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A standard deck of cards has 52 total cards divided evenly into 4 suits — there are 13 clubs, 13 diamonds, 13 hearts, and 13 spades. Clubs and spades are black cards, while diamonds and hearts are red cards. Gustavo and Katya are trying to decide who will get to choose where they go for dinner. They decide to draw cards without replocement until they get a red card. If the first card drawn is red, then Gustavo gets to choose. If a red card comes on the second drawn card or later, then Katya gets to choose. Is this a fair way to decide who goes first? Why or why not?
A. No, there is a higher probability that Gustavo gets to choose dinner.
B. No, there is a higher probability that Katya gets to choose dinner.
C. Yes, they both have an equal probability of choosing dinner.

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Answer:

B. No, there is a higher probability that Katya gets to choose dinner.

Reason is given below

Step-by-step explanation:

  • Let R represent the event that a red card is drawn and B represent the event that a Black card is drawn
  • Since there are 26 red cards in a deck of 52 cards, P(R) = 26/52
  • If the first card drawn is red. Gustavo gets to choose and the game is over
    So P(Gustavo gets to choose) = P(R)=26/52

  • P(second card being drawn is red given that the first card drawn is black) is the conditional probability P(R|B)
  • If the first card drawn is black, that will leave 26 red cards and 25 black cards out of a total of 51 cards
  • So P(R|B) = 26/51 which is the probability that Katya will get to choose
  • If the first two cards are black then P(R|BB) = 26/50
  • 26/50 > 26/51 > 26/52 since the lower the denominator the higher the fraction for the same numerator

  • Therefore the probability of selecting a red card on the second or successive draws keeps increasing
  • Therefore the probability that Katya gets to choose is higher than the probability that Gustavo gets to choose

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