Math question

Hi,
Can anyone help me with the following question:
You have a dice with 12 sides (so numbers 1 to 12), you may throw it 5 times. What is the chance you will throw 3 times the number 6?
I thought it was 1/12*1/12*1/12 but then I don’t think about the other 2 times you can throw the dice, correct?
Thanks!
Jelle
Answer #1
I have no idea on working this out. But it would have to be far more complex than 5 multiplications surely. As you’re saying you want the odds for throwing 3 6’s out of 5 throws correct? You have 11 other numbers that could possibly come up in those 5 throws without a 6 being rolled at all. I’m sure someone will work it out for you though.
Answer #2
Found a rather difficult solution, don’t understand all of it myself yet but maybe it can help:
http://www.wikihow.com/Calculate-Multiple-Dice-Probabilities
http://www.ehow.com/how_5858157_calculate-dice-probabilities.html
or maybe this
http://anydice.com/
http://topps.diku.dk/torbenm/troll.msp
Answer #3
I think the hardest part to figure out is you need 3 6’s out of 5 throws maximum. I think is good at maths
Answer #4
I think the hardest part to figure out is you need 3 6's out of 5 throws maximum. I think is good at maths Thanks for the vote of confidence! If you find the right equation, I think it won’t be that much of a problem… Btw, another Belgian? They seem to be dropping out of the woodwork today… Small world!
Answer #5
OP, before I answer the question, what is this for? Is it for fun or…? Do you study statistics and probability?
Answer #6
Well there are 248,832 total combinations to start with. But i don’t know how you’d work out getting at least 3 times a chosen number (6 in this case)
As i’m guessing odds would go out of this world if you wanted all 5 throws to be the same number. @
You probably are better at math than me, but i thought since you said you didn’t quite understand info from link you posted, someone who was better at it then us would
Answer #7
I remember studying this for stats and probability,
I was never good in it, they want to throw number 6, three times.
chances of getting a 3 in the die. = 1/12
chances of getting other numbers in the die = 11/12
If the person is throwing the die 5 times, you have to consider the other 2 cases where the die is not a 3 too.
Check out this link, I think your question goes under “Combinations with Repetition”
http://www.mathsisfun.com/combinatorics/combinations-permutations.html[code]
Answer #8
it could be:
(11*10*10) / 12^5
because 11*10 is the total combinations when the 3 sixes are present at the first 3 positions, multiplied by the total amount of positions (in the five rolls in total) the number six can be rolled at, over the total amount of combinations (12^5). Could be wrong.
Answer #9
Not trying to doubt what you’re saying as i don’t have a clue at all, but you’re basing figures on rolling a 6 with first 3 throws. Is that the right way to work it out though. Because that’s where i’m totally confused. As the odds of getting a 6 on first 3 rolls has got to be a lot higher than getting 3 out of 5 rolls. If that makes sense.
Answer #10
Thanks all! i found the formula though it is:
5!/(3!*(5-3)!) * (1/12)^3 * (11/12)^2
= 10* (1/12)^3 * (11/12)^2
= 0.005
= 0.5% chance of having 3 6'es