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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.

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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 $\dfrac6{15}$

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