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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?


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KEYWORDS: math
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To: MNDude
Six rolls of a die have 66 (46656) possibilities.

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).

21 posted on 05/07/2015 7:43:58 AM PDT by DuncanWaring (The Lord uses the good ones; the bad ones use the Lord.)
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To: DuncanWaring
Only six outcomes have one five? I think you're way off on that one.

5 1 1 1 1 1
5 1 1 1 1 2
5 1 1 1 1 3
5 1 1 1 1 4...

think you see where this is going.

22 posted on 05/07/2015 7:46:12 AM PDT by Teacher317 (We have now sunk to a depth at which restatement of the obvious is the first duty of intelligent men)
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To: MNDude

jewbacca and pelican have the correct formulation, if I understand your question correctly.


23 posted on 05/07/2015 7:47:48 AM PDT by Teacher317 (We have now sunk to a depth at which restatement of the obvious is the first duty of intelligent men)
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To: MNDude
Dude -- your question is unclear. So,

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!

24 posted on 05/07/2015 7:50:57 AM PDT by SAJ
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To: MNDude
So I’m trying to calculate how many people we will reach if we make calls out to 1000 people, and we call each one up to six times. (stop calling those once we’ve made contact of course).

Ooooh! That's a very different problem! Now order matters, because a successful "hit" ends the stream of "rolls"!

1000 iterations of rolling up to 6 times, and stopping when you get a "five". Let me think on that one for a bit. Definitely tweaking the dustier parts of my brain for old college courses.

25 posted on 05/07/2015 7:52:08 AM PDT by Teacher317 (We have now sunk to a depth at which restatement of the obvious is the first duty of intelligent men)
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To: MNDude

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.


26 posted on 05/07/2015 7:53:00 AM PDT by LambSlave
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To: Teacher317

That must be why you’re the teacher and I’m not.


27 posted on 05/07/2015 7:53:14 AM PDT by DuncanWaring (The Lord uses the good ones; the bad ones use the Lord.)
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To: MNDude

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


28 posted on 05/07/2015 7:55:10 AM PDT by Nifster
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To: gorush

you haven’t taken a math class recently have you?


29 posted on 05/07/2015 7:56:03 AM PDT by Nifster
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To: Fester Chugabrew

Which is known as 1 in 6. Odds are not shown as percentages because that is misleading.


30 posted on 05/07/2015 7:57:15 AM PDT by Nifster
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To: LambSlave
You're right, he only wants the number of expected connected calls out of 1000... the odds per caller do not change, so once we have the odds, we just multiply by 1000 to get the expected number of live contacts.

And since the calls stop once he gets a single "five" (a live contact), the odds of subsequent rolls are irrelevant (because they won't waste time calling again).

So the odds of getting at least one "five" in 6 rolls is the 66.52 percent chance LambSlave gave...

Multiply that by 1000 target clients, and it gives you 665.2 likely live phone contacts if you stop at 6 tries, and you have a 16 percent chance of a live contact per attempted call.

31 posted on 05/07/2015 7:58:35 AM PDT by Teacher317 (We have now sunk to a depth at which restatement of the obvious is the first duty of intelligent men)
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To: Jewbacca

Wrong. You apparently missed the posts that state very clearly for dice rolls the answer is independent of previous results. This is not a closed box with balls that are drawn and reducing the opportunity case ( at least not as described by the poster)


32 posted on 05/07/2015 7:58:56 AM PDT by Nifster
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To: Teacher317
Sorry, Teach, you fail. If you're going to make a list of the serial OUTCOMES of a series of rolls, a 5 may occur as the outcome of any roll. SERIAL OUTCOMES with exactly one 5 in 6 rolls would be:

5 N N N N N
N 5 N N N N
N N 5 N N N
N N N 5 N N
N N N N 5 N
N N N N N 5

where N is some non-5 outcome of a roll...thus... ...exactly 6 such serial outcomes. Your list confuses permutations with combinations, ok?

33 posted on 05/07/2015 7:59:08 AM PDT by SAJ
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To: JPG
That 'can't' happen....but it did.

Deniro Fires Slots manager

34 posted on 05/07/2015 8:00:23 AM PDT by Bloody Sam Roberts ("It is never untimely to yank the rope of freedom's bell." - - Frank Capra)
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To: Teacher317

Awesome!! That’s they information I trying to figure out. Thank you everyone!


35 posted on 05/07/2015 8:00:45 AM PDT by MNDude
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To: MNDude

You are in so far over your own head it is not even funny.


36 posted on 05/07/2015 8:01:01 AM PDT by Nifster
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To: pelican001

That’s correct.


37 posted on 05/07/2015 8:03:42 AM PDT by grania
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To: MNDude

To be picky, though... your 16 percent and your rolling dice comparison is slightly off... 1/6 is actually a 16.66666(repeating) percent, which is what we used for our calculations. If it is really exactly 16.0 percent, and not 16.666666 percent, then your expected final contact number becomes (0.84^6)*1000... or closer to 648.7 contacts. I don’t know if those last 17 live contacts are important or not, but it is the difference between 1/6 and 16 percent.


38 posted on 05/07/2015 8:08:18 AM PDT by Teacher317 (We have now sunk to a depth at which restatement of the obvious is the first duty of intelligent men)
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To: SAJ
I don't like failing, LOL... and I think you misunderstand both his question and my answer.

You are right, however... if looked at as a serial outcome, and he only cares about exactly one "five" and the other five results must all be non-fives, then there are only 6 outcomes out of 46656 outcomes: one where the sole five is in the first slot, the second, the third, etc.

39 posted on 05/07/2015 8:11:59 AM PDT by Teacher317 (We have now sunk to a depth at which restatement of the obvious is the first duty of intelligent men)
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To: Nifster

So then, we were asked to provide mathematical information that is intellectually misleading.


40 posted on 05/07/2015 8:13:52 AM PDT by Fester Chugabrew (Even the compassion of the wicked is cruel.)
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