Question 1200014
<font color=black size=3>
Problem 1: <font color=blue>A simple gambling game involves the throw of a die. 
Players are charged 40 cents if the die shows 1,2 or 3 they are charged nothing for 4 and they receive 30 cents or 60 cents respectively for 5 or 6. 
Let the random variable X be the payout to the player.
Calculate the expected value and how much profit would the casino expect to make on 100 games?</font>



X = payout, in cents, to the player
X can take on the following values: {-40,0,30,60}
The negative value is where the player loses money.


The probability of rolling a 1, 2 or 3 is 3/6.
This is the probability of X = -40 happening.


The probability of rolling a 4 is 1/6 because there's one side labeled "4" out of 6 sides total.
This is when X = 0


The probability of rolling a 5 is also 1/6.
This is the probability of X = 30 happening.


The probability of rolling a 6 is also 1/6.
This is the probability of X = 60 happening.


Let's set up a table with each X and P(X) value.
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>P(X)</td></tr>
<tr><td>-40</td><td>3/6</td></tr>
<tr><td>0</td><td>1/6</td></tr>
<tr><td>30</td><td>1/6</td></tr>
<tr><td>60</td><td>1/6</td></tr>
</table>


Now multiply across the rows to form the X*P(X) column.
Example: -40 times 3/6 = -120/6
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>P(X)</td><td>X*P(X)</td></tr>
<tr><td>-40</td><td>3/6</td><td>-120/6</td></tr>
<tr><td>0</td><td>1/6</td><td>0/6</td></tr>
<tr><td>30</td><td>1/6</td><td>30/6</td></tr>
<tr><td>60</td><td>1/6</td><td>60/6</td></tr>
</table>


Add the items in that third column to get the expected payout.
(-120/6)+(0/6)+(30/6)+(60/6)
= (-120+0+30+60)/6
= -30/6
= -5


A negative payout means the player lost money.
The player expects to lose, on average, 5 cents per game.
From the perspective of the casino, they are winning 5 cents per game on average.


Over the course of 100 games, the casino would expect to gain 100*5 = 500 cents = $5
Or the calculation would be 100*(0.05 dollars) = 5 dollars.



Answers:
Expected payout per game = -5 cents (perspective of the player)
Casino profit on one hundred games = $5



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Problem 2: <font color=blue>If the backer weighted the die so that it is 50% more likely to turn up each of the results 1,2,3 as to turn up 4,5 or 6.
What is the probability now of each outcome?
Find expected value now.</font>


50% more likely involves the multiplier 1.5, which converts to 3/2.


Rolling a 1,2, or 3 initially had probability 3/6 aka 1/2.


Increase that by 50% to get (1.5)*(1/2) = (3/2)*(1/2) = 3/4


Therefore, event X = -40 has probability P(X) = 3/4


On this weighted die, rolling a 1,2, or 3 has probability 3/4. 
That means 1-3/4 = 1/4 is the leftover amount for the other values


Split that remaining bit into three equal pieces: (1/4)*(1/3) = 1/12
Rolling a 4 has probability 1/12
Rolling a 5 has probability 1/12
Rolling a 6 has probability 1/12


The event X = 0 has probability 1/12
The event X = 30 has probability 1/12
The event X = 60 has probability 1/12
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>P(X)</td></tr>
<tr><td>-40</td><td>3/4</td></tr>
<tr><td>0</td><td>1/12</td></tr>
<tr><td>30</td><td>1/12</td></tr>
<tr><td>60</td><td>1/12</td></tr>
</table>


For the sake of consistency, let's get each denominator to 12.
3/4 = (3/4)*(3/3) = 9/12
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>P(X)</td></tr>
<tr><td>-40</td><td>9/12</td></tr>
<tr><td>0</td><td>1/12</td></tr>
<tr><td>30</td><td>1/12</td></tr>
<tr><td>60</td><td>1/12</td></tr>
</table>


Now form the X*P(X) column
<table border = "1" cellpadding = "5">
<tr><td>X</td><td>P(X)</td><td>X*P(X)</td></tr>
<tr><td>-40</td><td>9/12</td><td>-360/12</td></tr>
<tr><td>0</td><td>1/12</td><td>0/12</td></tr>
<tr><td>30</td><td>1/12</td><td>30/12</td></tr>
<tr><td>60</td><td>1/12</td><td>60/12</td></tr>
</table>


Add the X*P(X) items
(-360/12)+(0/12)+(30/12)+(60/12)
= (-360+0+30+60)/12
= -270/12
= -22.50


The player expects to lose, on average, 22.50 cents per game.
The casino expects to gain, on average, 22.50 cents per game.


Playing 100 games means the casino expects to make 0.225*100 = 22.50 dollars




Answers:
The probability of each outcome is shown in the table above
Expected payout for the player = -22.5 cents
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