1 in 6. Odds do not change based on number of times. The odds are the same for every roll of the dice.
Well for the dice, its exactly the same, prior rolls do not influence latter rolls.
multiplication. potentially factorials.
80% chance, I believe.
I’m no math whiz either, but I think in the case of dice it would be expressed in a percentage as 100/6 = 16.6667%
How about:
(number of times you roll the dice) * (1/6)
of course, this proves statistians are liars, in that you could theoretically roll it 6 times and never get the number you were looking for.
(There’s actually a real way to do this, but I don’t have any coffee in me.)
What’s the event?
Is it a human influenced event?
Your wording is a little unclear. If you mean "What are the chances that I will get a 'five' six times in a row?", that that would be one thirty-sixth.
Regards,
Just calculate the chance of NOT getting a 5 in 6 roles of the dice:
5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6 = 0.335
So the chance of getting at least one 5 is 1 - 0.335 = .665
or 66.5%
Think of it this way. What's the chance of rolling one die one time and NOT getting a 5? It's 5/6.
So the chances of rolling a die six times and not getting a 5 is: (5/6) to the power of 6. That's 0.335. Then subtract that from 1 (100%). Result: 0.665 or 66.5%.
Disclaimer: I did this on a untrustworthy calculator. Please do not use my result to build a rocket or anything like that.
Six of those total outcomes contain exactly one "five".
If you're looking for the odds of getting exactly one "five" in six rolls of a die, that would be 6/46656, or .0001286 (to the best of my recollection).
jewbacca and pelican have the correct formulation, if I understand your question correctly.
1) the probability of throwing a 5 one or more times in 6 rolls is: 1 - (5/6)^6, where '^' is the symbol for exponentiation.
2) the probability of throwing a 5 exactly once in 6 rolls is: 6 / 64, or, reduced, 3 /32, or approx. .0937
For questions like 2), the fastest (probably easiest, too) solution can be found with Pascal's triangle, which for 6 rolls (the 7th line of the triangle) is: 1 6 15 20 15 6 1
Hope that's of some use to you, and FReegards!
Odds of rolling (at least) one five in n rolls (assuming independence/equal likelihood of rolls) would be => 1 - (5/6)^n. So for six rolls you get 1 - 0.3348 = ~67% chance.
You are not looking to calculate percentage. You want to calculate the odds (or the chances) of something occurring. This involves the use of statistics and depending on the nature of the sample size and the occurrence can be quite complex to calculate. Moreover, you may want to know the correlation values and that gets even more interesting
There is a 5/6 chance of failure on each roll. Multiply that number by itself roll each successive roll.
EXACTLY 1 five in six rolls equals the prob of 1 success times the prob of fail/fail/fail/fail/fail.
that's (1/6)(5/6)(5/6)(5/6)(5/6)(5/6) X 6
the final X 6 is because you don't care which of the six rolls is a success.
apologies in advance for typos. I'm on my iPad. I hate typing on it.
Basically if you want to know the probability of rolling a “5” at least once when rolling a 6 sided die 6 times it goes like this:
The percentage for rolling anything except a 5 is 5/6 or 83.33...% probability. Now the probability of rolling every die roll NOT a 5 is 5/6 times 5/6 (6 times) or (5/6)^6 which comes to 33.49% probability that you will not roll even 1 “5” when rolling 6 times.
So the inverse (meaning you DO roll at least 1 “5”) is basically 100% minus the probability that you will NOT roll any “5”’s or 66.51% probability.
Remember though that there are lies, damn lies and then statistics. What I wrote above is considered statistics. :-)
using Common Core...I’m guessing two weeks for the answer
Your chances of getting a specific number at least once in 6 rolls of a die is 1-(5/6)^6 = ~66.5%
What is the common core method of figuring this out? :)