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?

  • $\begingroup$ 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

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 =$1-\dfrac{{18\choose 13}}{{52\choose 13}}$

  • $\begingroup$ 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

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.]


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.


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.