Re: [CFRG] Combinitorics probabilities

David Jacobson <david@dmjacobson.com> Mon, 08 August 2022 22:00 UTC

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Date: Mon, 08 Aug 2022 22:00:15 +0000
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To: David Jacobson <david@dmjacobson.com>
From: David Jacobson <david@dmjacobson.com>
Cc: Robert Moskowitz <rgm-sec@htt-consult.com>, cfrg@ietf.org
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Subject: Re: [CFRG] Combinitorics probabilities
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Oops.  The formula I gave for Beta is for a different Beta function that takes only 2 arguments.  The correct Beta function is called the incomplete beta function and takes 3 arguments.

   —David


> On Aug 8, 2022, at 2:36 PM, David Jacobson <david=40dmjacobson.com@dmarc.ietf.org> wrote:
>
> 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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