lulu

Take any $k-1$ primes $q_i$ not dividing $a$, then use Dirichlet to produce a prime congruent to $2(q_1\times \cdots \times q_{k-1})^{-1}\pmod a$.

Nonsensical question: math.stackexchange.com/questions/5080477/…

Nonsensical question math.stackexchange.com/users/1644470/andre-lin

A new user has posted two, long, incoherent non-answers to this question

User reposting nonsense question: math.stackexchange.com/questions/5061508/…

troll: math.stackexchange.com/users/1619161/user271865

This seems very vague and very broad. Of course, the trig functions have well known definitions and the properties are well established (by proof, not just by feelings). Any basic reference should be able to answer your questions.

We're just going in circles. In the vision of the game the accepted answer considers, the dealer stops at the first significant card. If it's an Ace, that's a win. If it is a King, Queen, or Jack, that's a loss. You might not prefer this way to view the game (it's not my preferred method either, as the probability is not uniform), but that doesn't make it wrong. In any case, it's obvious that my comments are not helping, so I will stop.

There are $16$ signficant cards. Therefore, there are $52-16=36$ insignificant cards. Thus, if you stop at the first significant card, the longest possible deal has length $37$.

Sorry, I don't want to keep going over this. I suggest going over the simpler versions I've mentioned (various sample spaces for two coin tosses and such).

Right. The sample space deals stop at the first significant card, any one of $\{A,K,Q,J\}$. Not just Ace.

It's the set of all possible deals, of any length from $1$ to $37$, with order taken into account. Or you can think of it as the set of all possible deals stopping at the first Ace, but then the probability is not uniform.

The author is computing the probability of winning on the $i^{th}$ draw. Of course the total probability of winning is then the sum over the possible $p_i$. To win on the $i^{th}$ draw you need to choose $i-1$ non-special cards and the denominator should be all the possible draw sequences of $i$ cards. Perfectly correct, though worth stressing (again) that it's computationally unhelpful. I note that the author simply punts as to the value of the sum, which I quite understand. A machine can do it, of course.

As I said, it is part of my response to your question. The statement "each of the four types of special cards is equally likely to be observed first" is equivalent to the statement "amongst the set of full draw sequences, those that start with $A$ are equi-numerous with those that start with $K$ (or $Q$ or $J$)".

I gave the detailed proof, by bijections between subsets of the full (uniform) sample space in my posted answer to your question. Granted people often omit this sort of thing or simply refer to "symmetry", but one should always be able to write it out in full detail.

@RFZ It's not that we discard them, it's that they don't matter. Of course, before you get any of the special cards, you'll probably see a bunch of the others. That's fine, just keep going. Nothing important happens until you see the first special card. And, again, that special card is equally likely to be any of the $16$ special ones, and $4$ of those are Aces.

I have posted something below, too long for a comment, which might help. I don't know that the point is isolated for discussion elsewhere...as I said, it is a common error but it occurs under many different guises. In this way, it is like the so called fence post error in counting...it's hard to isolate, but it definitely comes up a lot. Again, I suggest studying the $5$ card case to convince your self that assuming uniform distribution on the truncated sample space is simply wrong.

Once again, you need to consider them to compute the proper probabilities. You can certainly ignore them in the construction of the sample space, but if you want to collapse all the sequences which start with an $A$ into one point, you have to collapse the probability along with them.

Your error comes up all the time. People, say, use Stars and Bars to count the elements in a sample space, but you really can't, or at least you can't do it easily. The events counted by Stars and Bars are not equally probable. If you toss two fair coins there are three possible outcomes: $\{T, T\}, \{T, H\}, \{H,H\}$. yet the probability of getting one $H$ and one $T$ is $\frac 12$, not $\frac 13$.

In the $5$ card version, there are $4!=24$ sequences in which the unique $A$ is drawn first. And of course there are $5!=120$ sequence in $\mathscr S_5$. Thus the probability that the ace is drawn first is $\frac 4!$ is $\frac {4!}{5!}=\frac 15$, as it should. Note that we get $\frac 15$ instead of $\frac 14$ since for this calculation the $X$ is treated like any of the other non-A cards.

Same problem as in the $5$ card version, your probabilities are wrong. Look at the full sample space, $\mathscr S_{52}$, consisting of all draw sequences of the full deck. Now those sequences are equally probable, that's the assumption of fair shuffling. There are $51!$ sequences in which the $\spadesuit A$ is drawn first. That has to be accounted for in your probability.

There is no problem with using different sample spaces for a given problem, but you have to ensure that the probabilities work. You are assuming a uniform probability (every sequence in $\Omega$ has the same probability), but this is false. Study the $5$ card version. And, as I suggested, look at the simpler question "what is the probability that the $A$ comes up first?". Your distribution has got to yield $\frac 15$ for that, clearly. Yours doesn't.

Side note: I struggle to understand why this was downvoted. I'll upvote to counter. The OP describes a clear problem and includes a detailed attempt at a solution. The OP recognizes that it is flawed but can't find the gap, a situation we are all familiar with. I might suggest including the correct solutions/methodologies for comparison or at least remarking that the proposed method gives answers so close to $0$ as to obviously be wrong, but that would be my only correction to the post (well, that and the $A$ notation I already mentioned).

General comment: it was not a great idea to use the letter $A$ to denote two separate things (the Ace and some set of sequences). The Ace has priority, I suggest changing the other one.

As another way to magnify the issue, use your calculation to compute the probability that the first draw card is an $A$ (in your full deck problem). If I have followed, you believe that is $\frac {|\Omega_1|}{|\Omega|}$, yes? So that would be effectively $0$. Whereas the correct answer is $\frac 1{13}$ as each rank is equally likely to be drawn first.

In situations like this, it is often useful to apply the same methodology to a simpler problem...that tends to make issues more visible. So suppose you had $5$ cards: $\{A, K, Q, J, X\}$ where $X$ is some low card. Obviously the probability that $A$ is seen ahead of $K,Q,J$ is $\frac 14$, as it is in your version. In this case, what is your $\Omega$? Well, $|\Omega_1|=1$, $|\Omega_2|=4$, $|\Omega_3|=12$, $|\Omega_4|=24$,$|\Omega_3|=24$, right? Thus $|\Omega|=65$, And there are exactly $2$ elements of $A$, in this case. Namely, $A$ and $XA$. So your methodology gives $\frac 2{65}$.

I've looked at every value from $20$ digits out to $50$, and the error is completely stable. Beyond that, one would have to write and check the code very carefully, and that's what I do not want to do. Shouldn't be hard to establish some lower bound on the error, though it might be messy.

No, I think I've done enough work on this. Good luck!

Well, it isn't an identity, just an approximation. I am seeing an error of about $5\times 10^{-8}$.

However, if you use the true first solution of $\tan x =-x$, namely $x_2=2.02875783$, you do get close to $\gamma$.

Also, using your $x_1$, however you got it, $\exp \left(-\frac 1{x_1\sin x_1}\right)\sim .393$ Something is off here.

@Randall Well, it is certainly true that searching for the critical points of $x\sin x$ leads one to look for solutions of $\tan x = -x$. Beyond that...

Can you clarify your calculation? As it stands, it does not make sense.

What you have written is wrong. As I say, $\tan x_1>2.33$. More simply, $\tan x$ is obviously positive between $0$ and $\frac {\pi}2$. Hence the first positive solution to $\tan x = -x$ must be greater than $\frac {\pi}2$.

The least positive solution to $\tan x = -x $ is $2.02875783\cdots$.

That doesn't appear to be even close to true. at that point, $\tan x_1\sim 2.330728753$

Are you claiming that, with $x_1\sim 1.16556$ we have $\tan x_1=-x_1$?

In the sense that both of them equal $f(x)g'(x)+g(x)f'(x)$, yes.

Once again: your notation is pointlessly confusing. Why rename everything? Just use $f(x), f(k)$ etc. But more importantly, hand-waving treatment of limits is very often wrong, and that's all that's happening here. Study the proper proof of the product formula, which carefully treats the limits, to see the difference. I do recommend the informal argument I sketched in the comments as I find it intuitive. It can be made rigorous, though that requires some effort. But, rigorous or not, I think it aids intuition. It's not a substitute for the rigorous proof, of course.

In any case, I feel that I am just repeating myself, which is not helpful. So I will stop commenting. Good luck. I suggest: use my linear form to see how the product form drops out intuitively. Maybe that addresses your question?

I don't understand the point of the notation. Why not stick to $f(x), f(k)$ and so on? But, even with the algebra corrected, the question makes no sense. We can compute the limits and see that the two forms are not the same. Taking $f(x)$ identically $1$ makes it intuitive that the naive relation can't hold. Beyond that, what are you asking?

Again, it's your question that makes no sense. You are asking why some identity is false, but of course it is false because we can demonstrate that it is false, there is no "why" beyond that. Besides that, I can't follow your notation. What happened to the $x-k$ in the denominator? $(f(x)-f(k))+(g(x)-g(k))$ simply goes to $0$ as $k\to x$. It does not approach the derivative.

Yes, but it makes no sense. If you want an informal, intuitive, argument use the linear form of the deriviative: $f(x+\delta)=f(x)+\delta f'(x)$. You can both the addition formula and the product rule out of that.

No idea what you are saying. Of course the standard product rule is correct.

Why isn't $1+1=3$? Identities either hold or they don't. I don't see where "why" enters into it. The point of my example...try your "intuitive" argument with $f=1$, but $g$ non-constant. We know it already fails in that case

Not sure what you are hoping for here. It's clear that you couldn't have $(fg)'=f'\times g'$ since taking $f(x)=1$, the constant, would then tell us that every function had derivative $0$.

@James As I have said, convergence isn't helpful here. There isn't any sort of obvious way to get around that, even for quadratics (nobody knows if $n^2+1$ takes infinitely many prime values, for instance, though it is widely conjectured that it does). It seems that fundamentally new ideas are required.

@James Well, not really. Finite sums all converge, of course. As do some, but not all, infinite sums.

@James Primes in arithmetic progressions