Cascade, on 2011-November-01, 20:29, said:
Probabilities of the longest and shortest suits in a four-hand deal:
Longest Suit
4 0.0292944677905751
5 0.402336782879317
6 0.422075366415914
7 0.126442683507377
8 0.0183051683243252
9 0.00147821697795984
10 6.58450191573011E-5
11 1.45628565027458E-6
12 1.27745286982186E-8
13 2.51963123452264E-11
Shortest Suit
0 0.183766327186213
1 0.609953117477308
2 0.205349135658076
3 0.000931419678402451
These numbers were computed by considering every possible four hand distribution and calculating the probability of those four-hand distributions. The longest and shortest suits were noted and the probabilities summed accordingly.
Cascade, on 2011-November-02, 18:52, said:
The bolded (my editing) number is slightly different than what I calculated. I have double checked my calculation and think it is correct. Can anyone confirm?
FWIW, I calculated the same basic statistics a while ago and I'll confirm the Longest Suit proportions that Cascade quoted (and in particular that 1.83% rather than 1.87% is the probability that the longest suit in a deal is of exactly 8-card length).
For those that are interested in the full, gory details, of the total of
53,644,737,765,488,792,839,237,440,000 possible deals, there are:
1,571,494,042,604,960,223,750,000,000 where the longest suit is 4 cards (in any of the 4 hands)
21,583,251,210,971,361,009,130,800,768 where the longest suit is 5 cards
22,642,122,348,654,241,172,787,919,872 where the longest suit is 6 cards
6,782,984,599,117,957,218,132,857,856 where the longest suit is 7 cards
981,975,954,511,555,218,232,092,504 where the longest suit is 8 cards
79,298,562,143,148,725,113,050,600 where the longest suit is 9 cards
3,532,238,785,856,985,879,197,952 where the longest suit is 10 cards
78,122,061,820,624,147,552,512 where the longest suit is 11 cards
685,286,242,093,646,948,664 where the longest suit is 12 cards
1,351,649,568,417,019,272 where the longest suit is 13 cards
These lead, as he says, to his figure of 3.43% for the probability of 3 (or more) deals in a sample of 36 deals containing at least one suit of 8 (or more) cards (and to a probability of 2.88% of exactly 3 such deals: so there is a non-negligible probability of 0.55% of there being 4 or more such deals out of the 36).
To give this number (3.43%) some perspective, it's rather larger than the probability (2.99%) of picking up a 4-4-4-1 hand in any one deal. I don't suppose most of the club members would believe that could be true: it just shows how our intuitions (preconceptions? prejudices?) can lead us astray.