Question 1205333
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Let's say one single person buys all 3500 tickets.


They would spend 3500*2 = 7000 dollars in total.
They would have 100% probability of getting the prize of $4800. 
But they walk away with a net loss of 4800 - 7000 = -2200 dollars.


Divide this net loss over the number of tickets.
-2200/3500 = -0.62857 approximately
This rounds to <font color=red>-0.63</font> which is the approximate expected value.


The person should expect to lose, on average, around 63 cents per ticket.


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Another approach is to set up a table as shown below
<table border=1 cellpadding=5><thead><tr><th>Event</th><th>X</th><th>P(X)</th><th>X*P(X)</th></tr></thead><tbody><tr><td>Win</td><td>4798</td><td>1/3500</td><td>4798/3500</td></tr><tr><td>Lose</td><td>-2</td><td>3499/3500</td><td>-6998/3500</td></tr></tbody></table>
where X = net earnings = amount you walk away with
Note that 4800 - 2 = 4798
P(X) = probability of getting those earnings


The X*P(X) column is the result of multiplying each X and P(X) value together. 


Add up those X*P(X) values:
4798/3500 + (-6998/3500)
4798/3500 - 6998/3500
(4798 - 6998)/3500
-2200/3500
-0.62857 approximately
That leads to <font color=red>-0.63</font> when rounding to the nearest cent.


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Answer: <font color=red>-0.63 dollars</font>
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