r/probabilitytheory

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

If you throw the same coin multiple times with the exactly same force in the same surface, it won't give you the same side every toss, right? But why? If repeated with the precisely identical conditions, why the results are still unpredictable? Or it is actually predictable?

Sorry if it looks confusing, English is not my first language and I'm not that expert in physics lol, just curious and has this though randomly.

Edit: maybe considering that it lands on a surface, and it bounces or not, changing the coin's side.

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?

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 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

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Hello all! I have this problem that I tried to solve but confused at some point.

For some context, I am playing Dungeons & Dragons with some friends. In D&D, whenever you want to do some actions, you roll a 20-sided die. Rolling 1 means absolute fail while rolling 20 means absolute succes in most cases. Various success levels changes based on the rest of the numbers, but unrelated to my problem.

Sometimes you can have advantage on those rolls, which means you roll the die twice and take the higher one.

I have this gambling themed item that allows me to roll 3d8 instead of a 1d20, whenever I have an advantage. Which is 3 8-sided die.

While rolling this 3d8 item, I have severeal possibilities:

• Rolling 3 uneven rolls
• Rolling 3 even rolls
• Rolling sequential numbers (ex: 1,2,3)
• Rolling 3 identical rolls
• Rolling 3 7s
• Rolling 3 8s
• Additionally if the total of the rolls equals to 13, I can roll any one of the dice and possibly get a better roll

Rolling any of the above set of rolls automatically makes my attempt absolute success, with additional effects that are unrelated.

I tried to calculate my possibilities of winning compared to regular 1d20. Please let me know if I calculated wrong (It's been a while since highschool)

For the regular d20, my probability of absolute success with advantage is (1/20+1/20)= 1/10 or 0.1

With this item, my possibilities are

3 uneven rolls, 4/8 x 4/8 x 4/8

3 even rolls, 4/8 x 4/8 x 4/8

sequential roll, 6/8 x 5/8 x 4/8 (I can roll 6 as highest to have increasing sequantial)

3 identical rolls, 1/8 x 1/8 x 1/8

3 sevens, 1/8 x 1/8 x 1/8

3 eights, 1/8 x 1/8 x 1/8

If I did math correct, the outcome is 251/512, or roughly 0,4.

Now here where it gets complicated:

• I couldn't calculate reverse-sequential,
• I think I calculated the regular d20 calculations wrong.
• having an option to reroll on of the die, if the total 13 should affect the final outcome but I don't know how to calculate that.

In the end, I am almost sure that rolling 3d8 is always better than rolling 2d20 bu I want to understand exactly how and have a better understanding of maths!

Thanks for everyone for help!

Hi, I need a good source to study Stats and Probability for Quantitative Finance. Please suggest books, courses, online material, YouTube, anything works.

Hi, I’m a software engineer with very little background in probability and statistics.
Recently, I’ve been playing Ludo on YouTube Playables, and I’ve started to wonder whether the dice rolls are actually fair.
In particular, when one of my pawns is a certain number of spaces ahead of an opponent’s pawn, (4 player game, 3 computer vs me) it feels like the opponent rolls the exact number needed to capture my pawn more often than I would expect from a fair six-sided die.
I realize this could simply be confirmation bias, so I’d like to test it properly rather than rely on intuition.
I can collect data from my games, but I’m not sure how to analyze it. How would I formulate and statistically test the hypothesis that the dice rolls are biased in situations where an opponent has an opportunity to capture my pawn?
Could someone explain what data I should collect, what statistical test would be appropriate, and the probability/math behind it in a way that’s accessible to someone without much statistics background?