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RE: [CFRG] COMBINITORICS PROBABILITIES

Dan Collins <dcollinsn@gmail.com> Mon, 08 August 2022 21:57 UTCShow header

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From: Dan Collins <dcollinsn@gmail.com>
Date: Mon, 08 Aug 2022 14:57:24 -0700
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To: Robert Moskowitz <rgm-sec@htt-consult.com>
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Subject: Re: [CFRG] Combinitorics probabilities
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Robert,

In your scenario, you want "with replacement", meaning that the probability
of receiving a later packet is not modified by the result of earlier
trials. (As opposed to drawing colored marbles from a bag "without
replacement", where drawing a blue marble reduces the probability that the
next marble will be blue.)

Specifically, with your scenario where your packet reception rate is 95%
(p=0.95), and the total number of packets sent is 5 (n=5), and you want to
know the probability that at least 3 packets were received, you want to
evaluate the CDF at k=2 (get the chance that two or fewer packets were
received), and then subtract that from one. This can be done using the
summation formula for the CDF shown here with k=2, n=5, p=0.95.

https://en.wikipedia.org/wiki/Binomial_distribution#Cumulative_distribution_function

(The summation formula iterates over i, the number of successful trials
(received packets), and multiplies n choose i (the number of permutations
of successful trials) by p^(i)*(1-p)^(n-i) (the probability of any
individual permutation being found)).

The Binomial CDF is available as an Excel function:
https://support.microsoft.com/en-us/office/binom-dist-function-c5ae37b6-f39c-4be2-94c2-509a1480770c
In your example, I think you would do this:
=1-BINOM.DIST(2,5,0.95,TRUE)
Which comes to 99.88% for receiving at least 3 out of 5 packets.

I think the result in your original post is not correct. You wanted the
probability that at least 2 out of 3 packets would be received. There are
four ways that this could happen: all 3 packets are received, or the first
one is dropped, or the second one is dropped, or the third one is dropped.
This uses the following probabilities:
= (0.95*0.95*0.95) + (0.05*0.95*0.95) + (0.95*0.05*0.95) + (0.95*0.95*0.05)
This comes to 99.275%, which agrees with the BINOM.DIST function with these
parameters:
=1-BINOM.DIST(1,3,0.95,TRUE)

Hopefully I have correctly understood the question.
Regards,
Dan


On Mon, Aug 8, 2022 at 2:37 PM Robert Moskowitz <rgm-sec@htt-consult.com>
wrote:

>
>
> On 8/8/22 17:16, Robert Moskowitz wrote:
> >
> >
> > On 8/8/22 17:07, Dan Brown wrote:
> >> I think you want:
> >> https://en.wikipedia.org/wiki/Binomial_distribution#Tail_bounds
> >
> > Pretty heavy lifting and it does not read like my problem.  Then I
> > read the intro:
> >
> > The binomial distribution is frequently used to model the number of
> > successes in a sample of size n drawn with replacement from a
> > population of size N. If the sampling is carried out without
> > replacement, the draws are not independent and so the resulting
> > distribution is a hypergeometric distribution, not a binomial one.
> > However, for N much larger than n, the binomial distribution remains a
> > good approximation, and is widely used.
> >
> >
> > I believe this is "without replacement" and N is close to n.
> >
> > e.g.: 5 messages are sent.  You want to receive at least 3 of them;
> > any 3 and more is ok.  The probablity of receiving any one message is
> > p...
> >
> > So off to look at hypergeometric distribution?
>
> No not hypergeometric.  Back to binomial.
>
> Fun!  This is stuff I learned back around '69 or '70!  Where are those
> brain cells hiding?
>
> I had enough stat then that I could have degreed in it, but I did not
> consider it fun...
>
> Maybe I burned those cells after getting my comp sci degree?
>
> :)
>
> >
> >
> >> Best regards,
> >> Dan
> >>
> >>> -----Original Message-----
> >>> From: CFRG <cfrg-bounces@irtf.org> On Behalf Of Robert Moskowitz
> >>> Sent: Monday, August 8, 2022 4:59 PM
> >>> To: cfrg@ietf.org
> >>> Subject: [CFRG] Combinitorics probabilities
> >>>
> >>>   CAUTION - This email is from an external source. Please be
> >>> cautious with
> >>> links
> >>> and attachments. (go/taginfo)
> >>>
> >>> Well I spent the afternoon googling, but my search foo is weak.
> >>>
> >>> I want the formula for the probablity of receiving at least m out of n
> >>> messages
> >>> given the probablity of receiving any message is p.
> >>>
> >>> I did find:
> >>>
> >>>
> https://urldefense.com/v3/__https://www.statology.org/probability-of-at-
> >>>
> >>> least-
> >>> two/*:*:text=P(X**B2)*20*3D,(X**B2)*20*3D*200.3673__;I37iiaUlJeKJpSUlJQ
> >>> !!JoeW-IhCUkS0Jg!cUlR8MdsZ0VvH1GymBznGvOigS-
> >>> vQjTeU2LxJmllO1oVh8_GNKrvuam52NbSOIT2KzNggbgkbpzkfqyWTupj$
> >>>
> >>> But this is a series to find the final answer, not the 'final' formula.
> >>>
> >>> So for example to receive at least 2 out of 3 messages where the
> >>> probablity
> >>> of
> >>> any message at 95% comes out to 97.2%
> >>>
> >>> But what about 3 out of 5?  etc.
> >>>
> >>> Pointer is greatly appreciated.
> >>>
> >>> I took stat just too many decades ago, and I have not kept that
> >>> knife sharp.
> >>>
> >>> thanks
> >>>
> >>>
> >>> _______________________________________________
> >>> CFRG mailing list
> >>> CFRG@irtf.org
> >>>
> https://urldefense.com/v3/__https://www.irtf.org/mailman/listinfo/cfrg__;!!Jo
> >>>
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 * [CFRG] Combinitorics probabilities  Robert Moskowitz
 * Re: [CFRG] Combinitorics probabilities  Taylor R Campbell
 * Re: [CFRG] Combinitorics probabilities  Dan Brown
 * Re: [CFRG] Combinitorics probabilities  Robert Moskowitz
 * Re: [CFRG] Combinitorics probabilities  Robert Moskowitz
 * Re: [CFRG] Combinitorics probabilities  Robert Moskowitz
 * Re: [CFRG] Combinitorics probabilities  David Jacobson
 * Re: [CFRG] Combinitorics probabilities  Robert Moskowitz
 * Re: [CFRG] Combinitorics probabilities  Dan Collins
 * Re: [CFRG] Combinitorics probabilities  David Jacobson
 * Re: [CFRG] Combinitorics probabilities  Robert Moskowitz

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