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Bayes' formula probability problem

  1. Dec 3, 2008 #1
    1. The problem statement, all variables and given/known data
    A stack of cards consists of six red and five blue cards. A second stack of cards consists of nine red cards. A stack is selected at random and three of its cards are drawn. If all of them are red, what is the probability that the first stack was selected?

    2. Relevant equations
    let X be the event of drawing three red cards, A be the first deck and B the second.
    then P(A) = P(B) = .5
    P(X|A) = (6/11 * 5/10 * 4/9) = .121212
    P(X|B) = 1

    3. The attempt at a solution
    P(X|A)P(A) / [P(X|A)P(A) + P(X|B)(B)] = (.1212 * .5) / [(.1212 * .5) + (1 * .5] = .195

    I think I did this right. Can anyone confirm or give me a hint as to what might be wrong? Intuitively the 1/5 answer bothers me since it seems to me that the chance of choosing one of the two decks seems like it would still be 1/2 since drawing three red cards does not imply one deck or the other, but the chapter is on Bayes' Formula and not on independence.
  2. jcsd
  3. Dec 3, 2008 #2


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    You have a 1/2 chance of picking each deck, but the information that you're given clearly alters the probability (that's the whole point of Bayes' rule... to account for information like this). What if I told you that you drew eight cards and all were red? Would you continue to stick to your "but each deck has a 1/2 chance of being picked anyway" argument anyway?
  4. Dec 3, 2008 #3
    If information was given that excludes the possibility of choosing one of the decks, such as your example of > 6 red cards or > 0 blue cards, then that would obviously rule out one of the two decks. But if the information given does not rule out one of the two decks in this way, then it seems like it does not really affect the 50/50 outcome. I understand what Bayes' rule is doing, but I just can not quite fully accept why it works in this way. I guess I just need to turn it around in my head a bit more.

    Aside from that, does my solution look good?
  5. Dec 3, 2008 #4


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    It looks fine. Ok, consider we take this to the extreme.

    You have two decks of 52 cards.... one of them has 100,000 red cards (the "red" deck), one of them 1 red card and 99,999 blue cards (the "blue" deck). You pick a deck at random, and look at the top card. It's red. Which deck did you pick? Since we're just looking at intuition, let's run a thought experiment to find out. You pick a deck at random one million times. Each time if the top card is blue, you don't count the result (since we need to have the top card red). If the top card is red, we see which deck we got. Assuming the experiment is ideal, we'll get something like:

    500,000 times the blue deck is chosen
    500,000 times the red deck is chosen

    So of instances in which the top card is red (there will be 500,005 times this occurs, every time we picked the red deck, and one out of every 100,000 times we picked the blue deck):

    5 times the blue deck is chosen
    500,000 times the red deck is chosen.

    Now, if you want to take even odds on the deck being the blue deck the next time the top card is red, I'll take that bet.

    You can see how being the odds of getting the blue deck have decreased, since we thought it was a 50/50 chance when we believed the top card could still be blue, but the extra information here has changed the probability as it's thrown away all the times the blue deck came up with a blue card on top
  6. Dec 3, 2008 #5
    it still all seems a little backward to me, but I worked some of the non-assigned problems and my understanding of how the formula really works is definitely growing. By backwards, I mean I am having a hard time wrapping my head around thinking that if something has already happened, you can then figure out the probabilities of things that had to happen to get the result. The idea that it has already happened keeps getting in the way. Thanks for the excellent discussion!
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