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Bridge probability of card distribution

Summer is here and some days are more rainy than others. What to do? What to do? Take three of your friends (why not one of yours best friends and two 'enemies') and play a bridge game, for example the contract bridge which is a trick-taking game using a standard 52-card deck. Bridge is one of the world's most popular partnership card games, may be dated from the early 16th-century invention of trick-taking games such as whist. If you don't know how to play it then why not learning? There are plenty of videos on YouTube about that although I found that the Australian Bridge Federation video lessons are one of the best.

The game is using a standard 52-cards deck, 13 of each colour: hearts, spades, clubs and diamonds.

Anyway, this post is not about the game itself but about the probability of getting a certain hand where the cards' colours are distributed in a certain way like 10 of hearts, 1 of diamonds, 1 of spades and 1 of clubs (bridge probability of card distribution).

So the problem sounds like this:

In a classic bridge game each player gets 13 cards. What is the probability that the colour distribution is:

  1. 10-1-1-1 (e.g. 10 of hearts, 1 of diamonds, 1 of spades and 1 of clubs)
  2. 5-5-3-0 (e.g. 5 of hearts, 5 of diamonds, 3 of spades and none of clubs)

1.) Firstly let us remember some basic notions about probability:

  • the probability that the event A occurs is noted with P(A) and is P(A)=\frac{\text{number of ways event A can happen}}{\text{number of all possible outcomes}}
  • the probability that an event A succeed r-times from n-experiments is P(A)=\binom{n}{r}p^r\times (1-p)^{n-r} where p is the probability that the event A occur and (1-p) is the probability that the event A will not occur.

2.) Secondly let us digest this problem (a) for a while:

  • What is the number of all possible outcomes? With other words in how many ways can we arrange 52 cards taken by 13? Yes, that's right, there are combination of 52 cards taken by 13 which is noted as \binom{52}{13} or nCr on your scientific calculator. So if we would try to arrange these 52 cards in hands of 13 cards in all the possible ways we would find that there are 52C13 different way of arranging them. 52C13 is a big number like few hundred billions (635 013 559 600 to be more precise).
  • In how many ways we can arrange 13 cards of one colour (like hearts) if we would taken them in pairs of 10 cards of hearts? Yes, that's right, there are combination of 13 cards taken by 10 which is \binom{13}{10}=286 different ways.
  • In how many ways we can arrange 13 cards of diamonds if we would taken them in pairs of 1 card of diamonds? Yes, that's right, there are combination of 13 cards taken by 1 which is \binom{13}{1}=13 different ways. The same as for spades or for clubs.
  • Now, in how many ways we can arrange these \binom{13}{10} hearts with these \binom{13}{1} diamonds and with these \binom{13}{1} spades and with these \binom{13}{1} clubs? Well, if you have a 6-sided dice and another 6-sided dice then how many different pairs you can get? Yes, that is 6x6 different pairs. The same applies here. We can arrange these different pairs in \binom{13}{10}\times\binom{13}{1}\times\binom{13}{1}\times\binom{13}{1} different ways. But wait! Nobody said that the first must be hearts, the second diamonds and so forth. In fact they can be of whatever colour as long as they are distributed like 10-1-1-1. So in how many ways we can choose 1 colour of 4 such that it is used to make a set of 10 of a colour? Well, we could choose 10 cards of hearts and the rest (1-1-1) of the other colours, we could choose 10 of diamonds and the rest (1-1-1) of the other colours, and so on. So in total would be 4 different ways.

So the the number of ways we can arrange these 10-1-1-1 of a colour (like 10 hearts, 1 diamonds, 1 spades and 1 clubs) is \binom{13}{10}\times\binom{13}{1}\times\binom{13}{1}\times\binom{13}{1} but there are 4 different ways to do it so we have to multiply that by 4. But remember that the total number of ways in which all these 52 cards can be arranged in pairs of 13 cards is \binom{52}{13}. It means that the probability that the cards will be distributed by colours like 10-1-1-1 is \frac{4\times\binom{13}{10}\times\binom{13}{1}\times\binom{13}{1}\times\binom{13}{1}}{\binom{52}{13}}\approx 4\times10^{-6} i.e. 4 chances in one million. Wow!

3.) What are the odds for the (b) to happen?

  • As I've mentioned earlier there are \binom{52}{13} ways to arrange 52 cards in pairs of 13 cards. That's a lot!
  • To arrange 13 cards of the same colour such that you get any 5 of them there are \binom{13}{5} ways to do it.
  • To arrange 13 cards of the same colour such that you get any 3 of them there are \binom{13}{3} ways to do it.
  • To arrange 13 cards of the same colour such that you get any 0 of them there are \binom{13}{0} ways to do it.

So to arrange these 52 cards such that you you can get 5 cards of one colour (e.g. hearts) and 5 of other colour (like diamonds) and 3 of other colour (like spades) and none of the clubs there will be \binom{13}{5}\times\binom{13}{5}\times\binom{13}{3}\times\binom{13}{0} ways to do it. But wait, nobody said that the first colour must be hearts or something else so with 4 available colours there are many other ways to arrange these cards such that the colours are distributed 5-5-3-0. We know that there are 4 colours which can combine in \binom{4}{2} different ways to get a pair of 5 of a colour (#1) and 5 of another colour (#2), i.e. 4C2=6 different ways. Because two colour were already been used there will be only 2 different colours available to choose for the pair of 3. So in total will be {6}\times[2]=12 different ways to mix those \binom{13}{5} respective \binom{13}{3} arrangements. So the odds that the cards will be distributed by colours like 5-5-3-0 is \frac{12\times\binom{13}{5}\times\binom{13}{5}\times\binom{13}{3}\times\binom{13}{0}}{\binom{52}{13}}\approx 9\times10^{-3} i.e. 9 chance in one thousand (which can be rounded to almost 1%).

Scientia potentia est - the knowledge is power!

Now, if you think that this article was interesting don't forget to rate it. It shows me that you care and thus I will continue write about these things.

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Eugen Mihailescu

Founder/programmer/one-man-show at Cubique Software
Always looking to learn more about *nix world, about the fundamental concepts of math, physics, electronics. I am also passionate about programming, database and systems administration. 16+ yrs experience in software development, designing enterprise systems, IT support and troubleshooting.
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By Eugen Mihailescu | June 25, 2014 | Discrete mathematics, Maths | No Comments |

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