Question 1197108
<font color=black size=3>
1 in 100 = 1/100 = 0.01
1 in 1000 = 1/1000 = 0.001
1 in 50 million = 1/(50 million) = 1/(50*10^6) = 0.00000002


Those probability values add to:
0.01+0.001+0.00000002 = 0.01100002
Subtract that from 1
1-0.01100002 = 0.98899998
This represents the probability of winning $0


X = net winnings = (amount won) - ($2 cost)
P(X) = probability of getting those net winnings<table border = "1" cellpadding = "5"><tr><td>X</td><td>P(X)</td><td>X*P(X)</td></tr><tr><td>7</td><td>0.01</td><td>0.07</td></tr><tr><td>78</td><td>0.001</td><td>0.078</td></tr><tr><td>999998</td><td>0.00000002</td><td>0.01999996</td></tr><tr><td>-2</td><td>0.98899998</td><td>-1.97799996</td></tr></table>Add up the values in the X*P(X) column
0.07 + 0.078 + 0.01999996 + (-1.97799996) = -1.81


The expected earnings for each lottery ticket, on average, is -1.81 dollars.
This means Robert expects to lose on average about $1.81 per ticket.


It doesn't matter how many tickets he buys. He won't earn a profit since the profit per ticket is negative.
</font>