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Math question?

Posted on 05/07/2015 7:27:31 AM PDT by MNDude

I'm trying to figure something out for work, but my math skills are not very good. How do I calculate percentages if an event reoccurs multiple times?

Suppose there is an event that has the chance of occurring on in 6 times (like rolling a dice and getting a "five").

What are the chances I will get a "five" if I roll the dice six times?


TOPICS: Chit/Chat
KEYWORDS: math
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To: Nifster

Wrong. You apparently misread the question.

And where I actually posted the correct formula (my last post on this thread).


41 posted on 05/07/2015 8:14:23 AM PDT by Jewbacca (The residents of Iroquois territory may not determine whether Jews may live in Jerusalem)
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To: MNDude

I just sent you a PM. Call center calculations are NOT based on probability. They are best calculated using an erlang calculator that assumes talk time, work time, and contact time. You have to figure out those numbers, then figure out how you dial and the contact rate.

Using probability to staff call centers is a BIG mistake, as you will ALWAYS come up short, and under budget your expenses.

Read my PM and drop me a note with some specifics.

I used to run call centers that did both inbound and outbound contacts—about 13 million contacts a month. Projecting this stuff is an art—and you need to make sure you are going at it from the right direction.


42 posted on 05/07/2015 8:15:01 AM PDT by Vermont Lt
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To: Teacher317

Actually the odds of contact are reduced with every iteration of the list.

This is a replacement problem. If you are calling at the same time every day you cannot expect the customer to be home tomorrow night if they were not home tonight.

So, you need to calculate the entire process, rotating contact times, etc. Also you need to clear your list from the do not call lists in your state. The issues brought about from calling when you shouldn’t are not worth the effort for a temporary measure.


43 posted on 05/07/2015 8:17:54 AM PDT by Vermont Lt
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To: MNDude
AT LEAST 1 five in six rolls is (5/6)^6

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.

44 posted on 05/07/2015 8:18:14 AM PDT by Tanniker Smith (Rome didn't fall in a day, either.)
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To: Teacher317
Teach -- 6 outcomes out of 46656?? Rubbish.

There are 46656 total outcomes, WHEN CONSIDERING ALL 6 SIDES of the die. There are only 64 total outcomes when considering only 5s vs. non-5s...which we are here. Of those 64, exactly 6 meet our specification, to wit, exactly one 5, as I illustrated in a previous post.

Don't trust me, but I **should** trust Blaise Pascal, were I you. There is an excellent discussion of (essentially) this very problem in letters between Pascal and the Chevalier du Mere in Peter Bernstein's dandy book, Against The Gods, which is a fine treatise on the history and development of probability theory and risk analysis.

45 posted on 05/07/2015 8:28:34 AM PDT by SAJ
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To: Tanniker Smith

Sigh....


46 posted on 05/07/2015 8:29:31 AM PDT by SAJ
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To: DuncanWaring

It’s called a Bernoulli trial.

If you look on wikipedia,

the probability of success is 1/6 and the number of trials is 6.

perhaps you can understand other variations by looking at the article.


47 posted on 05/07/2015 8:33:04 AM PDT by menotomy
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To: MNDude

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. :-)


48 posted on 05/07/2015 8:33:16 AM PDT by Grombrindol
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To: MNDude

using Common Core...I’m guessing two weeks for the answer


49 posted on 05/07/2015 8:34:21 AM PDT by stylin19a (obama = Eddie Mush)
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To: MNDude

Your chances of getting a specific number at least once in 6 rolls of a die is 1-(5/6)^6 = ~66.5%


50 posted on 05/07/2015 8:39:10 AM PDT by Kirkwood (Zombie Hunter)
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To: Kirkwood

As specifically noted in #24.


51 posted on 05/07/2015 8:41:21 AM PDT by SAJ
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To: SAJ

I read the OP’s question and not any of the responses. Sorry if that bothers you.


52 posted on 05/07/2015 8:59:44 AM PDT by Kirkwood (Zombie Hunter)
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To: Jewbacca

The “36” is in superscript (ie to the 36th power)

1-(5/6)36 = 99.86%


Write it like this, with the carat to signify superscript

(5/6)^36


53 posted on 05/07/2015 9:01:32 AM PDT by ifinnegan
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To: stylin19a

...two weeks.

54 posted on 05/07/2015 9:03:56 AM PDT by RckyRaCoCo
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To: MNDude

What is the common core method of figuring this out? :)


55 posted on 05/07/2015 9:29:04 AM PDT by minnesota_bound
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To: MNDude

“I’m trying to figure something out for work, but my math skills are not very good. How do I calculate percentages if an event reoccurs multiple times?

Suppose there is an event that has the chance of occurring on in 6 times (like rolling a dice and getting a “five”).

What are the chances I will get a “five” if I roll the dice six times?”

If the rolls are independent and by “get a five” you mean you will get at least one five then the answer is:

1-(5/6)^6 which is the probability of the complement of getting no fives.

If the rolls are independent and by “get a five” you mean you will exactly one five then the answer is:

6*((1/6)*(5/6)^5)= (5/6)^5 which 6 times the probability of getting just one five on any specific roll.


56 posted on 05/07/2015 10:04:22 AM PDT by Catphish
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To: MNDude

16% chance independent of time of day? or is the 16% the average for the entire day?


57 posted on 05/07/2015 10:06:09 AM PDT by reed13k (For evil to triumph it is only necessary for good men to do nothings)
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To: ifinnegan

Freeping in an iPhone. No clue where that symbol is.


58 posted on 05/07/2015 11:24:28 AM PDT by Jewbacca (The residents of Iroquois territory may not determine whether Jews may live in Jerusalem)
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To: MNDude

Are you rolling one die six times or a pair of dice six times?

If its a pair of dice, there are 36 total possible combinations in the first roll, only four of which will add up to ‘five’.

Your odds of getting a ‘five’ in that first roll are 4/36 or 11.1 percent.

Your odds of getting a ‘five’ in each subsequent roll are the same.

Your odds of getting at least one ‘five’ after six rolls are 24/36 or 66.67 percent.


59 posted on 05/07/2015 11:56:39 AM PDT by chopperman
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To: MNDude

Depends on what factors control the situation you want to predict.


60 posted on 05/07/2015 11:58:22 AM PDT by A CA Guy ( God Bless America, God Bless and keep safe our fighting men and women.)
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