Question

Consider the following card game with a well-shuffled deck of cards. If you draw a red card, you win nothing. If you get a spade, you win $5. For any club, you win $12 plus an extra $15 for the ace of clubs. Let Xi denote the possible winnings in this scenario

How much should one pay to play so this game breaks even?

Answer 1

X_i = 0 ; If a red card is drawn ; Probability of drawing a red card = Number of red cards / Total number of cards = 26/52

X_i = $5 ; If a spade card is drawn ; Probability of drawing a spade card = Number of spade cards / Total number of cards = 13/52

if a club is drawn you get $12 for club card and extra $15 for ace of clubs therefore if club of ace is drawn X_i = 15+12= 27 ;

i.e

X_i = 27 ; If a club ace is drawn ; Probability of drawing a club ace = Number of club ace cards / Total number of cards = 1/52

X_i = $12 ; If a Club card is drawn but not a club ace ; Probability of drawing a club card but not a club ace = Number of club cards and not a ace / Total number of cards = 12/52

Therefore following is the probability distribution of X_i

Xi	P(Xi)
0	26/52
5	13/52
12	12/52
27	1/52

For this game to break even ; one play Expected value of the game i.e E(X)

Therefore ;

For one should pay $4.54 to play for this game to break even