HW 10

Gambler’s Ruin

Smith has $1 is betting to win $8. He wins as much as he bets with the probability to win = .4. ###a) If Smith bets $1 each time

s <- 1
M <- 8 
q <- .6
p <- .4

qz <- (((q/p)^s)-1) / (((q/p)^M)-1)
round(qz,4)
## [1] 0.0203

Using the template from the Gambler’s Ruin problem we see the probabilty of Smith getting $8 if he bet $1 each time to be around 2%.

b) If Smith bets as much as possible

x <- 3 
s <- 3
p <- .4
dbinom(x, s, p) 
## [1] 0.064

If Smith bets as much as he can he will need to win 3 times in a row to get to $8 ($1,$2,$4,$8). This resembles as binomial distribution with a probability of .4. Using the bold method Smith has a 6% chance of getting $8.

C) Timid or Bold?

While Smith has a higher probability of getting $8 by going bold and betting as much as he can each time, he has to be exactly right 3 times this way. The timid method would take longer and also has a lower probability of getting $8. Smith should probably go bold and bet as much as possible since the probability is higher.