You and your friend decide to invent a new gambling game. The game involves roll
ID: 3206614 • Letter: Y
Question
You and your friend decide to invent a new gambling game. The game involves rolling a single die (dice), and drawing a card from a standard deck of cards. The die has six sides, three black, and three red. The standard deck of cards contains 52 cards. The cards are evenly divided among -1 different "suits" (hearts, diamonds, clubs, and spades), and among 13 different "types" {2, 3, 4, 5, 6, 7, 8, 9, 10 jack, queen, king, ace). Each card is of one suit, and is of one type. Accordingly, there are 4 cards of each "type" and 13 cards of each "suit". The colors of cards are as follows: hearts and diamonds are red, spades and clubs are black. You win your bet if the color on the die matches the color on the card. You lose your bet if the color on the die does not match the color on the card. If you play the game once, what is the probability that you win? If you played this game 100 times, what is the expected value of your winnings if you bet $1 each time? You decide to add a joker to the deck of cards. A joker has no value, no suit, and no color. If the card drawn is a joker, you lose. What is the probability of drawing a joker? What is the probability that you will win, with the joker added to the deck? If you played this game, with the joker, 100 times, what is the expected value of your winnings if you bet $1 each time? Are you more likely to win with or without the joker added to the deck?Explanation / Answer
probabilty of winning =red on die*red on cards+black on die*black on cards =(1/2)*(26/52)+(1/2)*(26/52)=1/2
expected value of winning =expected value on single game*100=(1*probabilty of winning -1*probability of lose)*100=0
including joker
probability of drawing a joker =1/53
probabilty of winning=(1/2)*(26/53)+(1/2)*(26/53)=26/53
hence expected value =(26/53 -27/53)*100 =$-1.88
we are more likely to win without joker.
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