Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Probability of Card Draw

  1. Nov 5, 2012 #1
    With a regular deck of 52 cards, how would you find the probability of getting the Ace of Spaces if you took turns until someone got it? This is without reshuffling.

    If it was a 1/52 chance each time, I would be able to do it like
    (1/52) * (inf. sum ((51/52)^2)^n)

    But this would have a limit of 52 trials, and I never learned how to do that. Also, the probability of getting it would change for every card drawn. I also know that it would be different depending on who draws the first card.

    Could I get any help?
    Thanks
     
  2. jcsd
  3. Nov 5, 2012 #2
    This is an interesting problem. It turns out that all probabilities are what you expect. For example, if you and a friend took turns finding the ace of spades, both of you have exactly a 50% chance of finding it first. Let's look at this in a little more detail:

    First let's say there are 52 people in a line waiting to draw the ace of spades. They each have a 1/52 chance of doing so. The first person has a probablility of (1/52) of drawing the ace of spades. The second person now has a (1/51) chance of drawing an ace of spades from the appended deck, but he also has a (1/52) chance of never seeing the deck if the first person draws the ace of spades. So his chances of drawing the ace of spades are (51/52)(1/51)=(1/52). You can carry out this progression to show similar results for the rest of the line.

    Now let's say it's you and a friend taking turns. If every trial has a (1/52) chance of being successful, the probability is evenly split and it is a fair game between you and your friend.

    The only way that this game is not fair is if you have n people playing and 52 (mod n) ≠ 0. This logically makes sense because some players would have more possible turns than others.
     
  4. Nov 6, 2012 #3

    tiny-tim

    User Avatar
    Science Advisor
    Homework Helper

    Hi IceXaos! :smile:
    Simplest way:

    Number the cards 1 to 52.

    You choose 26, your opponent chooses 26.

    It doesn't matter which order you choose them (you could choose all your 26 cards first, for example) …

    if the Ace of Spades is in your half, you win, if it isn't, you don't! :wink:
     
  5. Nov 6, 2012 #4
    I didn't even attempt the simple way of looking at it. That's kind'a unexpected.

    Thanks guys.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook