GamblingStrats-Practice.Rmd

---
title: "Strategies Testing"
author: "Logan"
date: "`r Sys.Date()`"
output: pdf_document
urlcolor: blue
---
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knitr::opts_chunk$set(
  collapse = TRUE,
  fig.align="center",
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```

---
## Beginning Problems
## Problem 1, Consider a 20-sided die, what is the expected number of rolls before obtaining a minimum value of X:

Using Monte-Carlo Methods we can find that the average number is about 20 rolls for a minimum of 20, about 10 for a minimum of 19, and 6.67 for a minimum of 18. Next we should try to derive these solutions analytically.
```{r}

possible = seq(1:20)

diceroll <- function(){
  return(sample(possible, 1, replace = T))
}
  
timeTaken <- function(minVal){
  time = 0
  inputs = 0
  while (inputs < minVal) {
    time = time + 1
    inputs = diceroll()
    #print(inputs)
  }
  return(time)
}


options = seq(1:10000)
#options2 = seq(1:10000)
#options3 = seq(1:10000)

for (i in options) {
  options[i] = timeTaken(17)
  #options2[i] = timeTaken(19)
  #options3[i] = timeTaken(18)
}
#hist(options, probability = T, breaks= 50)
print("For a minimum of 20:")
mean(options)
#print("For a minimum of 19:")
#mean(options2)
#print("For a minimum of 18:")
#mean(options3)

```

## Analytically,
Consider a fair 20-sided die. The probability, $p = \frac{1}{20}$, of obtaining any side. The time taken to reach a certain value can then be modeled using $T_{20}$.

$$P[T_{20} = t] = \frac{1}{20}(\frac{19}{20})^{t-1}$$
Thus, the expected value of $T_{20}$ is the following infinite sum.
$$E[T_{20}] = 1(\frac{1}{20}) + 2(\frac{1}{20})(\frac{19}{20})+3(\frac{1}{20})(\frac{19}{20})^{2}+4(\frac{1}{20})(\frac{19}{20})^{3}... $$
Note that this is an increasing geometric series with $r = \frac{19}{20}$,

$$ E[T_{20}] = \sum_{t=1}^{\infty}[t(\frac{1}{20})(\frac{19}{20})^{t-1}] = a $$

Now consider,

$$ \frac{19}{20}E[T_{20}] = \sum_{t=1}^{\infty}[t(\frac{1}{20})(\frac{19}{20})^{t}] $$
Thus, $$ \frac{1}{20}a = a - \frac{19}{20}a $$ is a geometric infinite sum after doing a little algebra.
$$ \frac{1}{20}E[T_{20}] =  \sum_{t=1}^{\infty}[t(\frac{1}{20})(\frac{19}{20})^{t-1}] - \sum_{t=1}^{\infty}[t(\frac{1}{20})(\frac{19}{20})^{t}] = \sum_{t=1}^{\infty}[(\frac{1}{20})(\frac{19}{20})^{t-1}] $$

This is a closed form sum:
$$\frac{1}{20}E[T_{20}] = \frac{\frac{1}{20}}{1-\frac{19}{20}}=\frac{\frac{1}{20}}{\frac{1}{20}}$$
Thus, we find that $E[T_{20}] = 20 $
We then find that the general formula for the $P[X \geq a]E[T_X] = \frac{P[X \geq a]}{1-P[X<a]}$
This may also be written as:
$$E[T_{X}] = \frac{1}{P[X \geq x]} $$
Thus we see that $E[T_{19}] = 10$, $E[T_{18}] = \frac{20}{3} \approx 6.66$, $E[T_{17}] = 5$
So, the Expected number of rolls is equal to the inverse of the probability of obtaining those rolls.

I now wonder if that can be said about any sized dice and any probability dice.
Theoretically the math should follow a similar written proof, but may we find these results through monte-carlo?

```{r}



diceroll <- function(possible){
  return(sample(possible, 1, replace = T))
}
  
timeTaken2 <- function(minVal, possible){
  time = 0
  inputs = 0
  while (inputs < minVal) {
    time = time + 1
    inputs = diceroll(possible)
    #print(inputs)
  }
  return(time)
}

meanRollsBasicDie <- function(sides, minVal, numberOfTimes){
  options = seq(1:numberOfTimes)
  possible = seq(1:sides)
  for (i in options){
    options[i] = timeTaken2(minVal, possible)
  }
  mean(options)
  
}

meanRollsBasicDie(10, 9, 1000)

```

Using these defined functions, let's consider a fair 100-sided die. According to our rule, the number of rolls it should take to obtain at least a 100 is $E[T_{[100,100]}] = 100$, the number of rolls to obtain at least an 91 is $E[T_{[100,91]}] = 10$, and finally at least a 51 is $ E[T_{[100,51]}] = 2 $
```{r}
meanRollsBasicDie(100, 100, 1000)
meanRollsBasicDie(100, 91, 1000)
meanRollsBasicDie(100, 51, 1000)
```
Now, consider other sided dice, we may use a 1000 sided die, 2 sided coin, or we may attempt to vary the probability vector of the sampling.

```{r}
meanRollsBasicDie(1000, 501, 1000)
meanRollsBasicDie(2, 2, 1000)
meanRollsBasicDie(6, 4, 1000)

```





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