Re: [CFRG] Combinitorics probabilities

[Robert Moskowitz <[e-mail]>]

thanks Dan,  This looks right!

And it allows for playing with the probablity of receiving any one message.

BTW, this is for applying FEC to a message over a path with packet loss 
of 1-p.

chop the message up, add some FEC pieces and see what you get at the 
other end.

If you can turn a bad path into successful transmission, you don't have 
to go into retransmission.

Tradeoffs abound, but that is of course.

again, this looks right.  Thanks

On 8/8/22 17:57, Dan Collins wrote:
> 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 
> <[e-mail]> 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 <[e-mail]> On Behalf Of Robert Moskowitz
>     >>> Sent: Monday, August 8, 2022 4:59 PM
>     >>> To: [e-mail]
>     >>> 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
>     >>>
>     >>>
>     >>> _______________________________________________
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>     https://urldefense.com/v3/__https://www.irtf.org/mailman/listinfo/cfrg__;!!Jo
>
>     >>>
>     >>> eW-IhCUkS0Jg!cUlR8MdsZ0VvH1GymBznGvOigS-
>     >>> vQjTeU2LxJmllO1oVh8_GNKrvuam52NbSOIT2KzNggbgkbpzkfuM8oXQq$
>     >>
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