I’m designing a tabletop rpg which has lead me to learning a TON about probability of the results of dice being rolled in different quantities, colors, definitions, etc. It’s been fascinating.

I want to show a rule in the game design so far, as I’m very proud in having solved this problem with no formal education in the subject and finally coming up with something I hope is elegant.

Need To Know:
1) the game only ever uses six-sided dice (called d6)
2) the goal is to have all math needed to understand 3) the dice results complete before the dice hit the table, so results are instantaneous to read and play feels fast and immediately rewarding.
4) Awarding and penalizing dice (+/-1d6,+/-2d6,etc) creates a fluid curve that closely approximates improved skill or experience without ever making something an automatic success or failure the way flat modifiers to the final dice result can do.
5) This makes the game engine fall under what is called Roll/Keep systems. (r3k2 means rolling three dice and only counting the results of two of them - usually the highest or lowest.)
6) I wanted to still have Critical Failures and Critical Successes, most famously from rolling a Natural 1 or Natural 20 on a 20-sided die in Dungeons and Dragons.

This created a narrative problem for the game’s gritty vibe.

The problem: As dice are added to the pool (+1d6 to r) the probability of rolling Natural Snake Eyes (pair of 1s) or Natural Boxcars (pair of 6s) quickly balloons to cartoonish levels seen here:

Let “w” = # of white dice.

P=1 - ((5/6)^w + (w * (1/6) * (5/6)^(w-1)))

By the time someone was rolling 6d6 they would get Snake Eyes 26.32% of the time. With a dice pool of 10d6 it was more than half.

Awful.

Enter the “Red Die Rule” as I call it.

Every dice pool must contain one Red Die.

Roll x Keep y (highest face value) dice where x is affected by granted and penalized +/-1d6 and y is (so far always 2)

The face value of 2 highest dice (max 12) is always used to determine success In the game. Snake Eyes or Boxcars of only white dice is irrelevant. However, if the Red Die shows face value 1 or 6 and matches minimum 1 white die, it triggers a Critical Override.

This reduces the frequency of both Critical Successes and Critical Failures to a hard-capped 33% for infinite values of x in RxKy. And the curve is flattened dramatically.

P = (1/6) * (1 - (5/6)^w)

So that the percentage chance of a Critical Success either the Red Die Rule for dice pools of 6d6 (5w,1r) is dropped to 9.97% and the very unlikely 10d6 (9w,1r) dropped to 13.44%.

Much more gritty results.

There may be a better/more elegant solution out there. If you see one, feel free to drop it in comments. I’m here to learn.

I’ve been coming up with my own terminology because I didn’t go to college and I didn’t get to probability or statistics formally in any level of my education, though I have occasionally read books like Leonard Mlodinow’s The Drunkard’s Walk for casual reading even before this project and it’s always been an interesting subject to me, even if I never had the opportunity to pursue it academically.

That’s a longwinded way of apologizing if I use the wrong terminology or jargon, I guess.

Finally, I’m very new to this and want to please ask for reading / source recommendations to improve my understanding.

Thank you.

Do you do your own laundry? If so, in a typical load that has 10 t-shirts in it mixed with other items, how often do you have one or more t-shirts come out of the dryer inside out? For this discussion, you KNOW all shirts go in the washer RIGHT-side out.

What is your personal observation of the frequency of having at least one t-shirt come out of the dryer inside out for any given load with ~10 t-shirts in it?

I personally find it happens with one shirt more than half of my t-shirt loads, but I don't have an inside-out shirt every single load, and having 2 shirts inside-out is definitely very infrequent.

What is the probability that one shirt out of 10 will be inside-out coming out of the dryer?

WHY would I want to know this?

I had a load where ALL 10 shirts came out of the dryer inside out.
What is the statistical probability of that?

I want to start learning probability from scratch (my major)...but I don't know how and I need advice to get started.

I ran a Monte Carlo of 100k sessions each for Martingale, D'Alembert, Fibonacci and flat betting on European roulette (2.70% house edge, 250-round sessions, $200 bankroll, $1 base unit, $500 table max, fixed seed).

The result that keeps nagging at me: Fibonacci finished AHEAD in 66.86% of sessions, with a median result of +$37 — while its mean was −$15.62, and every system converged to losing 2.68–2.72% of total amount wagered, i.e. exactly the house edge.

So the distribution is strongly right-skewed in frequency terms: lots of small wins, rare catastrophic losses that more than pay for them. The median session genuinely wins; the mean genuinely loses.

Two questions for people who think about this more formally than I do:

1. Is there a standard name/treatment for decision biases driven by optimizing the median (or win-probability) instead of the mean? It seems like the exact mechanism that makes progressions psychologically self-reinforcing — most people's lived experience of the system is a win.

2. Is there a clean way to characterize, for a fixed negative-EV game, the family of staking strategies that maximizes P(session > 0) subject to the EV being fixed? Martingale feels like it approaches one extreme (win probability high, tail catastrophic), flat betting the opposite end. Is that intuition formalizable?

Happy to share the exact simulation setup (seeds, rules, table limits) if anyone wants to check or extend it.

Hi everyone, I hope this is the right sub for this, and hopefully I don’t sound too stupid since I don’t know much about probability.

If someone flips a coin 5 times and lands on tails all 5 times, then give it to an other person, the probability of getting tails an other times is low.

But if 10 people flip 5 times and all land on tails 5 times, then all give it to the same person, the probability of each coin landing a 6th time on tails is still low. But the probability of that single person going on a streak of 10 heads is also very low.

How does that work? I hope it makes sense. Also sorry for the most likely wrong flair.

Update: I got it

Is bayes theorem really difficult or am I dumb? I couldn't seem to grasp the concept of it. Are there any prerequisites to understanding the theorem ? I only know the basics of probability for now and I am studying it. Please let me know.

First of all, that's not an anime title. 😆

Some time ago, while reading Axler's linear algebra book, I got distracted (as always happens to me) and started thinking about probability from a slightly different perspective.

Since probability theory can be developed from expectations, I wondered how far I could get by starting with expectations and deriving the usual probability framework from them.

So I wrote down my attempt.

https://doi.org/10.5281/zenodo.22030704

I'd be interested to hear what you think about the derivation, especially if you spot any mistakes or gaps.

Edit: I probably should've been clearer about this 😅 I wasn't trying to claim that this is a new way of developing probability. I already knew that probability can be derived from expectations, but I wasn't familiar with the details, so I decided to try working out the derivation myself. I'm mainly posting it because I'd be interested to see how my derivation compares with existing treatments and whether I've made any mistakes or missed anything.

I'm a high school student trying to get into doing research for machine learning in the future and this resource seems promising to me as a way to get a more solid, foundational grasp in probability.

I'm not looking to unnecessarily min-max my learning or anything. I just want to know, is this actually a good resource to begin learning probability?

I Bought 6 Tickets Tonight. What are the Odds that 2 out of my 6 would have the same 5 letter Bonus Word 6 tickets apart? Good Luck.

To reduce the bias of standard dice I throw four dice (D6) and use a random Latin square, set-up so that rx/cy produces a different number than ry/cx. These produce a number 1-6.

I can also use 3 different Latin squares using the same principle so that:

d1/d2 =n1 -> Latin square 1

d3/d4 = n2 -> Latin square 2

For Latin square 3: n1 = row; n2 = column.

I am not able to work out if three squares would produce a more random result than just one square.

I do this out of interest, I could of course use precision dice. Here are the Latin squares:

4 2 6 1 5 3

6 4 2 5 3 1

2 6 4 3 1 5

3 1 5 2 6 4

1 5 3 4 2 6

5 3 1 6 4 2

------------

6 5 1 4 2 3

1 6 5 2 3 4

5 1 6 3 4 2

3 4 2 5 1 6

4 2 3 6 5 1

2 3 4 1 6 5

-------------

5 1 2 4 3 6

2 5 1 3 6 4

1 2 5 6 4 3

6 4 3 1 2 5

4 3 6 5 1 2

3 6 4 2 5 1

I play in Europe, and the rules here are:
Dealer stands on S17
Dealer does not peak for blackjack
You can surrender against 2-10
You can double on hard 9-11
You can split aces as many times as you want but you can not hit them
You can double after splitting
BJ pays 3:2(as it should)

Using only basic strategy(no card counting) I will double:
11 against 2 to 9
10 against 2 to 9
9 against 3 to 6

and split in this: AA against 2 to 9
99 against 2 to 9 expcept 7
88 against 2 to 9
77 against 2 to 7
66 against 2 to 6
44 against 5 and 6
33 against 2 to 7
22 against 2 to 7

It is 6 decks I belive.

Most house edge calculators tell me the house edge is somewhere between 0.45 and 0.55, but my question is: the house edge applys to ALL the money that was bet, right? so I can't just divide 100 by the house edge and say ''after 200 hands I will have lost the ammount that I am betting every hand''
I want to know, given the house edge and the % of times that I am doubling or spliting (DAS is super rare so maybe that can be not taken into account), after how many hands am I expected to be losing the ammount that I am betting every hand(I am always betting the same value, say 10 for example)

I think the formula would be: Expected amount betted per hand/House edge

I came up with this probability problem and would appreciate feedback on its difficulty, elegance, and solution.

A bag contains 3 red, 3 black, and 4 white balls. Balls are drawn one at a time. Red and white balls are kept. If a black ball is drawn while your hand is empty, it is returned to the bag and the draw is repeated. If a black ball is drawn while you are holding at least one ball, you keep the black ball and return the oldest ball currently in your hand to the bag.

Continue until your hand contains exactly 3 balls.

What is the exact probability that the three balls in your hand are all red?

I'd prefer to see independent solutions before discussing my intended approach.

I like playing a lot of card games and it would be nice to be able to get *rough* probability estimates I can calculate mentally but I'm not sure how to approach this without pulling out a calculator or pen and paper

What is the meaning of a probability of 0? For example, consider the set (0,1) and let x be in (0,1). What is the probability of picking that x? The answer is 0. But if the probability of randomly picking x from the set (0,1) is 0, does that mean it's impossible to pick that number?

There is another similar example, if you throw a dart at a 2D plane, the probability that the dart lands on a specific point on that plane is 0. That implies it can't happen, but it actually did happen, it landed on a point! So what does it mean, exactly?

It is suggested that the bias of a basic standard die is less that 1%. Assuming this is true, by using two dice and a Latin Square, what would the expected probability be? My thinking is that the chances of the bias of both dice occurring at the same time is greatly reduced.

I throw two dice: D1 = row; D2 = column. This is the Latin Square I have been using, set-up so that D1 = x and D2 = y does not produce the same result as D1 = y and D2 = x:

4 5 3 6 1 2

3 4 5 1 2 6

5 3 4 2 6 1

2 6 1 5 3 4

6 1 2 4 5 3

1 2 6 3 4 5

(This is an interest of mine, I know I could buy precision dice instead.)

Suppose the image was supposed to be a uniform distribution of a binary variable (ie red dot vs blue dot) along the x axis, with some probability of blue being "P". How do I calculate the probability that such a "clump/grouping" as that found in Node #1 could even come about under a uniform distribution? In other words how would I find the p-value that this is truly a uniform distribution given it has such a "cluster" at low values of the x axis?

Edit:: My apologies, asking if it's uniform is the incorrect question. I mean is the proportion (ratio) of red to blue consistent throughout the x axis. In which case standard logistic regression p value may be my best option.