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Re: [CFRG] Combinitorics probabilities

David Jacobson <david@dmjacobson.com> Mon, 08 August 2022 21:37 UTC

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Date: Mon, 08 Aug 2022 21:36:47 +0000
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To: Robert Moskowitz <rgm-sec@htt-consult.com>
From: David Jacobson <david@dmjacobson.com>
Cc: Taylor R Campbell <campbell+cfrg@mumble.net>, cfrg@ietf.org
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Subject: Re: [CFRG] Combinitorics probabilities
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I played with this in Mathematica and got this result

(Beta[p, m, 1 - m + n] Gamma[1 + n])/(Gamma[m] Gamma[1 - m + n])    This assumes that m >= 1 && m <= n.

And Beta(a, b) = (a-1)! * (b-1)!  /  (a+b-1)!


Here is the Mathematica input that got me the result:

Assuming[{x, m, n} \[Element] Integers && 0 < p < 1,
 Probability[x >= m, x \[Distributed] BinomialDistribution[n, p]]]

   —David Jacobson




> On Aug 8, 2022, at 2:17 PM, Robert Moskowitz <rgm-sec@htt-consult.com> wrote:
>
>
>
> On 8/8/22 17:06, Taylor R Campbell wrote:
>>> Date: Mon, 8 Aug 2022 16:58:45 -0400
>>> From: Robert Moskowitz <rgm-sec@htt-consult.com>
>>>
>>> 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.
>> This sounds like the coupon collector's problem?
>>
>> https://en.wikipedia.org/wiki/Coupon_collector%27s_problem
>
> Close, but I can't map this to my problem in packet reception.
>
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