Probability Distribution with a resettable count to win

[latot]

TL;DR Summary

Is there a way to get the probablity distribution of the number of wins in a game, when you have a "x" probability to win, but at certain number of rolls you win, with a resettable count that force win.

Hi hi, I was thinking about this, all of this starts playing a game, I'll show a simplification:

We ca win several times.

We have a count $n$, where is the max number of rolls until you win, let's say we can win a $m$ amount.

In every roll we can win $m$ with a probability of $p$.

If we win in a roll, or the count reach 0, the count will start again.

We have one variable, the number of rolls we can use, this variable should set a new function that is the amount that we can win, and the probability, we can write this in other way too, we can question what is probability to win a certaim amount with a specific number of rolls.

I don't know if there is a way to calculate this, the part that can be maybe more hard is the count, bacause depending if we win or not, it can start again and again.

Fixed, sorry.

Thx.

 

[PeroK]

It's not clear what game you're describing here.

 

[latot]

Okis, an example, let's say we can win coins, if we win we will get 10 coins per win:

We have 50% of probs to win in every roll.

coins: 0
count: 5

roll 1 time, let's say we loose:

coins: 0
count: 4

roll again, and we loose:

coins: 0
count: 3

Here I'll split this

case 1: roll again and we win

coins: 10
count: 5 (reset because we win)

case 2: roll and we loose every time until count reach 0

roll

coins: 0
count: 2

roll

coins: 0
count: 1

roll

coins: 10
count: 5

In the last the count reach 0, so we win, and the count start again.

This is an infinite game, but there should exist a distribution $Probability(Number of rolls, coins)$

I don't know if there exist a known way to get the distribution.

Thx.

 

[FactChecker]

In this example, what is a "win"? (are there two types of "win"?) What is n? What is m? What is "coins", what is "count"? I think you should explain the first few lines in more detail.

 

[latot]

$n$ is the start number of the count, when it reach 0 you win automatically and start again, will decrease by 1 in every roll.

$coins$ the something that we want to get, more coins better!

$m$ the amount of coins we get by winning 1 time.

This a game where the only options is win or loose, in every roll we have a $p$ probability to win, like a dice with 2 faces.

The game don't have an end, so we can continue rolling for ever.

 

[PeroK]

Do you mean that you keep playing until you lose five rolls in a row? And, at that point you keep the coins you have won? Ten coins for every time you win a roll?

 

[latot]

You can playing forever.

Just when the count reach 0 it start again and we get the 10 coins, you can continue playing.

The coins are accumulative, don't start again, even if the count reset.

Yes, 10 coins when we win rolling the dice :3

 

[PeroK]

What's the point of all this? The game seems to go nowhere.

 

[latot]

The question is how to get the distribution of this...

The game it self is pretty more complicated this is just a simplified version, if you put cost and a lot of more things it takes sense, but the game it self is not the focus..., there is a lot of costs and other things.

The point is not winning the game, is solve the distribution of the game.