 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Two Packs Of Cards (Posted on 2004-02-05) Two identical pack of cards A and B are shuffled throughly. One card is picked from A and shuffled with B. The top card from pack A is turned up.
If this is the Queen of Hearts, what are the chances that the top card in B will be the King of Hearts?

 See The Solution Submitted by Ravi Raja Rating: 3.8571 (7 votes) Comments: ( Back to comment list | You must be logged in to post comments.) re: Agreement with nobody | Comment 12 of 14 | (In reply to Agreement with nobody by Juggler)

Comments on several things:
"the chances of seeing 2 king of Hearts in pack B is increased by 1/52 to (1/52)+1"

The probability of this was given as 1/51, and you are actually changing it to 1/52. In doing so, you are agreeing with stan's original post. The figure (1/52)+1 can't be a probability as it exceeds 1. It is a numerator as it represents the sum of a couple of conditional probabilities -- there is always at least one king of hearts, and a certain probability of a second.

"or 1/52.00000000000000208"

The difference from 1/52 is illusory, due to rounding errors in the calculation. ((1/52)+1)/53 = (53/52)/53 = 1/52 exactly, as stan had said, so you really agree with stan.

"the chances of picking the king from back A is not 1/51, you select the 1st card from pack A before you know that the Queen of hearts was the top card,"

Having seen the queen of hearts, we know now that the chosen card was not the queen of hearts. If the probability were 1/52 of the card having been the king of hearts, that would be the probability for each of the 51 possibilites, and the total would be only 51/52 of any card having been chosen. But we know for a fact at probability 1 (not 51/52) that some card was chosen.

We need to use Baye's formula for the probability that the king of hearts was chosen, given that the queen of hearts was not chosen:

p(k|~q) = p(k&~q)/p(~q) = p(k)/p(~q), since k and q are mutually exclusive.

Then,
p(k|~q) = p(k)/p(~q) = (1/52)/(51/52) = 1/51

Here, the | within p(k|~q) is read as "given".
 Posted by Charlie on 2004-02-06 14:20:34 Please log in:

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