Question 1197245
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X = amount won (before the cost is factored in) 
X either equals 58 or 0


H = heads
T = tails
Sample space = {HH, HT, TH, TT} = set of all possible outcomes
Event space = {HH} = subset of the sample space in which the player wins
If the player gets the result in the event space, then they win.


P(winning) = (1 item in the event space)/(4 items in sample space)
P(winning) = 1/4 = 0.25
P(losing) = 1-P(winning) = 1-0.25 = 0.75<table border = "1" cellpadding = "5"><tr><td>Event</td><td>X</td><td>P(X)</td><td>X*P(X)</td></tr><tr><td>Win</td><td>58</td><td>0.25</td><td>14.5</td></tr><tr><td>Lose</td><td>0</td><td>0.75</td><td>0</td></tr></table>E[X] = expected value
E[X] = sum of the X*P(X) values
E[X] = 14.5 + 0
E[X] = 14.5


The expected winnings is 14.5 cents per game.
This is before the cost is factored in.


To have a fair game, the cost per game must be the same as the amount expected to win per game.


The fair price per game is 14.5 cents
For 2 games, that cost will be 2*14.5 = 29 cents.


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Another approach


c = cost to play one game
X = net winnings
In this scenario, X is the amount of money the player walks away with after the cost is factored in.


The player walks away with 58-c cents if they won, or they walk away with 0-c = -c cents if they lose (i.e. they lose c cents).<table border = "1" cellpadding = "5"><tr><td>Event</td><td>X</td><td>P(X)</td><td>X*P(X)</td></tr><tr><td>Win</td><td>58-c</td><td>0.25</td><td>(58-c)*0.25 = 0.25*58-0.25c= 14.5-0.25c</td></tr><tr><td>Lose</td><td>-c</td><td>0.75</td><td>-c*0.75 = -0.75*c</td></tr></table>
E[X] = sum of the X*P(X) values
E[X] = ( 14.5-0.25c ) + ( -0.75c )
E[X] = 14.5 - c


To have a fair game, the expected net winnings must be 0
E[X] = 0
14.5 - c = 0
14.5 = c
c = 14.5
It should cost 14.5 cents to play the game once, if we wanted a fair game.


Therefore, 2 games should cost 2*14.5 = 29 cents.
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