Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What's the probability of drawing a card of a standard poker card deck, if 10 cards are removed and opponent have drawn 5 cards?

share|cite|improve this question
Don't you mean a specific card? The probability of drawing one of 37 cards is 100%, but the probability of drawing one of those is $\displaystyle \frac{1}{37}$. – El'endia Starman May 27 '11 at 20:23
up vote 6 down vote accepted

If $10$ cards were removed and someone else has already drawn $5$, there are $37$ cards left. If you are asking "What is the probability of drawing a particular card which I know is still remaining in the deck?" the answer is $1/37$. If you are asking "What is the probability of drawing a particular card, given that I don't know whether it has been drawn yet?" the answer is $1/52$, as just knowing that cards have been removed doesn't give you any usable information. You can think about it more clearly as "What is the probability that the 16th card in the deck is something specific?" That does not depend at all on what the first 15 cards are, unless you actually know what the first 15 cards are.

If the question is, "If I am playing poker, what is the probability that I will draw a card" the answer is $0$ or $1$, depending on whether you have anted up or not.

share|cite|improve this answer
The moral (for a poker player) is that if you need to draw one of 9 remaining spades to make your flush from a deck with 42 cards remaining (5 in your hand currently and 5 in your opponent's), you should figure your odds of success to be 9/47, NOT 9/42. The cards your opponent holds could be anything (maybe spades), but in the absence of information, you should just regard them as still being in the deck. – Austin Mohr May 27 '11 at 22:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.