What is the probability that a random bridge hand contains cards of exactly two suits?

My Attempt At A Solution Bridge hands consist of $13$ cards, and a suit contains $52$ cards, so the way to pick a random bridge hand would be $$\frac{{26\choose 13}-2}{{52 \choose 13}} $$

As user @Lord Shark the Unknown hinted: there are $26\choose 13$ ways to choose hands of two suits but one of those suits is only say hearts, and another only spades, so we must compensate for those.

Thank you for any corrections/hints.

  • 3
    $\begingroup$ There are $\binom{26}{13}$ hands only having spades and hearts. But one of these is all spades, and another all hearts. $\endgroup$ – Lord Shark the Unknown Oct 7 '18 at 17:29
  • $\begingroup$ would it be correct to say $$ \frac{{26 \choose 13}-2}{{52 \choose 13}}$$ $\endgroup$ – elcharlosmaster Oct 7 '18 at 17:31
  • $\begingroup$ Btw, unrelatedly, providing no previous questions are deleted, this is currently the 1,000,000th question on MSE! $\endgroup$ – Sam T Oct 7 '18 at 21:16

The total number of hands is $$\binom{52}{13}$$

Now we count the number of hands with cards from exactly two suits.

There are $$\binom{4}{2}$$ ways to choose the two suits.

Given this, there are $$\binom{26}{13}$$ ways to choose the cards in the hand.

However, we've counted four hands that we shouldn't count: the four hands that have all cards of the same suit (e.g. all $13$ spades).

For each of these $4$ hands, we've counted them $3$ times. For example, $13$ spades and $0$ hearts is counted, but so is $13$ spades and $0$ clubs, and $13$ spades and $0$ diamonds. The number of hands we've overcounted is therefore$$4\cdot 3$$

This means that the total number of hands with cards from exactly two suits is

$$\binom{4}{2}\binom{26}{13} - 4\cdot 3$$

The probability you want is the number of successful possibilities divided by the number of total possibilities:

$$\boxed{\,\displaystyle\frac{\displaystyle\binom{4}{2}\displaystyle\binom{26}{13} - 4\cdot 3}{\displaystyle\binom{52}{13}}\,\,}$$

  • 4
    $\begingroup$ +1 for this. But slightly simpler is to say that for each choice of two suits, there are $\binom{26}{13}-2$ ways to choose the cards. This gives you the same answer. $\endgroup$ – TonyK Oct 7 '18 at 17:41
  • 2
    $\begingroup$ Good point, thanks $\endgroup$ – Zubin Mukerjee Oct 7 '18 at 17:42

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