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Compute the probability that a bridge hand is void in at least one suit.

First I will select the suit I want to exclude from the $13$ cards, which is done in $4\choose 1 $ $=4$ ways, then I will choose the $13$ cards from the remaining $39$ cards.

So the probability should be equal to

$4\cdot$$\frac{39 \choose 13 }{52\choose 13} $

Is this correct?

N. F. Taussig
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So Lo
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    No, you are over counting. For instance, you count the hand with 13 spades three times (once as a void in hearts, again as a void in diamonds, and a third time as a void in clubs). – lulu Mar 21 '19 at 13:17
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    Note; of course your answer ought to be a decent approximation...hands with multiple voids being quite rare. – lulu Mar 21 '19 at 13:19
  • Possible duplicate of A bridge hand void in one suit – lulu Mar 21 '19 at 13:19
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    @lulu I am trying to understand your reasoning, please tell me if I got it right. According to my method, you are saying that suppose at first if I take my excluded suit to be the Diamonds, then this will also contain hands that are void in Spades, which will result in over-count when I will take my excluded suit to be the Spades? – So Lo Mar 21 '19 at 13:26
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    Exactly. Phrased (slightly) differently: look at the hand with $13$ spades. That hand is void in hearts, so you count it when you count the hands which are void in hearts. But then you also count it when you count the hands which are void in diamonds, or in clubs. If you want an exact answer you have to correct for that. As I say, your answer ought to be a very strong approximation though. – lulu Mar 21 '19 at 13:33
  • Note: the other question I linked to is not an exact duplicate. In that question, we are asked for the probability that a hand has a void in exactly one suit. But the analysis is extremely similar. – lulu Mar 21 '19 at 13:34

1 Answers1

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Let $E_1, \;E_2,\; E_3, \;E_4$ denote the events that your hand is void in $\clubsuit$, $\diamondsuit$, $\spadesuit$, $\heartsuit$ respectively.

We are looking for the probability P $= P(\cup_{i=1}^{4} E_i)$

By the inclusion-exclusion identity, we have

P = $p_1-p_2+p_3-p_4$ where $p_i$ denotes the probability that your hand is void in $i$ number of suits and contains $4\choose i$ terms each with the probability

$$\frac{\binom{52-13\cdot i} {13}}{\binom{52} {13}}$$ except for the case for $p_4$ since $p_4=0$ is evident as the hand cannot be void of all the $4$ suits.

The required probability P will then be $$4\cdot\frac{\binom{39}{13}}{\binom{52}{13}} - 6\cdot\frac{\binom{26}{13}}{\binom{52}{13}} + 4\cdot\frac{\binom{13}{13}}{\binom{52}{13}}$$

zhoraster
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vishu
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