Probability – Find the expected number of dice rolls

Accept $ A $ and $ B $ Dice one after the other $ A $ roll first. Suppose the roles are independent. $ A $ want to get a sum of $ 6 $ and $ B $ a summary of $ 7 $, The game ends when one of the two players reaches the goal and this player is declared the winner.

(1) Find the expected number of dice

(2) Find the variance of the number of dice rolls

My idea is as follows:

Leave random variable $ X $ means $ A $ wins the game and $ Y $ means $ B $ win the game. $ N $ is the number of dice rolls.

I can get the chance for that $ A $ wins in the first round, that is,
$$ mathbb {P} (X | N = 1) = frac {5} {36} $$
and the probability of $ A $ wins in the $ 2k + 1 $it's on
$$ mathbb {P} (X | N = 2K + 1) = ( frac {31} {36}) ^ k ( frac {5} {6}) ^ k frac {5} {36} $ $
and the probability that $ A $ win that
$$ mathbb {P} (X) = frac {5} {36} sum_ {k = 0} ^ { infty} ( frac {155} {216}) ^ k = frac {30} { 61} $$

But the expectation of $ N $ I'm not sure. Is that to use
$$ mathbb {E} (N) = sum_ {k = 0} ^ { infty} k mathbb {P} (N = k) ??? $$

Are you throwing dice when the result of a save throw is an automatic hit / miss?

There are several functions like that Bones of the earth Spell that says:

[…] When a pillar is created under a creature, that creature must pass a saving throw for skill or be lifted off the pillar. A creature can choose to fail the save

The metamorphic option of Sorcerer's Careful Spell, which states:

If you cast a spell that forces other creatures to save, you can protect some of these creatures from the full force of the spell. […] A selected creature automatically succeeds if it makes a saving throw against the spell.

And the force Spell that says:

[…] A target automatically succeeds in this saving throw if it can not be enchanted …

In such cases, does the creature of your choice roll the dice? Did they even make a saving throw?

c ++ logic behind dice throw permutation problem not understood

I'm having a hard time understanding the logic behind the infamous question, "Find all the ways to reach a certain sum with n cubes of k faces." After searching extensively for explanations, I went out empty-handed.

My confusion stems from the fact that it does not seem to follow the same logic as the "number of ways to make changes to coins" problem that is well explained in a number of videos, especially this one.

My understanding of the problem of money exchange

That's pretty good, I think. When we pass the first row, we are essentially looking at the number of ways to find the target from the first and second coins, not just the second coin (generally, this is the current coin plus all previous coins). , As explained in the linked video, we have to follow some rules:

     Simply copy the value from the cell above.  
     a) Determine the number of ways we can make the target when we EXCLUDE the current coin
     b) Determine the number of ways we can make the target when we INCLUDE the current coin
     c) Add the two together and store the result in the cell.

For step aWe can determine this by simply copying the value from the cell above: this, after all, is the number of ways we can achieve the same goal without the current coin. So we copy the 0 for 3, that's right; With a 2 coin, there are 0 ways to get the target to 3.

For step bwe have to subtract the current coin from the target, which gives us the rest. Then we have to figure out how many options there are to make that remainder out of the current coin. So for the change of value 3 we have (3-3), which is 0. For a zero goal, there is 1 way to make 0 with a 3 coin, as the table shows (highlighted). So we have 0 + 1 = 1.


I can see how to fill the entire table. The algorithm used is not important here.

My failure to understand the logic of the cube problem

The following algorithm comes from the link above:

long findWays(int f, int d, int s) 
    // Create a table to store results of subproblems. One extra 
    // row and column are used for simpilicity (Number of dice 
    // is directly used as row index and sum is directly used 
    // as column index). The entries in 0th row and 0th column 
    // are never used. 
    long mem(d + 1)(s + 1); 
    memset(mem,0,sizeof mem); 
    // Table entries for no dices 
    // If you do not have any data, then the value must be 0, so the result is 1 
    mem(0)(0) = 1; 
    // Iterate over dices 
    for (int i = 1; i <= d; i++) 
        // Iterate over sum 
        for (int j = i; j <= s; j++) 
            // The result is obtained in two ways, pin the current dice and spending 1 of the value, 
            // so we have mem(i-1)(j-1) remaining combinations, to find the remaining combinations we 
            // would have to pin the values ??above 1 then we use mem(i)(j-1) to sum all combinations 
            // that pin the remaining j-1's. But there is a way, when "j-f-1> = 0" we would be adding 
            // extra combinations, so we remove the combinations that only pin the extrapolated dice face and 
            // subtract the extrapolated combinations. 
            mem(i)(j) = mem(i)(j - 1) + mem(i - 1)(j - 1); // CONFUSION POINT A
            if (j - f - 1 >= 0) 
                mem(i)(j) -= mem(i - 1)(j - f - 1);   // CONFUSION POINT B
    return mem(d)(s); 

I've read this code over and over again, maybe too often, but I just do not understand it.

Number of cubes (d) = 3, number of faces on each die (f) = 3, target value (s) = 3. I've added some debugging issues and generated the following to show how the cells in my spreadsheet are calculated:

(1,1) = (1,0) + (0,0)
(1,2) = (1,1) + (0,1)
(1,3) = (1,2) + (0,2)

!! (2,1) is never calculated, CONFUSION POINT C !!

(2,2) = (2,1) + (1,1)
(2,3) = (2,2) + (1,2)

Having parsed the algorithm into plaintext shows that

the number of ways of making 3 with 2 dice (2,3) is
    the number of ways of making 2 with 2 dice (2,2) +
    the number of ways of making 2 with 1 die  (1,2)

At this point, it does not seem to follow the logic of the coin change problem; That is, the pseudo code block I posted above. I even tried to specify the numeric values ​​to apply this logic. i.e. the 1 is (1) and the 2 is (2,1) to follow a similar idea as "adding extra coins", but it just does not work. This is worrisome for me because, while I understand the (beautiful) concept behind dynamic programming, it does not seem to be able to apply the same logic to all similar problems of this kind.

My questions

Can someone please explain my three confusion points to me:

a) How it works and the logic behind it.

b) What on earth is this line doing?

c) Why is cell (2,1) never calculated? I'm pretty sure now that this is simply because we can never make 1 of 2 dice, but I'd like to confirm that if I missed something.

unity – Dice Motion with Touch In Mobile?

I just start a lesson.

Cube works perfectly left, right, up, down

Cube up and down: but if I comment on a code left and right, then work up and down

Cube left and right: but if I comment a code up and down, then work left and right


 Rigidbody2D rb;
 public float movespeed=3f;

 // Use this for initialization
 void Start()
     rb = GetComponent(); //Access A Rigidbody Component

     Debug.Log("Screen Width : " + Screen.width);     //1080
     Debug.Log("Screen Height : " + Screen.height);   //1920


 // Update is called once per frame
 void Update()
     if (Input.touchCount > 0)
         Touch touch = Input.GetTouch(0);  

         switch (touch.phase)
             case TouchPhase.Began:   //start touch

                 if (touch.position.x < Screen.width / 2)     //left
                     rb.velocity = new Vector2(-movespeed, 0f);
                 DebugPanel.Log("x", "first W:", touch.position.x < Screen.width / 2);

                 if (touch.position.x > Screen.width / 2)     //right
                     rb.velocity = new Vector2(movespeed, 0f);
                 DebugPanel.Log("x", "Second W:", touch.position.x > Screen.width / 2);

                 if (toufch.position.y < Screen.height / 2)    //Up
                     rb.velocity = new Vector2(0f, -movespeed);
                 DebugPanel.Log("y", "First H :", touch.position.y < Screen.height / 2);

                 if (touch.position.y > Screen.height / 2)    //Down
                     rb.velocity = new Vector2(0f, movespeed);
                 DebugPanel.Log("y", "Second H:", touch.position.y > Screen.height / 2);

             case TouchPhase.Ended:
                 rb.velocity = new Vector2(0f, 0f);      //when cube is ended then velocity is 0

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Vampire of the Dark Ages – Splitting Dice Pool between actions with and without dice roll

I'm a bit confused with the rules for splitting the cube pool in situations where you want to take multiple actions and do not roll any of the actions.

Suppose I am a Ravnos and want to use Ignis Fatuus to create an illusion AND simultaneously attack with a sword. The first action does not require a die roll. Only the output of 1 point willpower. The second action requires a die roll. But what would happen? Could I just assign 1 dice to "use" my discipline (not dice but use willpower instead) and then use the rest of my pool to attack?

For reference only, I play vampires: Dark Ages

Probability – dice games – when is greedy / short-term profit optimal?

In this question, I refer to two separate games. The first is a game in which you roll the dice and collect your points until a six appears:

You play a game with a normal six-sided dice. They start with 0
Points. Before each litter you decide, if you want to continue the litter
Play or finish it and keep your points. If you rolled 6 after every roll,
then you lose everything and the game ends. Otherwise, add the score
from the dice to your total points and continue / finish the game. When
should one stop playing this game?

The other is a game that I asked before, where you roll the dice and collect your score until a replay occurs:

I continue to roll and my score is the sum of all my litters.
However, if I roll a value that I rolled before, I lose everything. What is
the optimal strategy?

From what I learn in the first link, the greedy approach in the accepted answer is technically incorrect because the calculated gain does not take into account future litters, although it is optimal for some reason. In the second link (the question I asked), I did not take into account future litters that hindered my analysis of the time of the interruption (I have not yet figured out what the answerer means by giving the minimum of 1 + 2 + 3 " suggests "= 6). I am quite confused by these concepts and therefore have two questions:

1) Unfortunately, this is a repetition of part of the question I have asked before – what exactly is meant by long-term gain of future roles and how do you calculate it? Is it a recursive probability calculation that takes into account our decision after each litter? Do we let the number of rolls tend towards infinity?

2) More importantly, when is a greedy approach optimal and how do we prove it? I do not really understand him for the first link (despite the thorough debate about it).

Many Thanks!

dnd 5e – Do you have to spend hit dice to take a short break?

A short break is a downtime of at least one hour in which a character does nothing more strenuous than eating, drinking, reading and caring for wounds.

A character can At the end of a short break, roll out one or more hit cubes, up to the maximum number of hit cubes of the character that corresponds to the character's level. […]

Short break, Handbook of the Player, pg. 186

Emphasizing my. If you were required To spend Hit Dice during a short break to take a short break, the rules would say so explicitly, perhaps with language like "A character Got to Give up one or more hit cubes at the end of a short break to profit from it". But not her.

So if you're an assistant trying to use yours Arcane Restoration Function or a fighter that restores your data Action SurgeOr a sorcerer who restores spell slots, or another character who can gain additional benefits over a hack of hit dice by taking a short break.

Dice – How can I calculate the probability of reaching a certain difficulty value on 3d6 with critical success / failure modifiers?

In a Fuzion system variant, 3d6 is used to resolve capability checks. A task gets a difficulty level (DV) from 10 (easy) to impossible (30 and higher). Characters can have primary attribute levels from 1 to 10 and the same Skill Scale.

A skill check is done with Attribute + Skill + 3d6 against DV (you win in case of a tie). The tricky part is that when you roll 3d6, if you get all the dice as 6, you get an extra 2W6 added to your score. Likewise, you must deduct 2W6 from your score if all dice are 1.

How can I calculate the probabilities for a given Basic Rat + ability against a particular DV? For example, if I have a presence of 10 and a query of 10 and try to compete against a DV of 50, then:
10 + 10 + 3d6 (and possibly + -2d6) VS 50, what are my chances for at least 50?

Dice – What's the name of the tool that makes it less likely that the same number will be rolled consecutively?

I'm looking for an online tool that a friend once used in an RPG and that has adjusted the probabilities of an otherwise fair dice roll so that it's less likely to roll the same number multiple times. So, if you roll a 1d6 and get a 4, it's less likely that your next roll of 1d6 will be a 4 than roll 1, 2, 3, 5, and 6.

This was a pretty interesting game as it forced a wider value distribution for dice that we would not have seen otherwise. We still diced the same number on average, but the distribution was wider.

Does anyone know where I could find this tool?