MaplePrimes - Newest Questions and Posts

The Birthday Paradox

callumneily — Fri, 31 Jul 2026 19:28:43 Z

The Problem

Imagine walking into a room full of strangers. At first, everyone seems completely unrelated, with different names, different lives, and different birthdays. As the room gets more crowded, we of course know that sharing similarities with the strangers in the room becomes more likely.

But just how likely is it that, for example, two people in the room share a birthday?

The Birthday Paradox asks, more specifically: How many people must be in the room so that it is more likely than not that two people share a birthday?

Intuitively, we might imagine it must be well over 100 since there are 365 possible birthdays that each stranger can have. Probability, however, has a habit of ignoring our intuition.

Building the Math

Suppose there are only two people in the room: you, and one stranger. You have your own birthday, so there is a 1/365 ≈ 0.27% chance that the stranger next to you shares your birthday. If we add a third person to the room, we must consider the probability that either of the two strangers share your birthday, or the two strangers share a birthday with each other. This overlap makes our calculations more difficult, and it will only get worse as we add more people.

Instead, there's a cleaner way to find the solution. Let's find the probability P(n) that no two people share a birthday, then subtract it from 1 to get the probability that two people do share a birthday. Once P(n) < 0.5, then 1 - P(n) > 0.5, meaning at this value of n, there is a greater than 50% chance that two people in the room share a birthday.

For the case of only two people, there is a 364/365 chance that the second person has a different birthday than the first person. If we introduce a third person, there are 363 unclaimed birthdays left in the year, so a 363/365 chance that the third person's birthday falls on one of these days. Repeating this process for n people in the room, we get:

If we keep increasing n, we see that the probability P(n) that two people don't share a birthday decreases, meaning the probability 1 - P(n) that they do share a birthday increases.

Below are some values of 1 - P(n) I found using Maple Learn:

Based on these calculations, we see that when 23 people are in the room, it is more likely than not that at least two people in the room share a birthday.

Extending the Problem

Our question was very specific: we wanted to know what n needed to be in order for the probability of two people sharing a birthday to be > 50%. We found this number to be 23. But, we could extend this question to, say, the probability being 99.9%. This happens when there are 70 people in the room. Consider the animation below to see this visually, where the x-axis represents the number of people n, and the y-axis represents the probability that two people share a birthday, 1 - P(n):

Limitations

How many people must be in the room to guarantee that two of them share a birthday? By the shape of the graph above, it may be tempting to think that it is impossible to guarantee this, since the probability seems to asymptotically approach 1. However, we must think through this logically.

If 365 people were in the room, it is still possible that their birthdays would all fall on a different day (although it's extremely unlikely). But if we have more people than we have available birthdays, we can say at least two people will share a birthday with 100% certainty. That is, P(n) = 0 when n > 365, so our formula for P(n) above is limited to values of n that are 365 or less. This is a basic principle in mathematics called the "pigeonhole principle", which uses the analogy that if we have more pigeons than we have holes to put them in, then at least one of the holes must have at least two pigeons.

These ideas can be conceptually confusing, but once you go through the mathematical arguments a few times, mentioning the Birthday Paradox in a conversation is a great way to show off your random math knowledge and confuse your peers!

Attached is an interactive Maple worksheet from Maple's Math Apps gallery if you'd like to investigate the Birthday Paradox further: The_Birthday_Paradox.mw. If you're still interested in more info/applications, see this Maple Primes post by John May.

How to check if a parameter set is in SI base units

C_R — Thu, 30 Jul 2026 21:08:28 Z

For a numericial integration I want to check if all physical parameters are given in SI base units (before stripping of the units for the integration). Right now I am doing this

Parameters

(1)

H: length of rod
R: outer radius
r_sub: outer radius of heat flux
k: conductance
ρ: density
Cp: heat capacity
Q_flux: heat flux
T_amb: initial Tempearture
Nz: number of axial grid points
Nr: number of radial grid points
t_final: integration time
t_step: sampling interval for temperature fields
test_1: unit given in kilos
test_2: unit given in derived base units

(2)

α: thermal diffusivity

One way is to check whether the value of a physical parameter has not changed after simplification

(3)

(4)

Download check_for_SI_base_units.mw

I am interested in simpler ways to perform the test. The statement is quite long and not easy to understand.

Given the size of the unit package, there might be a command that I have overlooked to check whether units are in SI base units.

Have I missed test cases to make the check fool proof?

Update: A variant with split was added and attempts with functional programming

check_for_SI_base_units_02.mw

Agent Skill for Maple

OptimusMaplePrime — Wed, 29 Jul 2026 18:36:34 Z

Has anyone developed an Agent Skill for Maple compliant with the Agent Skills Specification?

I frequently use Claude Code to create and edit Maple Documents. It does fine alone, but I found adding a script to do a syntax check eliminated a lot of content errors. I also presume some instructions on how it should read the Maple documentation will improve its reliability. Before I go about writing my own Skill against the specification, I just want to check if there is already one in use within the community.

Simplifications and substitutions

Ronan — Wed, 29 Jul 2026 16:37:44 Z

I am trying to follow a series of substitutions used in a paper. I have given some examples of the substitutions. But I am having difficulty getting them to work. I know if all else fails I can manually recreate an expression with the substitution variables and then check backwards using eval(f,vals).
Would appreciate any insight on substitution techniques here.

restart

>	sidrels := [ac - b^2 = Delta, a + c - 2b = d, a - b = c__bar, a - c = b__bar, c - b = a__bar];

(1)

>	vals:='rhs=lhs'~(sidrels)

(2)

>	t1:=a + c - 2*b; simplify(t1,sidrels); #this should be d

(3)

>	t2:=Vector[column](3, [-a + b, c - b, 0]); simplify(t2,sidrels) ; # This should be <-c__bar,a__bar,0> or <c__bar,-a__bar,0,>r

(4)

>	t3:=normal(1 - b^2/(a*c));

(5)

>	simplify((5),sidrels); # This should be Delta/(a*c)

(6)

>	simplify(numer((5)),sidrels,'mindeg')/(denom((5)) ); # This works

(7)

>	t4:=[(-b^2 + bc)/(ac - b^2), (ab - b^2)/(ac - b^2)]

(8)

>	simplify(t4,sidrels,'mindeg');# this should be [(ba__bar)/Delta, (bb__bar)/Delta]

(9)

>	t5:=simplify(Vector[column](3, [3bc__bar - Delta, -3a__barb + Delta, b*(-c__bar + a__bar)]))

(10)

>	t5[3]:=simplify(t5[3],sidrels) ;#This should be b*b__bar

(11)

(12)

>	subs(sidrels[-2],b*(-c__bar + a__bar))

(13)

>	algsubs(sidrels[-2],b*(-c__bar + a__bar))

(14)

Download 2026-07-29_Q_Simplifications_and_Substitutions.mw

MapleV for emacs 30.2

Mac Dude — Wed, 29 Jul 2026 14:09:15 Z

I am a long-time user of Joe Riel's maplev mode in emacs. Upgraded emacs to version 30.2 and now it does not work anymore because of difficulties with something called abbrevlist. This is on MacOS. I am already on the last version of maplev.

Anyone have an idea what is going on and what to do? Somehow for me referencing a username no longer is possible (I get a blocking box with a spinner upon entering the at sign) so I can't create a notification for Joe.

TIA,

Mac Dude.

Recalcitrant differential equation

Alfred_F — Wed, 29 Jul 2026 09:10:23 Z

In the attached file, I'd like to solve an ODE as an exercise. Only the "series" option yielded a solution. How can the length of the result series be controlled? Is there a package that leads to a "closed" solution – even using well-known transcendental functions?

Download test.mw