Playing a card game in which 2 52 card decks are combined to create the pile and from this pile each player is dealt 14 cards. One rule of the game is if any player is dealt 3 doubles (a double being a card with same suite and value) a re-shuffle is permitted. However if a player is accidentally dealt 15 cards and is found to have 3 doubles what is the probability that the 15th card completed the three doubles.

  • $\begingroup$ There are two ways of interpreting this, depending on the player's ability to observe their hand as it's being dealt. Either you are asking for: $$P(\small\text{14 cards dealt with two doubles, and a 15th card from the remaining deck creates another double})$$ ...OR: $$P(\small\text{a 15th card from the remaining deck creates another double}\ |\ \small\text{14 cards dealt with two doubles})$$ The | is read as 'given' $\endgroup$ – enthdegree Feb 19 '16 at 21:05
  • 1
    $\begingroup$ @enthdegree I would read it as $P(\small\text{first 14 out of 15 cards dealt with two doubles not three | 15 cards dealt with three doubles})$ $\endgroup$ – Henry Feb 19 '16 at 21:37
  • $\begingroup$ @Henry the second one of mine is the same as yours $\endgroup$ – enthdegree Feb 19 '16 at 21:43

If there are $15$ cards dealt with exactly three doubles

then there are $6$ cards of the $15$ which are in the doubles

so the probability one of these $6$ was the last of the $15$ to be dealt is $\dfrac6{15}$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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