James placed a $10 bet on red and a $5 bet on the number 33 (which is black) on

Question

James placed a $10 bet on red and a $5 bet on the number 33 (which is black) on a standard 00 roulette wheel.

If the ball lands in a red space, he wins $10 on his ‘red’ bet but loses $5 on his ‘33’ bet- so he wins $5.

If the ball lands on the number 33, he loses $10 on his ‘red’ bet but wins $175 on his ‘33’ bet: He wins $165.

If the ball lands on a space that isn’t red and isn’t 33 he loses both bets, so he loses $15.

So for each spin; he either wins $165, wins $5, or loses $15.

The probability that he wins $165 is 1/38 or .0263

The probability that he wins $5 is 18/38 or .47374

The probability that he loses $15 is 19/38 or .5000

Let X = the profit that James makes on the next spin.

x

P(X=x)

x*P(X=x)

x2*P(X=x)

165

.0263

4.3395

716.0175

    5

.4737

2.3685

  11.8425

- 15

.5000

-7.5000

112.5000

1.0000

-0.792

840.36

  

The central limit theorem tells me that these winnings/losses will be normally distributed.

1) Using this information, we can calculate the probability of making money (winnings > 0) when playing 10,000 to be … _______ SHOW YOUR WORK OR CALCULATOR INPUT

2) The probability of having a profit over $1000 (win an average of $0.10 per game for 10,000 games) is _______

SHOW YOUR WORK OR CALCULATOR INPUT

3)

This means that we’d only expect one of every __________________ people repeating this bet 10,000 times to profit,

and we’d only expect one out of every __________________ to make over $1000.

4)

The probability of losing over $5000 is ______________ and the probability of losing over $8000 is ______________ .

SHOW YOUR WORK OR CALCULATOR INPUT

x

P(X=x)

x*P(X=x)

x2*P(X=x)

165

.0263

4.3395

716.0175

    5

.4737

2.3685

  11.8425

- 15

.5000

-7.5000

112.5000

1.0000

-0.792

840.36

Explanation / Answer

1. Mean winning for 10000 games = -0.792*10000 = -$7920

Standard deviation = $289780

So, probability of having winnings > 0 = P(winning > 0 )

= P((winning - (-7920))/289780 > 7920/289780) = P(z> 0.027) = 0.489

2. P(profit>1000) = P((profit - (-7920)/289780 > (1000+7920)/289780) = P(z>0.0308) = 0.488

3. So only one out of 1/0.488 = 2.05 people would repeat this bet

4. P(winning < -5000) = P((winning + 7920)/289780 < (-5000+7920)/289780) = P(z<0.01) = 0.496

P(winning < -8000) = P((winning + 7920)/289780 < (-8000+7920)/289780) = P(z<-0.00027) = 0.5001

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