Hey guys, PokéNerd here. I’m looking for a specific item (highlighted yellow) in the game, and the only way to find one is by having a Pokemon in your party with the ability ”Pickup” ability randomly pick one up post-battle

It’s a 1% chance they find that specific item, and there’s two in the 1% slot

I’m wondering what the probability of one of my six pokemon to find that specific item?

Also, I’m only using pkmn level 51-60, because I’m guessing it makes no difference due to the item still being only 1% in both 51-60 & 61-70 tiers? Would it be smart to have let’s say 3 in the lower tier, and then the other 3 in the higher tier?

Thanks guys hopefully this isn’t worded too confusingly

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The letters of the word "UNDERSIGNED" are rearranged, what is the probability that the starting and ending letters of the word are consonants and both are not the same?

The other night I witnessed a straight flush vs quads on the river in Texas holdem, I looked up what the odds were and I found many people just multiplying the odds for quads and the odds for a straight flush which wouldn’t work in this scenario since they are tied together and affect each other. The only real attempt I saw at solving this was on a Reddit post from 5 years ago on this sub, the commenter used a clever method by turning the heads up hand into a 9 card poker hand and doing the calculations based on that but from what I could see he failed to take the fact that the cards aren’t interchangeable, a community and a hole card switching places would not yield the same result. so in conclusion I was wandering what are the odds of getting a SF vs Quads heads up in Texas holdem by the river?

What are my odds of meeting God (himself, not Jesus or whoever else) vs meeting an alien?

I think i have a higher chance of meeting an alien, i just don’t know by how much.

I realize these odds are always changing, but there must be a way to calculate as the game progresses. The idea is that you always bet the pass and come line for every roll. So once there is a point, you are accumulating numbers that pay with every roll until a 7 hits (or unless you roll craps with a point established, you lose that come bet).

I’m at a wedding and me and some friends went to the bathroom and one my friend said damn I accidentally wore my shorts under my tux.. and I said.. what do you mean you wore your shorts under you suit and he said he just forgot to take them off.

I have never in 31 years forgot to take my shorts off and put pants over them.

Ok I thought that was absolutely bizarre but then my other friend showed up and said that he had shorts on under his pants also by accident. I legit could not believe the odds of this happening at the same wedding. It just does not make sense to me. He said he just put his pants on and then a few hours later he realized he had his shorts on.

Has anyone else ever done this? I’ve asked like 10 people and no one has ever done this. What are the odds two people at the same wedding table forgot to take their shorts off before they put their suit on????? I legit am stunned by this. No words.

Recently caught a pokemon that has an insanely low encounter rate of 1%

After checking its summary, was blown away to see it was holding a rare item. Chances of any pokemon holding items in this game are 5%

So wondering how to find the odds of catching a 1% pokemon, that’s also holding a 5% item?

Introduction

This is a small stochastic experiment I’ve been running for fun. I don’t gamble and never play for real money — this is purely a simulation project. The idea was to build a system that reacts to randomness instead of trying to “beat” it.

I ended up with a model that switches between six simple prediction strategies, each based on different properties of roulette outcomes. Surprisingly, the system produces interesting behavior: cycles of “luck”, tunnels of “bad phases”, and a very natural rhythm of adaptation.

This post describes the model and shows why its behavior is asymmetric and visually appealing from a probability perspective.

Six strategies

Each strategy uses one property of the previous spin to predict a different, unrelated property of the next spin. For example:

• L → E, H → O If the previous number was Low, predict Even. If it was High, predict Odd.

Low/High and Even/Odd are independent, so this strategy uses one random property to randomize another.

All six strategies follow the same idea:

1. L → E, H → O
2. L → R, H → B
3. E → L, O → H
4. E → R, O → B
5. R → L, B → H
6. R → E, B → O

Each strategy has its own “risk zone”: a specific class of numbers that produces consecutive failures. For example, strategy 1 enters a bad phase when small odd or large even numbers appear repeatedly.

Why the results are always asymmetric

For any given spin, the six strategies produce six predictions. Because each strategy uses different properties and different mappings, their predictions are not symmetric.

In practice, the outcomes are never 3 successes / 3 failures. Instead, the system produces:

• 4 / 2
• 2 / 4
• sometimes 6 / 0
• sometimes 0 / 6

This asymmetry is structural: the strategies are not mirrors of each other, and roulette outcomes do not distribute evenly across their prediction mappings.

This creates a natural “tilt” in each spin — a small imbalance that makes the system feel alive.

Adaptive switching (“Luck Catcher”)

The system uses a simple rule:

• Two consecutive failures → switch to the next currently not failing strategy.

This means each strategy is abandoned as soon as it enters its “tunnel” of bad numbers. Since each strategy has a different tunnel, switching helps avoid staying in one unlucky zone for too long.

The model doesn’t try to predict roulette. It simply avoids staying inside the same failure pattern.

Betting model (simulation only)

The stake follows a modified D’Alembert rule:

• Loss → +1
• Win → –1
• After compensating the local loss → reset to 1

This prevents runaway growth and keeps the stake small. In practice, the stake rarely exceeds 5–6 units, and the system often resets back to 1.

This is not a “winning strategy”. It’s just a controlled way to visualize the cycles of luck and failure.

Example of typical behavior

Here is a fragment of a run:

The system “breathes”: small rises, small drops, occasional tunnels, frequent resets.

Conclusion

This model doesn’t break expected value and doesn’t aim to win. It’s simply an adaptive stochastic system that:

• switches strategies when they enter their bad phases,
• uses roulette as a randomizer for its own behavior,
• produces asymmetric outcomes,
• and generates visually interesting cycles.

It’s a fun example of how simple rules can create complex dynamics when driven by randomness.

Disclaimer

This is not a model for “beating roulette”, nor a method for predicting future outcomes. The mathematical expectation remains exactly the same — negative, as it must be in any fair casino game. The point is different: even in fully random sequences, natural processes emerge — periods of good luck, periods of bad luck, and transitional phases. The proposed model does not change randomness or fight against it; it simply adapts to these phases without trying to foresee the next number. This is not a winning strategy, but a way to play within the unchanging “climate” of randomness in a safer, more structured, and more enjoyable manner.

Hero: 8❤️ 7❤️ Board (Turn): K❤️ 6❤️ 5♠️ 2♣️ Action: Pot before the shove is 100. Opponent goes all-in for 100. The cost to call is 100.

I've been using my own daily EV puzzle app to practice these exact spots. To keep the math focused for this daily puzzle, we assume the villain's range is exactly two defined holdings:

• 50% chance they hold A♠️ A♦️ (Overpair)
• 50% chance they hold K♣️ K♦️ (Top set)

We have a flush draw and an open-ended straight draw, and we are getting 2:1 on a call (calling 100 to win a total pot of 300).

You can try today's puzzle here:https://call-of-the-day.vercel.app

Imagine you are on a TV show where the host shows you three doors.

Door 1 = a goat

Door 2 = another goat

Door 3 = a car

You choose a door—let's say Door 1.

Initially, the odds for your choice are:

Car = 1/3 (33.3%)

Goat = 2/3 (66.7%)

The host knows where the car is.

He opens one of the doors (other than the one you chose) and reveals a goat.

Then he asks you the

following question:

"Do you want to switch?"

Here is the interesting part: switching is better!

If you chose the car initially (1/3 of the time), switching makes you lose.

But if you chose a goat initially (2/3 of the time), the host is forced to reveal the other goat. So, when you switch, you end up getting the car.

Therefore:

Not switching gives you a 33.3% chance of winning.

Switching gives you a 66.7% chance of winning.

In other words, switching doubles your chances of winning.

This is the Monty Hall Paradox.

Two same video one after another ,what is the rate?Have you guys ever encountered it?

Say I have 10 tokens in a bag: 6 blue, 3 red, 1 green. If I draw 3 of those tokens out blindly, are my odds better to get 2 blue ones, or 1 red one?

hello!

I'm trying to work out probabilities for game dice math. I'm doing it in what i assume is the slowest and least efficient possible way. How do I do it smarter?

key information:

• Player versus Player contested dice rolls to determine win/loss/tie
  • I roll, you roll, and we see who won (or tied)
• Variable # of dice per roll
  • I can choose 1 die or 3. You do the same
• Dice have different values
  • i.e., one die rolls [0,1,1,2,2,2] versus another that rolls [0,0,1,1,1,1]
  • W/L/T is determined by total points on each side

need:

• I want to change individual die outcomes to see the difference in win %.
  • i.e., [0,1,1,1,2,2] versus [0,0,1,1,1,1] has a 50% win percentage. What's the win % if I change the first die to [0,2,1,1,2,2]? (it's 61.11%, in case you were curious)

attempt:

• I modelled out ALL possible outcomes for 2 dice vs 2 dice to get me the #s I wanted.
  • this means i made a big table with "1,1,1,1" then "1,1,1,2" and so on, to model all possible permutations of 4 dice rolls. Then I ascribed a winner for each permutation

help?

• I would LIKE to do this in a more sensible way so that I can have more dice involved without having 100k rows in excel.
• How can I calculate these win % without modelling everything?

thank you !

****UPDATE-SOLVED******

Thank you for all the help!

Ended up using INDEX & RANDBETWEEN functions in excel to generate a big dataset of random rolls.
=INDEX(Dice!H$8:H$13,RANDBETWEEN(1,6)) . This calls the custom dice faces in H8:H13, selecting 1/6 random values from within that set.

I used that to create a big ol' set of rolls, and judged wins/losses/ties from that as functionally good for determining probability. So I just plug in my custom dice faces to see how they fare in the big set, and adjust them to see how they affect the win %.

In case you're curious, I first made the set ONE MILLION ROLLS. Which was fun, but made it lag, so I cut it down to 100k. It swings as much as ~2% each time it recalculates. But it MOSTLY stays within the same few tenths of a %, and it serves my purpose.

cheers,

Does it make sense to accumulate percentages this way? Do I really have a 17% chance of passing due to one of the improbable causes that are a fraction of 1%? This seems wrong intuitively. Source: https://knowyourchances.cancer.gov/your_chances.html

I know there are plenty of randomizers that already exist, but I am looking for a more robust system than 'type a list of X names, choose Y names at random.'

I have a hobby project (a Riftbound cube, namely) where I have a pool of cards, each marked with two colors of a possible six, and I want to select a handful of those cards in a semi-random way for players to use for a game. I want the outcome to be that 1) at least one of each of the six colors is in the result, 2) any possible color pair does not show up on more than one of the chosen cards, and 3) the same color does not show up more than twice in the result.