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

A bridge hand void in one suitA bridge hand void in one suitProbability of Bridge hand having at most two suitsProbability of drawing at least one card from each suitProbability that a random 13-card hand contains at least 3 cards of every suit?Probability that hand contains Ace and King of at least one suit?How many bridge hands have exactly two 5-cards suits and a void so that the remaining suit has a run of 3 cards?Is there an alternative intuition for solving the probability of having one ace card in every bridge player's hand?The probability of a hand being void in zero suitsPropability of drawing at least 5 cards of the same suit?Probability that at least one suit is missing in a player's hand

0

$begingroup$

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 $4choose 1 $ $=4$ ways, then I will choose the $13$ cards from the remaining $39$ cards.

So the probability should be equal to

$4cdot$$frac39 choose 13 52choose 13 $

Is this correct?

probability combinatorics combinations

share|cite|improve this question

edited Mar 21 at 18:07

N. F. Taussig

45k103358

asked Mar 21 at 13:15

So LoSo Lo

64429

$endgroup$

• 
  
  
  

  
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  $begingroup$
  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).
  $endgroup$
  – lulu
  Mar 21 at 13:17
  

  
  
  
  


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  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note; of course your answer ought to be a decent approximation...hands with multiple voids being quite rare.
  $endgroup$
  – lulu
  Mar 21 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of A bridge hand void in one suit
  $endgroup$
  – lulu
  Mar 21 at 13:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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?
  $endgroup$
  – So Lo
  Mar 21 at 13:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – lulu
  Mar 21 at 13:33
  

  
  
  
  

 | 
show 1 more comment

0

$begingroup$

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 $4choose 1 $ $=4$ ways, then I will choose the $13$ cards from the remaining $39$ cards.

So the probability should be equal to

$4cdot$$frac39 choose 13 52choose 13 $

Is this correct?

probability combinatorics combinations

share|cite|improve this question

edited Mar 21 at 18:07

N. F. Taussig

45k103358

asked Mar 21 at 13:15

So LoSo Lo

64429

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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).
  $endgroup$
  – lulu
  Mar 21 at 13:17
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Note; of course your answer ought to be a decent approximation...hands with multiple voids being quite rare.
  $endgroup$
  – lulu
  Mar 21 at 13:19
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of A bridge hand void in one suit
  $endgroup$
  – lulu
  Mar 21 at 13:19
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @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?
  $endgroup$
  – So Lo
  Mar 21 at 13:26
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  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.
  $endgroup$
  – lulu
  Mar 21 at 13:33
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

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 $4choose 1 $ $=4$ ways, then I will choose the $13$ cards from the remaining $39$ cards.

So the probability should be equal to

$4cdot$$frac39 choose 13 52choose 13 $

Is this correct?

probability combinatorics combinations

share|cite|improve this question

edited Mar 21 at 18:07

N. F. Taussig

45k103358

asked Mar 21 at 13:15

So LoSo Lo

64429

$endgroup$

share|cite|improve this question

edited Mar 21 at 18:07

N. F. Taussig

45k103358

asked Mar 21 at 13:15

So LoSo Lo

64429

share|cite|improve this question

edited Mar 21 at 18:07

N. F. Taussig

45k103358

asked Mar 21 at 13:15

So LoSo Lo

64429

edited Mar 21 at 18:07

N. F. Taussig

45k103358

edited Mar 21 at 18:07

N. F. Taussig

45k103358

asked Mar 21 at 13:15

So LoSo Lo

64429

asked Mar 21 at 13:15

So LoSo Lo

64429

1 Answer
1

active

oldest

votes

4

$begingroup$

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 $4choose i$ terms each with the probability

$$fracbinom52-13cdot i 13binom52 13$$ 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 $$4cdotfracbinom3913binom5213 - 6cdotfracbinom2613binom5213 + 4cdotfracbinom1313binom5213$$

share|cite|improve this answer

edited Mar 21 at 15:07

zhoraster

16k21853

answered Mar 21 at 13:47

Vishweshwar TyagiVishweshwar Tyagi

441311

$endgroup$

add a comment | 

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4

$begingroup$

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 $4choose i$ terms each with the probability

$$fracbinom52-13cdot i 13binom52 13$$ 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 $$4cdotfracbinom3913binom5213 - 6cdotfracbinom2613binom5213 + 4cdotfracbinom1313binom5213$$

share|cite|improve this answer

edited Mar 21 at 15:07

zhoraster

16k21853

answered Mar 21 at 13:47

Vishweshwar TyagiVishweshwar Tyagi

441311

$endgroup$

add a comment | 

4

$begingroup$

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 $4choose i$ terms each with the probability

$$fracbinom52-13cdot i 13binom52 13$$ 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 $$4cdotfracbinom3913binom5213 - 6cdotfracbinom2613binom5213 + 4cdotfracbinom1313binom5213$$

share|cite|improve this answer

edited Mar 21 at 15:07

zhoraster

16k21853

answered Mar 21 at 13:47

Vishweshwar TyagiVishweshwar Tyagi

441311

$endgroup$

add a comment | 

$begingroup$

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 $4choose i$ terms each with the probability

$$fracbinom52-13cdot i 13binom52 13$$ 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 $$4cdotfracbinom3913binom5213 - 6cdotfracbinom2613binom5213 + 4cdotfracbinom1313binom5213$$

share|cite|improve this answer

edited Mar 21 at 15:07

zhoraster

16k21853

answered Mar 21 at 13:47

Vishweshwar TyagiVishweshwar Tyagi

441311

$endgroup$

share|cite|improve this answer

edited Mar 21 at 15:07

zhoraster

16k21853

answered Mar 21 at 13:47

Vishweshwar TyagiVishweshwar Tyagi

441311

edited Mar 21 at 15:07

zhoraster

16k21853

edited Mar 21 at 15:07

zhoraster

16k21853

answered Mar 21 at 13:47

Vishweshwar TyagiVishweshwar Tyagi

441311

answered Mar 21 at 13:47

Vishweshwar TyagiVishweshwar Tyagi

441311