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Probability of dealing four perfect hands of cards in a world of random shufflers

A couple of months ago (really? Two years?! Man!) I posted about an extraordinary coincidence: in a game of whist at a village hall in Kineton, Warwickshire, each of four players had been dealt an entire suit each. My post ‘Four perfect hands: An event never seen before (right?)‘ discussed this story. What really interested me was that the quoted mathematical analysis — and figure of 2,235,197,406,895,366,368,301,559,999 to 1 — appears to be correct; what lets down the piece is poor modelling. The probability calculated relies on the assumption that the deck is completely randomly ordered. Apart from the fact that new decks of cards come sorted into suits, whist is a game of collecting like cards together, so a coincidental ordering must be made more likely. Still unlikely enough to be worthy of mention in a local paper, maybe, but not “this is the first time this hand has ever been dealt in the history of the game”-unlikely.

Anyway, last week I was asked where the quoted figure 2,235,197,406,895,366,368,301,559,999 to 1 actually comes from. Here’s my shot at it.

There are 52 cards in a standard deck, each belonging to one suit of thirteen cards each. The number of ways of dealing 13 cards from a deck is therefore $\binom{52}{13}$. The probability of one player being dealt 13 cards from 52 all of any one of the four available suits (we don’t care which) is then $$\frac{\binom{4}{1}}{\binom{52}{13}}\text{.}$$

Having dealt thirteen cards from one suit to one player, 39 cards remain in three suits. The probability of the second player being dealt 13 cards from 39 all of the the same suit is $$\frac{\binom{3}{1}}{\binom{39}{13}}\text{.}$$

Then 26 cards remain in two suits. The probability of the third player being dealt 13 cards from 26 all of the same suit is $$\frac{\binom{2}{1}}{\binom{26}{13}}\text{.}$$

Finally, the remaining 13 cards are all of one suit and are dealt to the fourth player with probability $1$.

Then the probability of all of this happening together is $$ \frac{\binom{4}{1}}{\binom{52}{13}} \times \frac{\binom{3}{1}}{\binom{39}{13}} \times \frac{\binom{2}{1}}{\binom{26}{13}}\text{.}$$

If you calculate this, you should find the quoted probability. Or, if you prefer, the magic ‘to one’ number we are looking for arises from $$\frac{\binom{52}{13}\binom{39}{13}\binom{26}{13}}{4!}\text{.}$$

8 Responses to “Probability of dealing four perfect hands of cards in a world of random shufflers”

  1. Avatar Chris Taylor

    I can read this article for about 10 seconds. After that, all the probabilities (which were legible) suddenly become [Math Processing Error]. Anything to do with the new MathJax?

    • Avatar Chris Taylor

      And magically, within seconds of my the comment, the probabilities are back. I would like to take credit for this.

    • Avatar Peter Rowlett

      I got the same error. I called Aperiodical tech support (CP) and he says it takes one refresh to prompt your cache to clear, then you’re fine from there on.

  2. Avatar Peter Stewart

    Hi Peter
    I came across your explanation of ‘Probability of dealing four perfect hands of cards in a world of random shufflers’ and would like to reply.
    Given that you list 7 other similar events in the last 90 years means either:
    1. the fresh unused cards came out of the box in this order, the jokers removed and the cards dealt in their unshuffled state with not one single card of the remaining 52 having therefore changed its position in the pack or
    2. Or the calculation to reach odds of 2,235,197,406,895,366,368,301,559,999 to 1 must be wrong – they aren’t – or
    3. The very ‘laws’ of probability are WRONG!!! There is no law!!! At least not as we currently understand the ‘laws’ of probability. Surely, if just two of the eight events you list in total were genuinely from shuffled packs it would prove that probability just does not work in the way we perceive it to ?
    And the next question: what are the current odds of there being life on other planets – are they greater or less than all four players being dealt the perfect bridge hand?


    Peter Stewart

    • Avatar Karen E Littlejohn

      We just had 4 perfect hands dealt at our bridge club. Have pictures .

      • Avatar Hiram

        Sorry to be 6 years late! Richard Dawkins points out in a book that whatever the odds are of dealing the perfect 4 hands, the odds of whatever hand you deal next are exactly the same.

  3. Avatar John Forder

    I once read an article in a bridge magazine a bout a guy who was ‘kibbutzing’ (watching others play). The four people he was watching had two packs of card in use, and the ‘dummy’ would shuffle the cards ready for the next hand. Anyway, our ‘kibbutzer’ decided to play a little game with this group; he wandered off, got an identical pack to one of the ones in use, and prepared it in such a way that when dealt each player would receive a complete suit. He then returned to the table and waited for an opportune moment (when everyone’s attention was concentrated on an interesting hand) and subsituted his pre-arranged hand for the one on the table. So when the dealer picked up this deck and dealt, all the players ‘knew’ that it had been shuffled properly . . . and were gob-smacked by the hands they were dealt.
    I wouldn’t be at all surprised if a little trick like that could explain all the eaxmples of ‘perfect deals’.

  4. Avatar Jonathan Hooton

    According to John Riddington Young in his book The Inns and Taverns of Old Norwich (Wensum Books, 1975 ISBN 0 903619 18 0) p 48 In 1973 at the Queen Victoria, (later re-named the Bread and Cheese and closed in 2015) in a game of solo whist “Vic Stephens dealt and all four players picked up a complete suit of cards.” He then quoted the odds figure given above.


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