Lawyer seeks help for a mathematical problem

AI Thread Summary
A litigation lawyer seeks to determine the mathematical probability of an opposing lawyer "overlooking" a critical letter while successfully receiving 48 others. The discussion highlights that the chance of overlooking the same letter twice, once by facsimile and once by post, is calculated as 1 in 2401 under the assumption of independent events. Participants emphasize that while the probability seems low, human factors complicate accurate calculations. The consensus suggests that the opposing lawyer's claim appears implausible, but legal implications may limit the effectiveness of this argument in court. Ultimately, certified mail is recommended for future correspondence to strengthen legal standing.
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May I request assistance for a mathematical problem. The problem arises this way. I’m a litigation lawyer in private practice. In one of my cases the opposing lawyer claims that he “overlooked” a letter that I sent him. However he didn’t overlook any of the other 48 letters that I sent him. It just so happens that he “overlooked” the one letter that has serious consequences.

My normal practice is to send letters by facsimile and then send the original by ordinary post. So the opposing lawyer “overlooked” this particular letter twice – once when it was sent by facsimile and again when he received it through the post.

I’d like to know the mathematical chances of his story being true. There are altogether 49 letters sent twice – once by facsimile and once through the post – being 98 receptions altogether. Out of those 98 receptions he overlooks two letters. But they happen to be the facsimile and original of the same letter. The chances of that seem remote to me.

I would like to know is there a formula that can calculate the chances of doing this? I haven't attempted this because I wouldn't know where to start.
 
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The problem is identical to this one: you have 98 marbles in an urn, 96 are white, 2 are black. You draw out two marbles at random. What are the chances that they are both black?

The chance that the first marble is black is 2 in 98, and (assuming it is), the chance that the second marble is black is 1 in 97, for a total of 2 in 98 times 97, or 1 in 4753.
 
@avodyne - I am not so sure..

See the lawyer had originally a set of only 49 letters (He is not working with a set of 98 objects at once, but 49 at a time). Which makes his chances of overlooking one initially 1/49. Finally he gets a second set, from which he must overlook the same letter - another 1/49 chance

Overall, the chance of overlooking the letter twice should be 1/49 * 1/49 = 1 in 2409
 
Here's my take on it: the opposing lawyer has 98 "deliveries" (either fax or post) which could potentially be overlooked. He has overlooked 2 of them. Based on this, let's form a hypothesis that on average, he overlooks 1 out of every 49 deliveries purely by accident. So any given delivery has a 1/49 chance of being overlooked accidentally, and a 48/49 chance of being received. Considering only the letter that has serious consequences: the chance that this letter would be overlooked when faxed is 1/49, and the chance that it would be overlooked when sent by post is again 1/49. Under the hypothesis that the overlooking was an accident, these are independent events, and the chance that both events would happen (i.e. that both the fax and post of that particular letter would be missed) is indeed 1/2401 = 0.00042, regardless of what happens with the rest of the letters. And the probability of that one letter being missed twice and all others being received is smaller, by a factor of (48/49)^96 = 0.138.

Regardless, it's a pretty small probability and I would consider the opposing lawyer's story exceedingly unlikely.
 
I know that I'm certainly not the most mathematically advanced person here, but I do know what I learned in Psychology and Calculus about statistics, and this sounds like a problem that one can't actually calculate the real probability of (due to human factors). However, a simple permutation should do the trick, right? Go with the more conservative mathematical models of this; no matter what, it looks bad for the other guy.

Unfortunately, I don't think this will hold up legally in anything more than small claims court. You need to send certified mail (as I'm sure you, as a lawyer, know).
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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