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Card game probability

  1. Jan 18, 2005 #1
    I have a situation that seems easy, I just keep confusing myself. Say theres a card game in which the first step is for the player to choose a suite. After the suite is chosen, they draw a card. If the card matches their suite, they win and the game is over. If the card does not match the suite, the game continues and the player chooses another card without replacing the first card. If this second card matches the suite, they win. If the second card has not matched the suite either, the game is over and considered a loss. What is the probability of winning? Would it simply be the probability of the first card's success probability plus the second card's success probability: (1/4)+(13/51)? Or would the winning probability be the probability of success on the first card plus the quotient of the probability of failure on the first card, times the probability of success on the second card: (1/4) + (3/4)(13/51)?
  2. jcsd
  3. Jan 19, 2005 #2


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    Very simple, since you are not replacing the cards that are drawn, the deck is fat with the cards that have not been drawn. The more cards drawn of a given suit, the less probable that suit will be the next drawn. Crunch the numbers and you will find by the time you get to the 52 card in the deck [assuming a poker deck], the odds are %100 it will be the suit from which only 12 cards have been drawn. The odds are strictly determined by how many of each suit remain in the deck. The suit of the last card drawn is irrelevant.
  4. Jan 19, 2005 #3


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    Let's take a quick look at an extreme case. What happens if you have a deck with two cards, an ace of hearts, and an ace of spades.
    Now, your chance of winning should be 1 if you pick hearts or spades, right?
    But, if you take (1/2) + (1/1) = 3/2 you get a number bigger than one, so that's incorrect.
    Using the 'quotient' version will give you the correct answer.
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