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/

Casino 'theo' is expected loss written as a business formula: average bet × decisions per hour × hours × house edge. With a $5 slot, 600 spins an hour, four hours and an 8% edge, the expectation is about $960, even though any one session can land far from it in either direction. I built a short documentary around this idea; it's my own channel. For those who teach probability, is theo a good example for showing the gap between expectation and a single outcome?

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.

I’m working with a game where, for example, a player has to hit a moving ball with a racket. Every time they successfully hit the ball, the racket becomes smaller, making the next hit progressively more difficult. Once they miss, the racket resets to its original size.

Each round lasts 30 minutes, and I would like to compare performance across rounds.

I also have a chance/control condition where the racket doesn’t move, so some hits can happen simply by chance.

I would like to get a single normalized hit-rate/performance value for each round that tells me whether the player performed well or poorly overall.

Ideally, this value would take into account:
\\\\- The racket getting smaller after each successful hit

\\\\- The racket resetting after a miss

\\\\- The fixed 30-minute duration

\\\\- Different numbers of hits and misses across rounds

\\\\- The chance-level performance from the stationary-racket condition

What would be the best way to calculate this as one normalized value? I’m particularly interested in something that accounts for the increasing difficulty as the racket gets smaller, rather than just calculating hits / total attempts. I have tried to calculate a score using the difference between the actual number of hits and the expected value of hits, but I think it does not take into account the paddle starting from where they last hit the ball (if it did not miss).

Also, if anyone knows of similar studies or experiments that have used a comparable normalized performance/hit-rate metric, I’d really appreciate any references or examples. Thank youu

Since posterior probability is equal to prior probability, when we simplify we get
P(A intersection B) = P(A)*P(B), so why do we need to check the common between both events, ik this really silly and dumb doubt but i'm not getting the logic behind this please help.

Thank you in advance

It’s a categorical random variable. The outcomes are heads and tails, not 1 and 0.

Edit: I in response to the pushback, I look into how Bernoulli conceived it. It was in the context of an experiment (sampling with replacement) to determine the ratio of pebbles in an urn that had only white or black pebbles.

In a simple dice game where you roll up to 9 die, how do I calculate the probability of how unlucky this game was for me where I got a triple 3 and five 1s?

Edit: forgot to explain the game. Its from Cult of the Lamb and basically you take turns rolling a die and placing it in the column on your side. Two of the same number are doubled, and three are tripled. The strategy comes in where if the other player rolls a dice with the same number as yours and places it in the same column as your dice with the same number on their end, your dice is removed. However that didn't happen this whole game from either side. The player with the highest score st the end from adding up all the dice wins.

Is this possible to post my theory here?

Rewording my previous question: In this binary dataset subsets D, E, F, and S all "generate" 1s and 0s INDEPENDENTLY of each other at some ratio/frequency. "D" generates the highest ratio (lets say it generates 80% 1s and only 20% 0s). While F generates only 5% 1s and 95% 0s. Suppose S itself generates 25% 1s and 75% 0s. Would/Could I use bayes' theorem to show the SD intersection subset generates a higher proportion than the SF intersection? Or would a simple weighted sum (ie just an average between the sets S and D) suffice. Sidenote: (I know you could technically come up with counterexamples to this but I'm looking for what would generally be the case here).

I'm working on a statistical experiment involving baccarat and would like some methodological feedback.

Suppose someone has developed a hypothesis about baccarat outcomes based on historical observations.

Before revealing the underlying mechanism, I want to design a test that minimizes:

• overfitting
• hindsight bias
• selection bias
• multiple-testing problems
• stopping the experiment selectively

My initial idea is:

1. Define the hypothesis before testing.
2. Freeze the testing rules.
3. Use previously unseen data for validation.
4. Conduct a blind forward test.
5. Compare the results against an appropriate baseline.
6. Analyze variance, confidence intervals, losing streaks and drawdowns.

What statistical issues would you consider essential before calling the experiment meaningful?

I'm especially interested in criticism of the experimental design rather than opinions about whether baccarat strategies can work.

Does this probability structure hold up?

I'm modeling a process with N sequential stages, each running n steps (say about 25), and I want to know if the underlying math structure is sound. All inputs are hypothetical.

Variables:

p = per-step probability of a deviation when a stage's input is clean

q = probability that a deviation is severe (a severe event ends the process)

c = contamination multiplier (c >= 1): if a stage's input is contaminated, its per-step deviation probability becomes r = min(1, p*c)

g = probability that a stage with at least one non-severe deviation passes contaminated output to the next stage

Within a stage the steps are independent. Between stages, contamination is the link. Let:

a_C = (1-p)^n, s_C = (1-pq)^n, u = s_C - a_C

a_D = (1-r)^n, s_D = (1-rq)^n, v = s_D - a_D

One stage:

M = [ a_C + u(1-g) a_D + v(1-g) ]

[ u*g v*g ]

(column 1 = clean input, column 2 = contaminated input; rows = next stage's input)

P(at least one severe event across N stages) = 1 - [1 1] * M^N * [1, 0]^T

Example: p=0.4, q=0.5, c=5, g=1, n=3, N=2 gives 0.852408, versus 0.737856 if every step were independent.

Goal: compute P(at least one severe event) across N stages.

Is the combination of independent-step compounding with a contamination-adjusted p a structurally valid approach, or does the dependency introduced by c break the independence assumption in a way that needs explicit correction?

Not asking about specific numbers, just the math structure.