# Probability formula for number of rolls to get a particular result

**URL:** <https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042>\
**Category:** Factual Questions\
**Created:** [August 21, 2022, 5:59pm UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042 "2022-08-21T17:59:52Z")\
**Posts on this page:** 10\
**Page:** 3

<div class="post-metadata">

**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\
**Post date:** [August 27, 2022, 1:29am UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/44 "2022-08-27T01:29:12Z")

</div>

> [@DPRK](#):
>
> like work in the field of 2-adic numbers

Sure, that’s one approach, and gives some intuitive feel for the result. But you could also consider the sum:  
1 +2x + 4x^2 + 8x^3 + \ldots = \frac{1}{1-2x}  
That only converges for x\<\frac{1}{2}, but you can use analytic continuation to get around that.

As for p-adic numbers, I thought this to be a good intro:

[![](https://img.youtube.com/vi/3gyHKCDq1YA/maxresdefault.jpg "1 Billion is Tiny in an Alternate Universe: Introduction to p-adic Numbers") ](https://www.youtube.com/watch?v=3gyHKCDq1YA)

I didn’t realize that some of them contained roots of -1 without having a special “i”.

---

<div class="post-metadata">

**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\
**Post date:** [August 27, 2022, 2:29am UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/45 "2022-08-27T02:29:07Z")

</div>

Well, there are definitely various methods of summing divergent series, some of which do have applications in analysis as well as physics, engineering, and so on.

I am no expert in probability, but I think many times it does not really matter if a series, like a generating function, actually converges, as long as you can work with it as a formal series.

On the other hand, now if X is a discrete random variable, its moment generating function \operatorname{E}(e^{sX}) — this is essentially like a Laplace transform — will not always converge.

---

<div class="post-metadata">

**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\
**Post date:** [August 30, 2022, 12:16pm UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/46 "2022-08-30T12:16:29Z")

</div>

N/2 is a good approximation for large values of 2.

---

<div class="post-metadata">

**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\
**Post date:** [September 2, 2022, 2:27pm UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/47 "2022-09-02T14:27:31Z")

</div>

TL;DR  
I’ve been thinking about the divergent series  
1-2+3-4+\cdots and how it might sum to 1/4. There is a process called Cesaro summation that can be used to assign a value to certain divergent series. If you are not interested, stop here, but I wanted to describe it for any who are.

First consider the series  
\frac1{1-x}=1+x+x^2+x^3+\cdots  
Letting x=-1, we “get” \frac1 2=1-1+1-1+1-\cdots  
which is of course nonsense. The sequence of partial sums is  
1,0,1,0,1,\cdots which obviously does not converge. Cesaro’s idea was to replace the n^{th} term of the partial sum by the average of the first n terms. When this is done, you get the sequence, called C\_1 (the reason for the suffix will come clear 1,1/2,2/3,1/2,3/5,1/2,4/7,\cdots, which evidently converges to 1/2. Coincidence? Probably not, as we will see.

The case at hand is the series (\frac1{1-x})^2=1+2x+3x^2+4x^3+\cdots which, by letting x=-1, gives 1/4=1-2+3-4+5-\cdots. In this case the sequence of partial sums is 1,-1,2,-2,3,-3+\cdots and C\_1=1,0,2/3,0,3/5,0,\cdots which does converge, so no help? But note that this sequence looks like it is made by splicing a sequence that is constantly 0 with one that converges to 1/2. Using this fact, it is not hard to show that if we repeat the Cesaro process, we get a new sequence C\_2=1,1/2,5/9,5/12,\cdots converges to 1/4. Looking like less of a coincidence.

Here’s a conjecture (I have not looked at it seriously). It is not hard to show that (\frac1{1-x})^3=1+3x+6x^2+10x^3+\cdots, the n^{th} term being t\_nx^n, where t\_n-\frac {n(n+1)}2 is the n^{th} triangular number. Again, letting x=-1 gives 1/8=t\_1-t\_2+t\_3-t\_4+\cdots. This time, I conjecture, you have to go to C\_3 to get convergence to 1/8.

Just one further comment. Benford’s law states that in any sequence of number chosen at random and unlimited in the number of digits, the fraction that (base ten) that start with a 1 is log 2, the fraction that starts with a 2 is log 3 - log 2,…, the fraction that starts with a 9 is log 10 - log 9. Cesaro summation gives an explanation of sorts.

Consider the sequence, for example, a\_1,a\_2,a\_3,\cdots, in which a\_n represents the fraction of the numbers between 1 and n that start with 3. a\_1=a\_2=0, a\_3=1/3,…,a\_{29}=1/29, a\_{30}=1/15,…,a\_{39}=11/39 and so on. You get a jagged sawtooth graph, that certainly doesn’t converge to anything. Neither does C\_1, nor C\_2, nor any C\_k. But _mirabile dictu_, if you let k\to\infty, you get a sequence, call it C\_\infty that converges to log 4 - log 3.

---

<div class="post-metadata">

**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\
**Post date:** [September 3, 2022, 2:41am UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/48 "2022-09-03T02:41:10Z")

</div>

Well, the faster the terms and partial sums of your divergent series grow, the higher k will have to be in order to possibly make it converge. But if C\_k converges for some k, then it can only be to \lim\_{x\to1^-}\sum\_na\_nx^n, where a\_n are the original terms.

---

<div class="post-metadata">

**Author:** ![Hari\_Seldon](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hari_seldon/32/5173_2.png) [@Hari\_Seldon](https://boards.straightdope.com/u/Hari_Seldon)\
**Post date:** [September 4, 2022, 1:23am UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/49 "2022-09-04T01:23:13Z")

</div>

I wondered about that. Can you give a cite?

Incidentally, I should have mentioned that Cesaro summation leaves a convergent series convergent and to the same value.

---

<div class="post-metadata">

**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\
**Post date:** [September 4, 2022, 1:50am UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/50 "2022-09-04T01:50:01Z")

</div>

Cesaro means are described in section 5.4 of Hardy’s book on Divergent Series. A simple limitation theorem is Theorem 46, which states that if \sum a\_n is C\_k-summable, then the Cesaro partial sums A\_n^{k'}=o(n^k) for every order k'\<k, also including a\_n=o(n^k).

Theorem 55 proves that if a series is C\_k summable for some k, then it is Abel summable to the same limit. (Hardy then given an example of an Abel-summable series for which a\_n is not O(n^k) for any k.)

Further limitations of some of these methods as well as necessary and sufficient conditions for summability are described in the following chapters; for instance if the terms of a series grow too _slowly_ then it cannot be Cesaro summable or Abel summable without being convergent in the ordinary sense.

---

<div class="post-metadata">

**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\
**Post date:** [September 4, 2022, 1:55am UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/51 "2022-09-04T01:55:00Z")

</div>

> [@Hari\_Seldon](#):
>
> First consider the series … which is of course nonsense.

Is it, though? I see no reason to believe that an infinite sum of integers should give an integer result. Any _finite_ sum, sure, but this isn’t finite.

\lim\_{x \to -1^-} \sum\_{k=0}^{\infty} x^k is obviously \frac{1}{2}. Why shouldn’t we just say that the value of the sum is equal to the limit? We do that everywhere else in math. It doesn’t seem any more unusual than assigning a value to \frac{0}{0}.

---

<div class="post-metadata">

**Author:** ![DPRK](https://avatars.discourse-cdn.com/v4/letter/d/4491bb/32.png) [@DPRK](https://boards.straightdope.com/u/DPRK)\
**Post date:** [September 4, 2022, 2:19am UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/52 "2022-09-04T02:19:11Z")

</div>

We seem to be simul-posting 🙂 That is called Abel summation by Hardy. The reason it is not “the” sum is that there are lots of different sums of divergent series defined; however both Abel and Cesaro sums are “regular” in the sense that they leave convergent series convergent to the same value, and even “totally regular” in the additional sense that if the ordinary partial sums diverge to infinity then so do the Cesaro/Abel sums. So if you want 1+2+3+\cdots you will have to employ some other summation method (and give up some nice property, e.g. linearity)

---

<div class="post-metadata">

**Author:** ![Dr.Strangelove](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/dr.strangelove/32/6613_2.png) [@Dr.Strangelove](https://boards.straightdope.com/u/Dr.Strangelove)\
**Post date:** [September 4, 2022, 3:22am UTC](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042/53 "2022-09-04T03:22:00Z")

</div>

Yeah, easy to see that you have to give up some nice property:  
S = 1 + 2 + 3 + \dots  
S = 0 + 1 + 2 + 3 + \dots  
S-S = 1 + 1 + 1 + \dots  
0 = 1 + 1 + 1 + \dots  
1 + 0 = 1 + (1 + 1 + 1 + \dots)  
1 = 0

Oops.

[Previous page](https://boards.straightdope.com/t/probability-formula-for-number-of-rolls-to-get-a-particular-result/970042.md?page=2)