Q.5.17 - A First Course in Probability

Question

A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. If you bet 1 on a specified number, then you either win 35 if the roulette ball lands on that number or lose 1 if it does not. If you continually make such bets, approximate the probability that 

(a) you are winning after 34 bets; 

(b) you are winning after 1000 bets; 

(c) you are winning after 100,000 bets 

Assume that each roll of the roulette ball is equally likely to land on any of the 38 numbers 

Step-by-Step Solution

Answer

1. The probability that you are winning after $34$ bets $0\times 6628$.
2. The probability that you are winning after $1000$ bets $0\times 409$.
3. The probability that you are winning after $100,000$ bets $0\times 0020$.

Step 1: Given information (Part a)

A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. If you bet 1 on a specified number, then you either win 35 if the roulette ball lands on that number or lose 1 if it does not. 

Step 2: Solution (Part a)

Here we need to find 

$P[36X-n>0]=P[36X>n]$

$=P\left[X>\frac{n}{36}\right]$

Since ' X ' is a Binomial random variable with parameters n & b where

$p=\frac{1}{38}$

So, $n=34$

$P\left[X^3\frac{34}{36}\right]=P[X>0\times 5]$

Since continuity correction

$=P\left[\frac{X-\frac{34}{38}}{\sqrt{\left(34\right)\left(\frac{1}{38}\right)\left(\frac{37}{38}\right)}}>\frac{0\times 5-34\left(\frac{1}{38}\right)}{\sqrt{\left(34\right)\left(\frac{1}{38}\right)\left(\frac{37}{38}\right)}}\right]$

$=P\left[Z>\frac{-0\times 3947}{\sqrt{0\times 8712}}\right]$

$=P[Z>-0\times 4229]$

$=1-\mathrm{\Phi }(-0\cdot 4229)$

$=\mathrm{\Phi }\left(0.4229\right)$

$=0\times 6628$

Step 3: Final answer (Part a)

The probability that you are winning after $34$ bets $0\times 6628$.

Step 4: Given information (Part b)

A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. If you bet 1 on a specified number, then you either win 35 if the roulette ball lands on that number or lose 1 if it does not. 

Step 5: Solution (Part b)

Here $n=1000$

$P\left[X>\frac{1000}{36}\right]=P[X>27\times 5]$

$=P\left[\frac{X-1000\left(\frac{1}{38}\right)}{\sqrt{1000\left(\frac{1}{38}\right)\left(\frac{37}{38}\right)}}>\frac{27\times 5-1000\left(\frac{1}{38}\right)}{\sqrt{1000\left(\frac{1}{38}\right)\left(\frac{37}{38}\right)}}\right]$

$=P\left[Z>\frac{1\times 842}{\sqrt{0\times 6925}}\right]$

$=P[Z>0\times 2339]$

$=1-\mathrm{\Phi }\left(0.2339\right)$

$=1-0\times 5910$

$=0\times 409$

Step 6: Final answer (Part b)

The probability that you are winning after $1000$ bets $0\times 409$.

Step 7: Given information (Part c)

A roulette wheel has 38 slots, numbered 0, 00, and 1 through 36. If you bet 1 on a specified number, then you either win 35 if the roulette ball lands on that number or lose 1 if it does not. 

Step 8: Solution (Part c)

Here $n=100,000$

$P\left[X>\frac{100,000}{36}\right]=P[X>2777\times 5]=P\left[\frac{X-100000\left(\frac{1}{38}\right)}{\sqrt{100000\left(\frac{1}{38}\right)\left(\frac{37}{38}\right)}}>\frac{2777\times 5-100000\left(\frac{1}{38}\right)}{\sqrt{100000\left(\frac{1}{38}\right)}\left(\frac{37}{38}\right)}\right]$

$=P\left[Z>\frac{145\times 9211}{\sqrt{2562\times 3269}}\right]$

$=P[Z>2\times 8827]$

$=1-P[Z<2\times 8827]$

$=1-\mathrm{\Phi }(2\cdot 8827)$

$=1-0\times 9980$

$=0\times 0020$

Step 9: Final answer (Part c)

The probability that you are winning after $100,000$ bets $0\times 0020$.