A card game is played as follows: There are 5 players and a dealer, each with 5
ID: 3196069 • Letter: A
Question
A card game is played as follows: There are 5 players and a dealer, each with 5 cards in hand numbered 1-5. Each person shuffles his or her deck. The dealer turns over a card. Then, the first player turns over the top card in his or her deck. If the number on Player 1’s card matches the number on the dealer’s card, Player 1 wins. If not, the second player turns over the top card in his or her deck. If the number matches either of the cards showing, the second player wins. If not, the third player turns over the top card in his or her deck. If it matches any of the previous cards played, player 3 wins. If it does not match either, the fourth player turns over a card. If it matches any of the previous cards, player 4 wins. If not, player 5 turns over a card. Necessarily, player 5’s card will match some card on the table, and Player 5 wins.
(b) Suppose that it costs $4 to play. If someone wins before its your turn, you get $2 back. If someone wins after your turn, you lose all $4. If you win, you get your initial $4 back and you get an additional $3. Compute the expected value for each player. How much money will each player win/lose if the game is played 100 times?
Explanation / Answer
Player 1 wins:
For 100 games he has to pay $4 each, total cost to play 100 games = $400
If Player 1 wins x times, then he will get $(4+3)*x = $7x back
Gain = $3x
He will loss (100-x) games for that he has to lose $(100-x)*4
Player 2 wins:
He pays $400 to play, if he wins for y times then he will in-return will get $7y and gain will be $3y. He will lose (100-y) games. He will lose $(2x+(((100-y)-x)*4), for inreturn he will get when he lose before he plays = $2x
Total return = $(4y+2x)
Player 3 wins:
He pays $400 to play, If he wins for z times then he will in-return will get $7z and gain will be $3z. He will lose (100-z) games. He will lose $(2x+2y+(((100-z)-(x+y))*4)), for in return he will get when he lose before he plays $(2x+2y)
Total return = $(4z+2x+2y)
Player 4 wins:
He pays $400 to play, if he wins for a times then he will in return will get $7a and gain will be $3a. He will lose (100-a) games. He will lose $(2x+2y+2z+(((100-a)-(x+y+z))*4)), for in return he will get when he lose before he plays $(2x+2y+2z)..... Total return = $(4a+2x+2y+2z).
Player 5 wins:
He pays $400 to play, if he wins for b times then he will inreturn will get $7b and gain will be $3b. He will lose (100-b) games. He will lose $(2x+2y+2z+2a+(((100-b)-(x+y+z+a))*4)), for in return he will get when he lose before he plays $(2x+2y+2z+2a) ..... Total return = $(4b+2x+2y+2z+2a)
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