Study 8J61 Main |
| by Richard Pavlicek |

What is the longest suit you’ve ever held in a bridge hand? If your answer is 13, you’re probably lying, or perhaps the victim of a set-up, as the odds are 158+ billion to 1 against it. Even 12 is highly suspect, with odds of 313+ million to 1 against. Think about it: If you played 100 hands a day for 80 years, this would be *fewer* than 3 million — and I’m even giving you the benefit of leap years.

More realistic is an 11-card suit, as the odds against it are “only” about 2.7 million to 1. Actually you’d only need to play about 1.9 million hands to be a *favorite* (probability greater than one-half) to get one. But that’s still a lot of hands, so don’t get your hopes up! My personal high happens to be 11 — in hearts headed by A-Q-J and I had to lose to the K — but I’ve hardly played 1.9 million hands, so I’m either lucky or a card rack.*

*Reminiscent of the late Jacq van Gelder of Fort Lauderdale, Florida, whose hobby

was to create anagrams of people’s names. About 40 years ago he christened me

“Heil! VIP Card Rack” so I should attribute my 11-bagger to him.

For practical expectations, most players will be due a 10-card suit or two, as the odds drop considerably to only 60,737 to 1 against. While it’s hardly a yearly occurrence, bridge addicts usually play many times that number of hands in a lifetime. And now that we’ve lowered our expectations, holding a *9-card* suit might even be considered common, with odds of only 2698 to 1 against. In fact I held one last night on BBO:

A-K-Q-J-8-7-6-4-3 A Q-8 K

Surprisingly the bidding began 2 (strong) on my left, pass, 2 (negative). I decided to be cool and *pass*, then 2 , pass, 4 . When I now “saved” in 4 , opener took the bait and doubled, but RHO pulled to 5 . Grrr… I continued on to 5 , of course, but no surprises there (down one doubled). At least they were making 5 .

I have a database of 40,069 deals from 72 major tournaments, 1996 to 2014. (I planned to continue this but lost interest when top-level bridge became a travesty in 2015. See Stats Discontinued.) The following table shows a breakdown of the longest suits held by each direction. There were quite a few 9-baggers, but only one 10 for a lucky North player. Each cell shows the actual number of hands and percent. The last column shows the expected percent in theory for all possible bridge hands.

Length | North | South | West | East | Expected |
---|---|---|---|---|---|

13 | 0 | 0 | 0 | 0 | 0.0000 |

12 | 0 | 0 | 0 | 0 | 0.0000 |

11 | 0 | 0 | 0 | 0 | 0.0000 |

10 | 1 0.0025 | 0 | 0 | 0 | 0.0016 |

9 | 19 0.0474 | 15 0.0374 | 11 0.0275 | 9 0.0225 | 0.0370 |

8 | 191 0.4767 | 169 0.4218 | 168 0.4193 | 207 0.5166 | 0.4668 |

7 | 1425 3.5564 | 1413 3.5264 | 1389 3.4665 | 1418 3.5389 | 3.5266 |

6 | 6443 16.0798 | 6664 16.6313 | 6657 16.6138 | 6633 16.5539 | 16.5477 |

5 | 17768 44.3435 | 17936 44.7628 | 17756 44.3136 | 17900 44.6729 | 44.3397 |

4 | 14222 35.4938 | 13872 34.6203 | 14088 35.1594 | 13902 34.6952 | 35.0805 |

Percents shown as 0.0000 are not zero but simply round that way to 4 decimal places.

Some years ago I created a database of 10 billion random deals. The purpose was to test the algorithm of my pseudorandom generator, which seemed to work well, producing results extremely close to theoretical values. I wasn’t lucky enough to catch a 13-bagger, but I got a lot of 12s. Warning: Stand back or you may be hit by shrapnel!

Length | North | South | West | East | Expected |
---|---|---|---|---|---|

13 | 0 | 0 | 0 | 0 | 0.0000 |

12 | 32 0.0000 | 28 0.0000 | 18 0.0000 | 23 0.0000 | 0.0000 |

11 | 3611 0.0000 | 3722 0.0000 | 3662 0.0000 | 3656 0.0000 | 0.0000 |

10 | 164119 0.0016 | 164149 0.0016 | 164880 0.0016 | 164403 0.0016 | 0.0016 |

9 | 3704677 0.0370 | 3701446 0.0370 | 3705140 0.0371 | 3707559 0.0371 | 0.0370 |

8 | 46673774 0.4667 | 46679100 0.4668 | 46678185 0.4668 | 46678246 0.4668 | 0.4668 |

7 | 352664347 3.5266 | 352667011 3.5267 | 352662601 3.5266 | 352663604 3.5266 | 3.5266 |

6 | 1654768350 16.5477 | 1654799943 16.5480 | 1654753628 16.5475 | 1654743326 16.5474 | 16.5477 |

5 | 4433951588 44.3395 | 4433978902 44.3398 | 4433985914 44.3399 | 4434000999 44.3400 | 44.3397 |

4 | 3508069502 35.0807 | 3508005699 35.0801 | 3508045972 35.0805 | 3508038184 35.0804 | 35.0805 |

A bumbling bridge player explained to his partner how he planned to improve his game:

“Every night when I go to bed I think about the mistakes I made that day at the bridge table.”

“Gee,” wondered his partner, “How do you get any sleep?”

There are 635+ billion distinct bridge hands. This is easily determined by the combinatorial “52 choose 13” which is exactly 635,013,559,600. The following table shows the breakdown according to longest suit length.

Length | Hands | Percent | Odds against | Favored in |
---|---|---|---|---|

13 | 4 | 0.0000 | 158753389899 to 1 | 110,039,471,545 |

12 | 2028 | 0.0000 | ~313123056 to 1 | 217,040,378 |

11 | 231,192 | 0.0000 | ~2746692 to 1 | 1,903,863 |

10 | 10,455,016 | 0.0016 | ~60737 to 1 | 42,100 |

9 | 235,237,860 | 0.0370 | ~2698 to 1 | 1871 |

8 | 2,963,997,036 | 0.4668 | ~213 to 1 | 149 |

7 | 22,394,644,272 | 3.5266 | ~27 to 1 | 20 |

6 | 105,080,049,360 | 16.5477 | ~5 to 1 | 4 |

5 | 281,562,853,572 | 44.3397 | ~5 to 4 | 2 |

4 | 222,766,089,260 | 35.0805 | ~13 to 7 | 2 |

Odds against = Rarity of hand (~ means approximate)

Favored in = Sample needed for probability > one-half

The number of hands to be “Favored in” is a challenge to calculate. Let T = Total number of bridge hands (635013559600), H = Hands with desired longest suit, and N = number of samples. Then evaluate the expression:

((T-H)/T)^{N} < 1/2

When the exponent N = 1, the expression is greater than one-half, and as N increases it becomes continuously smaller, approaching zero. Effectively this is the probability of *never* getting the desired hand in N samples. The goal is to find the smallest N for which it is *less than* one-half. That is, if the probability of *not* getting the desired hand is below one-half, then you’re a favorite to get it.

The problem is evaluating the expression for huge powers of N. Don’t try this at home! Many years ago I created a super calculator for tasks like this. For an 11-card suit it was necessary to evaluate almost 2 million powers to determine the cusp values:

N = 1903862 → 0.5000000904115022

N = 1903863 → 0.4999999083744116

Probabilities are rounded to 16 places, but 72 places were kept to ensure accuracy.

Max Shireson of Stanford, California, brought my attention to another interesting probability problem. For example, consider the extreme case of a 13-card suit. There are exactly four such hands in 635+ billion (T) so one would seem due every T/4 hands, which brings up the question: What is the probability of not getting a 13-card suit in T/4 hands? The expression is easy to formulate:

((T-4)/T)^{(T}/4)

But not so easy to evaluate, as the fraction must be raised to the 158,753,389,900 power! Mind boggling, but it’s still within the grasp of my super calculator using a shortcut that the power T/4 is equivalent to the 10535 power, raised to the 697 power, raised to the 470 power, raised to the 46 power. The first stage (10535 power) is calculated exactly, as the numerator or denominator does not exceed my 131,000+ digit capacity; then a division gives that probability, which I retain to 144 places. The probability is then raised to the 697 power; the result to the 470 power; and that result to the 46 power. My final answer, rounded to 16 places:

0.3678794411702837

The probability of *getting* a 13-card suit (at least once) is simply 1 minus the above, or 0.6321205588297163. So what, you might ask? The curiosity is in the connection with higher math. My calculus knowledge has waned over the years from lack of use, but I believe the answer found is supposed to be the reciprocal of Euler’s number e, which to 16 places is:

1/e = 0.3678794411714423

Pretty close! Matching through 11 places. This is where I’m fuzzy, and I hope someone will enlighten me. Should this be an *exact* match, or is it just an approximation? I’m sure the value of 1/e is correct, so if theory predicts an exact match, my calculator becomes suspect, despite using 144 digits of precision throughout. Clue me in, somebody.

Addendum: Charles Blair, retired professor (University of Illinois), informed me that it is indeed an approximation. The larger the exponent, the closer it gets, but it can never be exact. As evidence he pointed out that the expression must be a rational number, whereas e (or its reciprocal) is irrational — but so am I sometimes, so I need all the help I can get.

For my own investigation I redid the calculations using different power factors (10810, 410, 301, 119) and increased the precision from 144 to 288 digits. I was happy to see the answers agreed to 137 places, i.e., my first attempt only obscured the seven least-significant digits from being exact. So my super calculator is off the hook!

Thanks to Sandro Bellini of Milan, Italy, who found the same answer as mine by different means, so either it’s right or we’re *both* ready for the tar pits. He also confirmed, like Charles, that it can never equal 1/e. That’s disappointing to naive me — but what if we throw in a piece of pi?

Study 8J61 Main | Top Long Suit Story |

© 2019 Richard Pavlicek