Main Prompt

Smith is in jail and has 1 dollar; he can get out on bail if he has 8 dollars.

A guard agrees to make a series of bets with him.

If Smith bets A dollars, he:

Find the probability that he wins 8 dollars before losing all of his money if:

Part 1

“he bets 1 dollar each time (timid strategy)”

This is the “Gamblers Ruin Problem”, so

\(P(i) = \frac{1-(\frac{q}{p})^t}{1-(\frac{q}{p})^N}\)

So if he were to bet 1 dollar each time:

p <- 0.4
q <- 0.6
t <- 1
N <- 8

(1-(q/p)^t) / (1-(q/p)^N)
## [1] 0.02030135

Below is a simulation of the described scenario

# We will run the experiment 100K times. Each time, the player will bet 1 dollar.
# If the players total cash on hand reaches 0, we add one to "failed"
# If the players total cash on hand reaches 8, we add one to "released"
# Finally, we can divide released by failed to get our probability

released <- 0
failed <- 0

for (i in seq(1,100000)){
  start <- 1
  for (i in seq(1,1000)){
    if (runif(1)<=0.4) {start = start +1} 
    else {start = start - 1}
    
    if (start == 0){
      failed = failed + 1
      break
    }
    
    if (start >= 8){
      released = released + 1
      break
    }
  }
}
released / failed
## [1] 0.02097074

Part 2

“he bets, each time, as much as possible but not more than necessary to bring his fortune up to 8 dollars (bold strategy).”

So based on the conditions:

bet_number <- c(0, 1, 2, 3)
total_cash <- c(1, 2, 4, 8)

That means that he will need to win three separate times. We know that the probability of winning a given bet is 0.4… therefore, the probability of winning would be

p**3
## [1] 0.064

But we can also simulate this example:

# We will run the experiment 100K times. Each time, the player will bet however many 
# dollars he has on him.
# If the players total cash on hand reaches 0, we add one to "failed"
# If the players total cash on hand reaches 8, we add one to "released"
# Finally, we can divide released by failed to get our probability

released <- 0
failed <- 0

for (i in seq(1,100000)){
  start <- 1
  for (i in seq(1,1000)){
    if (runif(1)<=0.4) {start = start + start} 
    else {start = start - start}
    
    if (start <= 0){
      failed = failed + 1
      break
    }
    
    if (start >= 8){
      released = released + 1
      break
    }
  }
}
released/failed
## [1] 0.06791969

Part 3

“Which strategy gives Smith the better chance of getting out of jail?”

The bold stategy provides Smith with a 6.4% chance of escape.

The timid strategy provides Smith with a 2.0% percent chance of escape.

The bold strategy is the better option, although both are pretty bad!