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Consider the spinner above. If a prime number comes up, you win that number of dollars. If a nonprime number comes up, you lose that number of dollars. If you play the game 1000 times, what would be your expected winnings?

 Jun 4, 2024
 #1
avatar+1654 
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Analyzing the game setup, we can see the following:

 

Landing on a prime number (2 or 3) results in winning that number of dollars.

 

Landing on a non-prime number (4) results in losing that number of dollars.

 

We are given the arc size information (120 degrees for prime numbers, 60 degrees for non-prime), but that doesn't directly affect the expected winnings calculation as long as the landing probability on each section is proportional to

its arc size.

 

Here's how to calculate the expected winnings:

 

Probability of landing on a prime number:

 

There are two prime numbers (2 and 3) and their combined arc span is 120 + 120 = 240 degrees.

 

Since the total degrees is 360 (full circle), the probability of landing on a prime number is (240 degrees / 360 degrees) = 2/3.

 

Expected value for landing on a prime:

 

Winning amount for 2 is $2.

 

Winning amount for 3 is $3.

 

Expected value when landing on a prime = (2/3) * ($2 + $3) = $4/3

 

Expected value for landing on a non-prime:

 

Losing amount for 4 is -$4 (since it's a non-prime).

 

We can't directly calculate the expected value for non-primes as there's only one number (4) mentioned. There could be other non-prime numbers with different values.

 

A common approach is to assume an average loss for non-primes. Let's say the average loss for non-primes is also -$4.

 

Probability of landing on a non-prime:

 

Probability of non-prime = 1 - (probability of prime) = 1 - (2/3) = 1/3

 

Overall expected winnings after 1000 games:

 

Expected winnings = (# of games * (probability of prime * expected value for prime)) + (# of games * (probability of non-prime * expected value for non-prime))

 

Expected winnings = (1000 games * (2/3 * ($4/3))) + (1000 games * (1/3 * -$4))

 

Expected winnings = ($1000 * -$4/3) = -$1333.33 (approximately)

 

You expect to take a loss of 1333.33

 Jun 4, 2024
 #2
avatar+806 
+1

We cam first find the probability we get each number. We have degree numbers, so let's use that to help us. 

Since 1 has a degree measure of 120 degrees, the probability we get 1 is \(120/360 = 1/3\)

Since 2 has a degree measure of 60 degrees, the probability we get 2 is \(60/360=1/6\)

3 has the same degree measure as 1, so we have \(1/3\)

4 has the same degree measure as 2, so we have \(1/6\)

 

This means after 6 spins, we have an expected value that

- two 1s

- two 3s

- one 4

- one 2

 

\(3+3-1-1-1+2 = 5\)

 

Now, 1000/6 = 166.67, so let's first do

\(166*5=830\)

 

We have 4 throws remaining. \(2/3 * 5 = 10/3\)

 

\(833.33 \approx 833\)

 

Not sure if this is correct, but it should be close.

 

Thanks! :)

 Jun 4, 2024

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