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Puzzle 8S99 by Richard Pavlicek
Some things are difficult to comprehend, like the Einstein Cross Quasar (pictured above) 8 billion light-years from Earth, or the staggering number of bridge deals — or even partner making the right lead against 3 NT. While pondering that, suppose you are East on this deal that decided the Sector 9 championship of Owl Nebula (pictured right) only 2600 light-years away:
Board 1.948329 E+28 | 10 8 7 K 7 2 K 6 5 3 2 J 8 | E-W Vul | Partner West Pass Pass | North 3 NT | You EAST 2 Pass | South 2 NT Pass | |
J J 9 Q J 10 9 8 7 K 10 9 4 | A Q 5 4 3 2 Q 10 8 — 7 6 3 2 | ||||||
3 NT South | K 9 6 A 6 5 4 3 A 4 A Q 5 |
With an empty suit at adverse vulnerability (redshift +1.0127) opening 2 was nebulous, but then so was the atmosphere. South lands in 3 NT, and partner of course leads a diamond — arguably your fault for not opening 2 (then partner would have led a spade). You discard a spade, South wins the A and leads three rounds of hearts to your queen. Game over. If you lead a club, declarer ducks; if a low spade, he hops. Either way declarer is destined for nine tricks: one spade, four hearts, two diamonds and two clubs.
The defense to beat 3 NT is cute: West must lead the J, which East and South duck. Partner got one lead right, but will he get another? No, the straight flush (two in fact) in diamonds is too tempting. Only a club shift will foil declarer, allowing the defense to establish two club tricks, besides a heart and two spades, before declarer can develop nine.
An interesting deal but hardly spectacular, and a bit humbling when you consider that it’s only one out of octillions. North-South have a 5-3 heart fit but wouldn’t enjoy 4 ; down three as the defense crossruffs early and scores the K later. North-South also have a 5-2 fit in diamonds — don’t play there — nor would East-West enjoy their 4-4 fit in clubs or 6-1 in spades. Aggregately, each side has one 8-card fit and one 7-card fit. The best East-West can do is to win seven tricks in spades, so the par contract is 2 NT North-South (plus 120).
Other notable (or not) features are the individual suit lengths: five 3-card suits (most common by far), three doubletons, two each of 4-6 lengths, one singleton and one void. Not surprisingly, half of these are odd length, and half are even length. How many straight flushes do you see? The answer is three, which is high for a deal. Besides West having two in diamonds, East has a subtle one in spades (the ace can be low in poker). The deal also contains 17 touching cards (as defined in Parity Clarity).
Haunting perspectives: The number of unique bridge deals is 53.6 octillion (29 digits). |
Astrophysicists estimate all life on Earth will end in 31.5 quadrillion seconds (17 digits). |
Organizational incompetence indicates bridge will end in 2.4 billion seconds (10 digits). |
Trying to comprehend all this, you take an evening walk. Looking skyward to the constellation Taurus, you spot the bright orange star Aldebaran (pictured right) — a relative neighbor only 65 light-years away. But alas, it is still too overwhelming. You could visit Aldebaran, the Nebula and the Quasar a thousand times (call Elon Musk for tickets) before you could witness every bridge deal. Don’t even think about it! But wait! You will have to think about it to participate in this puzzle contest.
Challenge yourself with this octet of octillions — or make your best guesses!
In all 53,644,737,765,488,792,839,237,440,000 bridge deals, which is greater? | ||
1. Number of odd-length suits | or Number of even-length suits | Equal |
2. Number of 5-3 fits | or Number of 5-2 fits | Equal |
3. Number of 6-4 fits | or Number of straight flushes | Equal |
4. Number of 7-card fits | or Number of 6-card fits | Equal |
5. Number of void suits | or Number of hands with 3-4 aces | Equal |
6. Number of 5-5 fits | or Inches to Einstein Cross Quasar* | |
7. Number of 3-card suits | or Angstroms to Owl Nebula* | |
8. Number of touching cards | or Picometers to Aldebaran* |
*Assume distances given on this page are exact for comparison purposes.
This puzzle contest ran from October 3-31, 2021. There were 42 entries from 31 persons (multiple entries were allowed but only the latest one counted). Only six persons (ranked below) got at least five of the eight answers right; and the great majority (21) got fewer than half right. Evidently October was a bad month for guessing.
Congratulations to Peter Randall, England, who was the first of three to submit perfect scores. Furthermore, he doubled his perfect record: In August he won my Two Eight-Baggers contest, and now this. While dozens of my participants have won two or more times, Peter stands alone in never not winning. Two for two!
Rank | Name | Location | Correct |
---|---|---|---|
1 | Peter Randall | England | 8 |
2 | Charles Blair | Illinois | 8 |
3 | Jean-Christophe Clement | France | 8 |
4 | Richard Stein | Washington | 7 |
5 | Jim Munday | New Mexico | 6 |
6 | Len Helfgott | New Jersey | 5 |
Years ago I remember astronomer Carl Sagan describing the cosmos with his tag phrase “billions and billions.” How petty! I can top that with octillions and octillions. If that makes me a Cosmo Topper, I may be seeing ghosts — and if that reference doesn’t age me, nothing ever will. Stand back for some huge numbers!
The first solution is somewhat counterintuitive. The instinctive answer is probably equal, since every bridge hand must contain one even- and three odd-length suits, or vice versa. Or if there were a difference it would seem to favor odd, because the most common suit length by far is three cards. Never mind that; the answer is even.
The easiest way to count the number of suits of any length is first to consider hands only. Expanding to deals is based on the indisputable logic that every hand must occur an equal number of times. How many times? There are 52!/(13!)^{4} bridge deals. Multiply by four to get the number of hands, then divide by the number of unique bridge hands (52c13) to get 39!/(13!)^{3} ×4, or 337,912,392,291,465,600.
There are seven possible even-length suits (note that a void is a 0-card suit, hence even) and seven possible odd-length suits. Rather than count both parities, I will count the odd lengths (I’m an odd kind of guy). The number of even-length suits is simply the difference between the odd count and the total number of suits.
Length | Combinations | Ways | Suits in All Hands |
---|---|---|---|
1 | 13c1 × 39c12 | 4 | 203,361,466,672 |
3 | 13c3 × 39c10 | 4 | 727,292,733,024 |
5 | 13c5 × 39c8 | 4 | 316,724,254,704 |
7 | 13c7 × 39c6 | 4 | 22,394,644,272 |
9 | 13c9 × 39c4 | 4 | 235,237,860 |
11 | 13c11 × 39c2 | 4 | 231,192 |
13 | 13c13 × 39c0 | 4 | 4 |
39!/(13!)^{3} ×4 × Odd-length suits in all hands: 1,270,008,567,728 | |||
= Odd-length suits in all deals: 429,151,633,351,626,294,573,982,156,800 | |||
+ Even-length suits in all deals: 429,164,170,896,194,390,853,816,883,200 | |||
= 52!/(13!)^{4} ×16 suits in all deals: 858,315,804,247,820,685,427,799,040,000 |
Note that the last column counts the number of suits, not hands. For example, a hand with 5=4=3=1 shape would be counted three times: first for having one club, second for having three diamonds, and third for having five spades — exactly what we want.
Determining the number of suits in all bridge deals is simply to multiply by 16. Wow! 858+ octillion suits — closing in on a nonillion. Hey! My next contest could be November nonillions, followed by December decillions, and even undecillions by the time I reach the nut house. More than half the suits are even-length, which makes it the clear winner.
Another solution that seems counterintuitive. The average combined fit length is 6.5 cards (13 cards split between two sides) so 7-card fits should occur more often than 8-card fits, because they’re closer to the mean — and so they do. Nonetheless, the majority of 7-card fits are split 4-3. Conversely, the majority of 8-card fits are indeed split 5-3, which happens to exceed the number of 5-2 fits.
Fits involve two hands (one side) so the first step is to count the number of fits in all sides. To expand to deals, multiply by 26c13, which is the number of combinations to split the remaining cards into two hands.
Fit | Combinations | Ways | Fits in All Sides |
---|---|---|---|
5-3 | 13c5 × 39c8 × 8c3 × 31c10 | 4 × 4 | 3,146,619,034,525,978,851,840 |
× 26c13 = 5-3 fits in all deals: 32,726,725,930,490,895,646,447,104,000 | |||
5-2 | 13c5 × 39c8 × 8c2 × 31c11 | 4 × 4 | 3,003,590,896,592,979,813,120 |
× 26c13 = 5-2 fits in all deals: 31,239,147,479,104,945,844,335,872,000 |
The first two combinatorials select the first hand, and the next two the second hand for the side. The Ways column accounts for the fit being in any suit, and the four orientations of the side: North-South, East-West, South-North and West-East.
Note that a side with appropriate fits in two suits would be counted twice (or three suits, thrice) to obtain the number of fits, not sides.
To verify the accuracy of this counting method, I expanded each of the 393,173 generic dealprints into its number of deals then allocated that count to each of the eight fits it represented. All totals agreed.
This problem is like comparing apples to oranges, and the diversity obscures any apparent winner. I think we can rule out a tie — else I may have discovered the Theory of Bridge Relativity and win a one-way trip to Einstein Cross Quasar. Counting 6-4 fits is straightforward (analogous to Problem 2) so consider the challenge of straight flushes. As usual to simplify, start with hands only.
The first step is to break down each suit length into the number of holdings with each number of straight flushes, as done in the following table. For example, a 7-card suit comprises 1716 holdings, of which 162 have one straight flush, 47 have two (e.g., K-Q-J-10-9-8-2) and 8 have three (e.g., A-7-6-5-4-3-2). There are no entries for suit lengths 0-4, since a straight flush requires 5+ cards.
Suit Length | Total Holdings | Holdings with 1-10 Straight Flushes | |||||||||
---|---|---|---|---|---|---|---|---|---|---|---|
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | ||
5 | 1287 | 10 | |||||||||
6 | 1716 | 62 | 9 | ||||||||
7 | 1716 | 162 | 47 | 8 | |||||||
8 | 1287 | 230 | 100 | 34 | 7 | ||||||
9 | 715 | 188 | 111 | 56 | 23 | 6 | |||||
10 | 286 | 70 | 72 | 46 | 27 | 14 | 5 | ||||
11 | 78 | 6 | 12 | 22 | 16 | 10 | 7 | 4 | |||
12 | 13 | 6 | 2 | 2 | 3 | ||||||
13 | 1 | 1 |
From the above data I found a neat solution. For example, a 7-card suit (see Problem 1) occurs 13c7 × 39c6 × 4 times in all bridge hands. Instead of using 13c7, I use the counts in the above table times the number of straight flushes: 162×1 + 47×2 + 8×3, or 162 + 94 + 24. Intuitively this counts the number of straight flushes, including cases where straight flushes exist in two suits.
Length | Straight Flush Factor | Times | Straight Flushes | |
---|---|---|---|---|
5 | 10 | 39c8 × 4 | 2,460,949,920 | |
6 | 62 + 18 | 39c7 × 4 | 4,921,899,840 | |
7 | 162 + 94 + 24 | 39c6 × 4 | 3,654,137,760 | |
8 | 230 + 200 + 102 + 28 | 39c5 × 4 | 1,289,695,680 | |
9 | 188 + 222 + 168 + 92 + 30 | 39c4 × 4 | 230,302,800 | |
10 | 70 + 144 + 138 + 108 + 70 + 30 | 39c3 × 4 | 20,471,360 | |
11 | 6 + 24 + 66 + 64 + 50 + 42 + 28 | 39c2 × 4 | 829,920 | |
12 | 30 + 12 + 14 + 24 | 39c1 × 4 | 12,480 | |
13 | 10 | 39c0 × 4 | 40 | |
39!/(13!)^{3} ×4 × Straight flushes in all hands: 12,578,299,800 | ||||
= Straight flushes in all deals: 4,250,363,376,377,263,298,186,880,000 | ||||
Fit | Combinations | Ways | 6-4 Fits in All Sides | |
6-4 | 13c6 × 39c7 × 7c4 × 32c9 | 4 × 4 | 414,574,312,849,272,576,000 | |
× 26c13 = 6-4 fits in all deals: 4,311,821,598,220,144,353,945,600,000 |
So now we have proof that bridge is superior to poker. Why settle for 4.25 octillion straight flushes when you can have 61 septillion more great trump fits? I rest my case.
I was impressed by Peter Randall, not only in winning but for providing his count totals for each problem in the comment box. We agreed on all but one — straight flushes — and his count was right! I had infiltrated a typo into my data, but once fixed we agreed exactly.
This was an easy problem if you noticed the complementary aspect. For every 7-card fit there must be a 6-card fit in the same suit for the other side, hence the numbers must be identical. How many? Too many, but if you want to know:
Fit | Combinations | Ways | Fits in All Sides |
---|---|---|---|
7-card | 13c7 × 39c19 × 26c13 | 4 × 2 | 9,840,824,874,877,284,288,000 |
× 26c13 = 7-card fits in all deals: 102,350,483,193,648,682,965,772,800,000 | |||
6-card | 13c6 × 39c20 × 26c13 | 4 × 2 | 9,840,824,874,877,284,288,000 |
× 26c13 = 6-card fits in all deals: 102,350,483,193,648,682,965,772,800,000 |
The first two combinatorials select the 26 cards for the side, and 26c13 accounts for all distributions between the two hands. The Ways column accounts for the fit being in any suit, and for either side. Note there is no factor to specify the orientation of each side, because 26c13 covers all possibilities.
With over 102 octillion 7-card fits, the challenge of finding an 8-card trump fit is amplified. That’s a heap to avoid!
The equality of course holds for any two fits that add to 13, but counts are reduced as the fits diverge. For example, the number of 8-card fits and 5-card fits are equal, with about 69.5 octillion each.
Counting the number of voids is surprisingly easy. The Ways factor accounts for the void being in any suit, so hands with two voids are counted twice, and the fluke 13-0-0-0 is counted three times, thereby counting voids, not hands.
Hands with 3 aces are counted separately from hands with 4 aces to ensure no redundancy.
Length | Combinations | Ways | Voids in All Hands |
---|---|---|---|
0 | 13c0 × 39c13 | 4 | 32,489,701,776 |
× 39!/(13!)^{3} ×4 = Voids in all deals: 10,978,672,851,964,438,613,962,905,600 | |||
Aces | Combinations | Ways | Hands |
3 | 4c3 × 48c10 | 1 | 26,162,863,584 |
4 | 4c4 × 48c9 | 1 | 1,677,106,640 |
39!/(13!)^{3} ×4 × Unique hands with 3-4 aces: 27,839,970,224 | |||
= Hands with 3-4 aces in all deals: 9,407,470,939,715,009,433,320,294,400 |
So the number of void suits is greater by about 1.5 octillion. Surely you will find this knowledge useful in solving other problems. Let’s see: If an octillion and a half hens lay an octillion and a half eggs…
Counting 5-5 fits is similar to 6-4 fits (Problem 3) except the equal division eliminates the ways factor to reverse the hand orientation of a side; i.e., 13c5 and 8c5 already account for all possible distributions. Sides with two 5-5 fits are counted twice (once for each suit) so the total reflects the number of 5-5 fits, not the number of sides.
Fit | Combinations | Ways | 5-5 Fits in All Sides |
---|---|---|---|
5-5 | 13c5 × 39c8 × 8c5 × 31c8 | 4 × 2 | 279,837,661,173,258,988,800 |
× 26c13 = 5-5 fits in all deals: 2,910,479,578,798,597,438,913,280,000 | |||
8,000,000,000 light-years × 9,460,730,472,580,800 m/ly × 39.37007874 in/m | |||
= Inches to Einstein Cross Quasar: 2,979,757,629,147,388,056,097,536,000 |
And the winner is… the Quasar by a whisker, albeit a strain to call a difference of 69+ septillion a whisker; but we’re playing with big cats around here — and hey, it would be close enough for government work.
Another straightforward count, already done for Problem 1 and repeated here:
Length | Combinations | Ways | 3-Card Suits in All Hands | |
---|---|---|---|---|
3 | 13c3 × 39c10 | 4 | 727,292,733,024 | |
× 39!/(13!)^{3} ×4 = 3-card suits in all deals: 245,761,227,312,338,046,214,479,974,400 | ||||
2600 light-years × 9,460,730,472,580,800 meters/ly × 10,000,000,000 angstroms/m | ||||
= Angstroms to Owl Nebula: 245,978,992,287,100,800,000,000,000,000 |
There you have it… The total number of 3-card suits is huge (245+ octillion) but still falls short of the number of angstroms to Owl Nebula. Consider yourself educated!
Peter Randall: 2.45980 E+29 angstroms.
Sorry, the GBL (Galactic Bridge League) Rules Committee has disqualified you. Law 77 forbids scientific notation except for board numbers, and this is evidence why. You overshot Owl Nebula by about a septillion angstroms, or 4 light-days. Turn your ship around and head back! Or if it’s my calculation that falls short, well… then… Beam me up, Scotty!
Counting touching cards is complicated, but as usual consider only hands to start. As with straight flushes, the first step is to break down each suit length into the number of holdings with each number of touching cards, as done in the following table. For example, a 4-card suit comprises 715 holdings, of which 360 have one pair of touching cards, 135 have two (e.g., A-K-Q-2) and 10 have three (e.g., 9-8-7-6).
Suit Length | Total Holdings | Holdings with 1-12 Touching Card Pairs | |||||||||||
---|---|---|---|---|---|---|---|---|---|---|---|---|---|
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | ||
2 | 78 | 12 | |||||||||||
3 | 286 | 110 | 11 | ||||||||||
4 | 715 | 360 | 135 | 10 | |||||||||
5 | 1287 | 504 | 504 | 144 | 9 | ||||||||
6 | 1716 | 280 | 700 | 560 | 140 | 8 | |||||||
7 | 1716 | 42 | 315 | 700 | 525 | 126 | 7 | ||||||
8 | 1287 | 21 | 210 | 525 | 420 | 105 | 6 | ||||||
9 | 715 | 70 | 280 | 280 | 80 | 5 | |||||||
10 | 286 | 84 | 144 | 54 | 4 | ||||||||
11 | 78 | 45 | 30 | 3 | |||||||||
12 | 13 | 11 | 2 | ||||||||||
13 | 1 | 1 |
Touching cards are then counted just like straight flushes. For example, a 4-card suit occurs 13c4 × 39c9 × 4 times in all bridge hands. Instead of 13c4, I use the counts in the above table times the number of touching cards: 360×1 + 135×2 + 10×3, or 360 + 270 + 30. Intuitively this counts the number of touching cards — and much to my surprise, revealed an amazing symmetry. Take a gander at this:
Length | Touching Card Factor | Times | Touching Cards |
---|---|---|---|
2 | 12 | 39c11 × 4 | 80,450,690,112 |
3 | 110 + 22 | 39c10 × 4 | 335,673,569,088 |
4 | 360 + 270 + 30 | 39c9 × 4 | 559,455,948,480 |
5 | 504 + 1008 + 432 + 36 | 39c8 × 4 | 487,268,084,160 |
6 | 280 + 1400 + 1680 + 560 + 40 | 39c7 × 4 | 243,634,042,080 |
7 | 42 + 630 + 2100 + 2100 + 630 + 42 | 39c6 × 4 | 72,351,927,648 |
8 | 42 + 630 + 2100 + 2100 + 630 + 42 | 39c5 × 4 | 12,767,987,232 |
9 | 280 + 1400 + 1680 + 560 + 40 | 39c4 × 4 | 1,302,855,840 |
10 | 504 + 1008 + 432 + 36 | 39c3 × 4 | 72,380,880 |
11 | 360 + 270 + 30 | 39c2 × 4 | 1,956,240 |
12 | 110 + 22 | 39c1 × 4 | 20,592 |
13 | 12 | 39c0 × 4 | 48 |
39!/(13!)^{3} ×4 × Touching cards in all hands: 1,792,979,462,400 | |||
= Touching cards in all deals: 605,869,979,469,049,895,596,093,440,000 | |||
65 light-years × 9,460,730,472,580,800 meters/ly × 1,000,000,000,000 p/m | |||
= Picometers to Aldebaran: 614,947,480,717,752,000,000,000,000,000 |
I was shocked to see the pattern in the second column — an exact match for lengths 2-7 with lengths 8-13, except in the opposite order. Who would expect that? No doubt there’s some underlying truth about touching cards that eludes me. It certainly raised skepticism about my method, but verification by other means produced identical counts.
Once again the cosmic distance prevails. Yes, I planned it this way as a strategic move against those who guessed at the answers. Surely no one would guess all three, since I had flaunted the huge number of bridge deals — and from the scores this month, it worked.
Charles Blair: Are you counting A-K-Q-3-2 (around-the-corner) as a straight flush?
Call it what you want, but please come to my Tuesday night poker game and bet it against my full house.
Len Helfgott: Your 31.5 quadrillion seconds (about 1 billion years) seems too quick, as the Sun’s conversion to a red giant is expected in 4-5 billion years.
My source was a study by Nature Geoscience that Earth will become uninhabitable in 1.08 billion years. Still, that’s refreshing compared to studies that bridge will become unbearable before the end of this century — thanks ACBL, USBF, WBF.
© 2021 Richard Pavlicek