2/36 = (1+2) (2+1)
p(4) = 3/36 = (1+3) (2+2) (3+1)
p(5) = 4/36 = (1+4) (2+3) (3+2) (4+1)
p(6) = 5/36 = (1+5) (2+4) (3+3) (4+2) (5+1)
p(7) = 6/36 = (1+6) (2+5) (3+4) (4+3) (5+2) (6+1)
p(8) = 5/36 = (2+6) (3+5) (4+4) (5+3) (6+2)
p(9) = 4/36 = (3+6) (4+5) (5+4) (6+3)
p(10) = 3/36 = (4+6) (5+5) (6+4)
p(11) = 2/36 = (5+6) (6+5)
p(12) = 1/36 = (6+6)

So on a single roll:
p(7) = 6/36 = 1/6
p(not 7) = 5/6
p(not 7 for 10 straight rolls) = p(not 7)^10 = (5/6)^10 = 9765625/60466176 ~= 16.15%

so the probability that you get one or more seven rolls somewhere in those 10 is (1-.1615) = 83.85%
Click to expand...

:thumbsup: for using 1 - probability of getting no 7's, rather than finding the sum of the probability of 1 seven, 2 sevens, etc.

 
Originally posted by: DrPizza
Originally posted by: Kyteland
if you roll a pair of dice these are the probabilities that you get a certain sum
p(2) = 1/36 = (1+1)
p(3) = 2/36 = (1+2) (2+1)
p(4) = 3/36 = (1+3) (2+2) (3+1)
p(5) = 4/36 = (1+4) (2+3) (3+2) (4+1)
p(6) = 5/36 = (1+5) (2+4) (3+3) (4+2) (5+1)
p(7) = 6/36 = (1+6) (2+5) (3+4) (4+3) (5+2) (6+1)
p(8) = 5/36 = (2+6) (3+5) (4+4) (5+3) (6+2)
p(9) = 4/36 = (3+6) (4+5) (5+4) (6+3)
p(10) = 3/36 = (4+6) (5+5) (6+4)
p(11) = 2/36 = (5+6) (6+5)
p(12) = 1/36 = (6+6)

So on a single roll:
p(7) = 6/36 = 1/6
p(not 7) = 5/6
p(not 7 for 10 straight rolls) = p(not 7)^10 = (5/6)^10 = 9765625/60466176 ~= 16.15%

so the probability that you get one or more seven rolls somewhere in those 10 is (1-.1615) = 83.85%
Click to expand...

:thumbsup: for using 1 - probability of getting no 7's, rather than finding the sum of the probability of 1 seven, 2 sevens, etc.
Click to expand...
I do this every day for a living. I know how to make it nice and easy.

Dice are cake compared to slot machines. 😉

 
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