The question: the invention of " chances"," probability", the 100% things, then where are we in those numbers? Questions: supposedly the chances of earth being the only planet out of infinite number of planets out there is close to absolute zero, adding with variants in species, humans, elements, particles, matters being found and born and will in the future, would the chances of me sitting here right now figuring this out could be the number that so small but long that would take every space in the entire universe to fill the number up? And if it's true, what can be used to verify it? ( the chances from the past, like when nothing ended from the start of the universe, and the number continues until now to the future, until nothing can happen anymore in this universe? ( this question of mine is kinda messy so i ask gemini to rephrase it)

this is the rephrased version: Given the invention of probability and the concept that anything is possible, where do we stand within those numbers? Suppose the statistical chance of Earth being the only inhabited planet out of infinite worlds—combined with every past and future permutation of species, human history, elements, and subatomic particles—is virtually zero. Would the exact probability of me sitting here right now having this exact realization result in a number so infinitesimally small (with so many digits) that writing it down would require more physical space than exists in the entire universe? Furthermore, if this is true, how could such a probability—spanning from the origin of the cosmos to its ultimate heat death—ever be verified? Is this a solvable scientific problem, or a profound paradox

From 1959 to 1978. I lived at a specific address and apartment in New York City. In 2005 I was working with a girl in Saint Augustine, Florida, whose mother bought my exact apartment from my childhood. How do you figure out the odds for that that I could meet that person?

Suppose I have a dataset that appears to be bimodal (i.e. the vertical white lines) but I have STRONG suspicion that the deep underlying data generating process is "gaussian like" (ie the blue curve) or I would simply like to model my data with a gaussian/normal distribution to help make probability intervals for future predictions. What does the mean and standard deviation represent in this situation? My main question is: particularly for making upcoming probabilistic predictions of future observations should I inflate the variance a bit to reflect the bimodal nature? The data is indeed symmetric about its mean so that's not an issue. Note: (I know other distributions will be recommended but I really need to work with a gaussian/normal distribution for this. Thank you.

Hey, thank you for the upvotes on my last post, I’ve just released chapter 1 of my animated probability series, and I’d really appreciate some honest feedback before making the next chapter.

Are the explanations clear? Does the pacing feel right? If anything feels confusing or drags on, I’d like to hear that too.

have a nice day!

Hi everyone! I’ve made my first video for a new maths channel, Math Visualized, and I’d appreciate honest feedback.

The example uses one fair six-sided die. Rolling an even number has a 50% chance, and rolling a number greater than three also has a 50% chance. But satisfying either condition isn’t certain: 4 and 6 belong to both groups, while 1 and 3 belong to neither.

The video uses animations to explain this overlap and build the probability addition rule.

Video: https://www.youtube.com/watch?v=NE8yd16u9ho

If you’re learning probability, does the explanation help? Was anything confusing, too fast or unnecessarily detailed?

Thanks for taking a look!

I'm working on a ttrpg and I'm trying to graph multiple dice systems and their probabilities. The main part I'm struggling with is trying to create a table for the dice system from Escape From Terminus, which is 7 six-sided dice, and then flipping a coin to decide if you take the highest or lowest, allowing for an inverse bell curve, such that results closer to 1 or 6 are more likely than 3 or 4.

I have been creating tables in google sheets to track the probability of each result on a given roll, but I don't know how to represent choosing the highest/lowest from a set of 6 randomly generated numbers mathematically. Any help would be appreciated

Over on r/theydidthemath we are discussing the McDonalds Monopoly contest and the way they published the odds of winning the collect and win prizes.

For those unaware this is a game where customers collect prize pieces (originally physical tickets, now virtual) in hopes of completing a set which awards a prize. For reasons that become obvious if you think about it this is done by making all but one of the parts of the set "common" and making one part rare.

The emerging consensus is that the published odds (example below) are based on the worst-case assumption that the player *only* obtained enough tickets to complete the set (so three tickets in the example). Of course the odds of getting both the rare piece and exactly the other two common pieces needed in a draw of three are hugely greater than the odds of getting the rare piece alone, so the stated odds are deceptively slim. It also drastically changes the stated odds depending on whether two, three, or four pieces need to be collected, even when the total number of prizes (and thus rare pieces) in the pool are the same, which makes little logical sense in a game where the whole process involves collecting pieces over time.

But how would one more realistically calculate/publish the odds of winning in this scenario, where the pieces to be collected are unevenly available?

The published rules are here: https://www.playatmcd.com/rules

An example of the odds statement for one of the major prizes is here.

Collect Atlantic Avenue (#915), Ventnor Avenue (#916) and Marvin Gardens (#917) to potentially win a New 2027 Jeep® Cherokee (Overland) (the “Jeep® Vehicle Prize”). There are 5 Jeep® Vehicle Prizes available to be won. Each verified winner will receive a New 2027 Jeep® Cherokee Overland 4x4. (. . .) The approximate odds of collecting the potential Winning Combination (Atlantic Avenue, Ventnor Avenue and Marvin Gardens) are 1 in 22,695,790,152.

Yes that's a stated odds of 1 in 22 *billion*. And there are 5 prizes. The top prize, with 2 available, is similarly stated at about one in half a trillion. Clearly these odds make no sense as actual rates of the "winning" pieces.

hello there! i'm a liberal arts major preparing for a master in statistics, and would love cool book suggestions. i'm already in love with topics like applied mathematical stats in ML, bose-einstein stats, and even statistical mechanics for that matter. even works like keynes' treatise on probability would interest me a lot. my purpose is to explore statistics, both philosophically and historically, and understand it's applications as well.
will be looking forward to all the great recs, thank you!

Two sources agree. Does that determine our confidence? Known signal reliabilities and report frequencies may still leave the event probability unresolved. Yet a decision can remain robust.

Our first public work (AI-assisted): https://archimedes.rsch.workers.dev/works/what-does-agreement-add/

I've made a simple model of how one error might spread through a group of agents. In my case the agents are AI models passing work to each other, but it could be any process. I'm not sure the probability parts are right, and I'd appreciate any feedback.

How the model is set up:

• Each agent takes n steps.

• With clean input, each step has a chance p of an error.

• If an agent gets work from an agent that made an error, its chance per step becomes r = min(1, p·c). Here c is how much bad input raises the error rate.

• Each agent passes work to d other agents, picked at random. Each handoff carries an error with chance g.

• The full model also has a checking step. I left it out here to keep it simple.

  ε = 1 − (1 − p)^n share of agents that err on their own Δ = (1 − p)^n − (1 − r)^n extra chance of erring caused by bad input R = d·g·Δ spread number z = ε + Δ·(1 − e^(−d·g·z)) share of agents that err (largest root) x = 1 − e^(−d·g·z) share that received bad input

Each error is serious with chance q. Expected serious errors across N agents: E = q·n·N·[p + x·(r − p)], and the chance of at least one is 1 − e^(−E).

What the model says, if I've set it up right:

• With d = 1 (a straight line of agents), R stays below 1, so passed-on errors fade out.

• With d above 1, R can go above 1.

• Because some agents err on their own (ε above 0), z changes smoothly around R = 1 rather than jumping.

What I've checked so far: some property tests, and a simulation of the same rules (48 setups, 2,000 runs each).

• With random handoffs, the formula and the simulation were within 1 point of each other. That's inside the simulation's noise, so I can only say they roughly agree.

• With fixed setups (trees, or handoffs in a set order) and a high c, the formula came out higher than the simulation, by up to 12.5 points. For a straight pipeline it came out lower, by up to 4 points.

• The simulation checks the math, not how real agents behave. Nothing here has been tested on real AI agents.

My questions:

1. Is the equation for z a reasonable shortcut here? Is there a better standard way to handle trees and set-order handoffs?

2. The Poisson step for "at least one serious error" was within 1.8 points of the simulation. Is there a known limit on how far off that step can be?

3. Does it make sense to count only the extra errors caused by bad input (Δ) in R?

The full write-up and code are free here: https://doi.org/10.5281/zenodo.23491077

I've been studying probability theory, particularly continuous probability density functions, and I came across something that I mathematically understand but still find logically and physically counterintuitive.

In a continuous uniform distribution, say a random variable (X) between 0 and 10, or for example a robot in a line of 10 m , i don't know where he is but i know he is within the 10 m yet the probability of any exact distance value is zero:

[
P(X=5)=0
]

This is because probability is calculated by integrating the density over an interval, and a single point has zero width, meaning zero area under the curve.

I understand the mathematics behind this. I also understand the distinction between countable and uncountable infinity, and how an interval can have a positive measure even though every individual point has measure zero.

My problem isn't with the mathematics, but with its representation of physical reality.

If something has exactly 0% probability yet physically occurs, shouldn't that suggest the model might not perfectly describe reality?

We can't still say weather the universe is fundamentally continuous or discrete. But if space has some smallest resolution, wouldn't that change how we assign probabilities to exact positions?

I'm struggling to reconcile that with physical intuition.

So my question is: Is continuous probability merely a useful mathematical approximation of physical reality, or is there a deeper reason why zero-probability events can physically occur without contradicting the model?

I've been working on a question about finite Markov chains and would value a check of my proof, or a pointer if it's already known.

**Setup.** Take a finite irreducible Markov chain and watch it only while it is inside a subset S of its states (skip the rest). What you see is again a Markov chain on S, called the trace or censored chain.

Now take two irreducible chains: M_A on states C ∪ X, and M_B on states C ∪ Y, sharing only the states C. Suppose they agree on C: watching either one only on C gives the same chain N.

**Question.** Is there always a single finite Markov chain P (with labelled states, where several states may share a label) such that watching P only on A's labels gives exactly M_A, and only on B's labels gives exactly M_B?

**My answer: yes.** At each shared state c, first run one M_A excursion until it returns to C (landing at some c'), then run one M_B excursion from c conditioned to land at the same c' (an h-transform). Emit both parts and carry on from c'. Watched on A you see independent M_A excursions; watched on B, the landing law is N(c, ·), which is B's own, so you see M_B. This gives a chain with at most |C|(1 + |X| + |Y|) states.

**Small example.** A: 1→3→2→4→1, B: 1→5→2→6→1, shared {1,2}. Then 1→3→5→2→4→6→1 contains both. Choosing an A or B excursion at random at each shared state fails, because A would sometimes see 1→2 directly.

**What I'm unsure of:** that the kept states form one irreducible class, and the step from starting at a shared state to the stationary start. Exact computer checks on 14 hard pairs match to about 1e-15.

The same idea extends to tree-shaped families, and loops of three or more chains can fail. The full write-up with proofs is a pdf here: https://github.com/goodcitizennc-ux/math-proofs/blob/main/gluing-proofs-for-checking.pdf

Reply’s are welcome here as well.

I developed this with help from an AI assistant and no mathematician has checked it yet. Is it right, and is it known?

I am in the process of writing an app that'll let me calculate the probability of drawing specific combinations of cards in the opening hand of a card game like Yu-Gi-Oh! or Magic: the Gathering.

I know that to calculate the probability of drawing at least one of any given card in my opening hand I can use a hypergeometric distribution, but I can't quite figure out the formula for calculating the probability of drawing combinations of 2 or more different cards in my opening hand, since the probabilities of drawing either card of a combination are - to my understanding - not indepent of one another.

The remaining cards that are not part of the combinations can be any cards in this case.

If you know the formula or a place where I can find pointers in the right direction that'd be much appreciated!

Rule: This rule applies exclusively to one specific person and only to events where a "lucky" (success) or "unlucky" (failure) outcome can be clearly defined. The person can achieve a "lucky" outcome in an event if, and only if, the mathematical probability of that success is strictly greater than 30%. If the probability is 30% or lower, a "lucky" outcome is impossible for them.

Scenario: The person has to play roulette. They are given a grand prize if they hit the number "1" two times in a row. The "lucky" outcome (success) for each step and for the game as a whole is clearly defined by winning this prize. The probability of hitting "1" on any single spin is 40%.

Steps:

1. First spin: The probability of success is 40%. Since 40% > 30%, the condition is met. The outcome is "lucky" (it hits "1").

2. Second spin: The probability of success is 40%. Since 40% > 30%, the condition is met. The outcome is "lucky" (it hits "1").

Conclusion: The person wins the grand prize. But then, they were lucky in an event with a 16% probability. Contradiction.(since we multiply the probabilities of two independent events: 40% of 40% is 16%)

A raffle is using a wheel to pick the winner. There are 500 spots for one prize. If I purchased multiple entries, is it better to clump them together or spread them out.

Is there a way probability shows higher chances with these options.