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There is a 52 card deck and you are dealt 13 cards.

Question: What is the probability neither of the following cases happen: no spades and no card higher than 7

Thus that means you get at least one spade and the card is higher than 7. I'm not sure if it's and or or with that wording though.

I understand that the $P(\text{at least one spade}) = 1 - P(\text{no spades})$.

$$P(\text{no spades}) = \frac{\binom{39}{13}}{\binom{52}{13}}$$

But how do I factor in the second requirement of the card being higher than a 7?

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  • $\begingroup$ You only have $6$ cards higher than $7$ that are spades. You have $8,9,10,jack,queen,king$. $\endgroup$ – Numbermind Dec 6 '17 at 17:34
  • $\begingroup$ @Numbermind I don't think the high card had to be a spade. $\endgroup$ – Arthur Dec 6 '17 at 17:39
  • $\begingroup$ @Numbermind So I can look at it as 1 - P(No spades and card above 7)? I just wasn't sure if I could add that second part in since I made it 1 - P(no spades)? So would it be 1 - (28 choose 13)? Because there are 6 cards lower than 7, and 4 suits of each so 6*4 = 24. Then (52-24 choose 13)? $\endgroup$ – James Mitchell Dec 6 '17 at 17:40
  • $\begingroup$ @RobertZ the actual question is what is the probability neither of the following cases happen: no spades and no card higher than 7 $\endgroup$ – James Mitchell Dec 6 '17 at 17:42
  • $\begingroup$ So... everyone here assumes an ace is lower than a seven? It doesn't change the principal, but questions like this should clarify that, as games where an ace is considered lower than a two are exceptionally rare compared to games where an ace is higher than a king. $\endgroup$ – fleablood Dec 6 '17 at 17:57
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No cards greater than 7 would include 2,3,4,5,6 and 7. There are 4 suits Thus (6*4) = 24. Subtract 6 spades because it is no spades too. Thus you have 24-6 = 18 cards.These are the outcomes that should not happen and the complement is the required probabiity. Thus the required probability =$1-\dfrac{{18\choose 13}}{{52\choose 13}}$

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  • $\begingroup$ I don't want to remove spades from the sets, because I need to choose at least one spade $\endgroup$ – James Mitchell Dec 6 '17 at 18:32
  • $\begingroup$ No, you don't. You need to choose one spade OR one card higher than 7. This tells you the prob of doing NEITHER. So the prob of doing at least $1$ is $1 - this$. $\endgroup$ – fleablood Dec 6 '17 at 18:39
  • $\begingroup$ Neither is the catch $\endgroup$ – Satish Ramanathan Dec 6 '17 at 18:40
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I think it'd be easier to not separate into two events.

There are $18$ "good" cards, the $2-7$ of diamonds, clubs, and hearts and $32$ bad cards, everything else.

So there are $18 \choose 13$ positive outcomes for this to happens and $52 \choose 13$ total outcomes.

So $P$ is $\frac {18 \choose 13}{52 \choose 13}=\frac {18!39!}{52!5!} $

It's probably straightforward to combine the probabilities of no spades, and lower than $7$ but.... why bother?

So $1 - \frac {18 \choose 13}{52 \choose 13}$

[This assumes an ace is higher than a king and a two is the lowest rank.]

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If you want to consider the events as two events (but I strongly urge you don't)

P(nothing higher than seven and no spaces)= P(nothing higher than seven| no spades)P(no spades) = P(no spades|nothing higher than seven)P(nothing higher than seven)

P(nothing higher than seven| no spades)=$\frac{18\choose 13}{39\choose 13}$

P(no spades) = $\frac {39\choose 13}{52 \choose 13}$

P(no spades|nothing higher than seven)=$\frac{18\choose 13}{24 \choose 13}$

P(nothing higher than seven) = $\frac {24\choose 13}{52\choose 13}$

Clearly these three different ways will all give the same result.

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