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I've been struggling with this problem for a while, so I have to ask you guys for a little mental push. My problem involves a full deck of playing cards and 4 players, but I'll simplify it to 4 cards and 2 players with the hopes to be able to apply the logic behind the simple question to the bigger one.

Okay, here goes:

There are the following cards in a deck: (K of Hearts), (10 of Diamonds), (8 of Diamonds), (10 of Hearts) There are two players and each one gets 2 cards.

What is the probability of AT LEAST one of the players to have (at the same time) NO hearts AND at least one card of diamonds?

Can anyone walk me through the logic and calculations and get me to the answer?

Thanks in advance!

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Any player that has no hearts has two diamonds, so you want the chance that the two heart cards are in the same hand. If you deal the first heart to somebody, he gets one of the remaining three cards, so the chance is $\frac 13$. This approach may not help you with the larger problem.

Added: for a slightly larger problem, take a standard deck, a player drawing two cards without replacement and we ask the chance that he gets no hearts and at least one diamond. The first card can be a diamond, probability $\frac 14$, in which case the second card can be any non-heart, probability $\frac {38}{51}$ or the first card can be a club or spade, probability $\frac 12$ and the second a diamond, probability $\frac {13}{51}$. The total is then $\frac 14 \cdot \frac {38}{51}+\frac 24\cdot \frac {13}{51}=\frac {64}{204}=\frac {16}{51}$. It takes care to get all the possibilities, and once each. In this case it would be easy to double count the draws of two diamonds.

Added: to have two players have no hearts each and at least one diamond, you just keep going. The new twist is that the chance of the second getting this is influenced by whether the first has two diamonds or not. So now we figure the chance the first player has two diamonds: $\frac 14 \cdot \frac {12}{51} = \frac 3{51}$ Then the chance the first player has exactly one diamond and one non-heart is $\frac {16}{51}-\frac 3{51}=\frac {13}{51}$ Given two diamonds in the first hand, the second player can draw a diamond then a non-heart with probability $\frac {11}{50}\cdot \frac {36}{49}$ and a club or spade then a diamond with probability $\frac {26}{50}\cdot \frac {11}{49}$. Given only one diamond and no hearts in the first hand, the second player can draw a diamond then a non-heart with probability $\frac {12}{50}\cdot \frac {36}{49}$ and a club or spade then a diamond with probability $\frac {25}{50}\cdot \frac {12}{49}$. So the total is $\frac 3{51}\cdot (\frac {11}{50}\cdot \frac {36}{49}+\frac {26}{50}\cdot \frac {11}{49})+\frac {13}{51}(\frac {12}{50}\cdot \frac {36}{49}+\frac {25}{50}\cdot \frac {12}{49})$. I'll leave it to you to gather all this together and to do the three card case.

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  • $\begingroup$ Sincere thanks for your reply, Ross! While your answer is no doubt correct, you're right - this approach can't help me with the larger problem. What I really need is to see it solved the more general way - by utilizing the combinations formula and having in mind that the given example could be more complex - with more cards and more players. Thanks again! $\endgroup$ – John Smith Jun 24 '13 at 20:37
  • $\begingroup$ Unfortunately, when problems are too small, the combinatorial approach often gets lost. See if my addition helps. $\endgroup$ – Ross Millikan Jun 24 '13 at 20:44
  • $\begingroup$ Thanks for your help, Ross! Now, if you can spare some more time, can you tell me what is the combined chance of any of the two players to have no hearts and have at least one diamond? And what if they could draw 3 cards each? $\endgroup$ – John Smith Jun 25 '13 at 8:08

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