Puzzle 8S99   Main

October Octillions

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 1.947E+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 that all life on Earth will end in 31.5 quadrillion seconds (17 digits).
Organizational incompetence indicates that 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, or Equal

2. Number of 5-3 fits, or Number of 5-2 fits, or Equal

3. Number of 6-4 fits, or Number of straight flushes, or Equal

4. Number of 7-card fits, or Number of 6-card fits, or Equal

5. Number of void suits, or Number of hands with 3-4 aces, or Equal

6. Number of 5-5 fits, or Number of inches to Einstein Cross Quasar*

7. Number of 3-card suits, or Number of angstroms to Owl Nebula*

8. Number of touching cards, or Number of picometers to Aldebaran*

*Assume distances given on this page are exact for comparison purposes.

Peter Randall Wins

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!

Winner List

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

Solutions

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!

1. More even-length suits than odd

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 odd-length suits and seven possible even-length suits (note that a void is a 0-card suit, hence even). 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 in all hands (52c13 × 4).

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
Odd-length suits:			1,270,008,567,728
Even-length suits:			1,270,045,670,672
Total suits:			2,540,054,238,400

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. More than half the suits are even-length, which makes it the clear winner.

Multiplying the above by 39!/(13!)³ × 4 (number of hands in all deals) gives the final humongous tally. Stand back!

Suit Length	Occurrence in All Deals
Odd	429,151,633,351,626,294,573,982,156,800
Even	429,164,170,896,194,390,853,816,883,200
Total	858,315,804,247,820,685,427,799,040,000

Wow! 858+ octillion suits — closing in on a nonillion. Hey! My next contest could be November Nonillions, followed by December Decillions, and maybe even Undecillions by the time I reach the nut house.

2. More 5-3 fits than 5-2 fits

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.

Fit Type	Combinations	Ways	Fits in All Sides
5-3	13c5 × 39c8 × 8c3 × 31c10	4 × 4	3,146,619,034,525,978,851,840
5-2	13c5 × 39c8 × 8c2 × 31c11	4 × 4	3,003,590,896,592,979,813,120

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 expand the above to deals, multiply by 26c13, which is the number of combinations to split the remaining cards into two hands. Octillions, here we come!

Fit Type	Occurrence in All Deals
5-3	32,726,725,930,490,895,646,447,104,000
5-2	31,239,147,479,104,945,844,335,872,000

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.

3. More 6-4 fits than straight flushes

This problem is like comparing apples to oranges, so 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. First consider the difficult task of counting straight flushes, and begin by considering 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.

Length	Holdings	1	2	3	4	5	6	7	8	9	10
5	1287	10	0	0	0	0	0	0	0	0	0
6	1716	62	9	0	0	0	0	0	0	0	0
7	1716	162	47	8	0	0	0	0	0	0	0
8	1287	230	100	34	7	0	0	0	0	0	0
9	715	188	111	56	23	6	0	0	0	0	0
10	286	70	72	46	27	14	5	0	0	0	0
11	78	6	12	22	16	10	7	4	0	0	0
12	13	0	0	0	0	6	2	2	3	0	0
13	1	0	0	0	0	0	0	0	0	0	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
Straight flushes in all hands:			12,578,299,800

Counting the number of 6-4 fits is analogous to the previous problem:

Fit Type	Combinations	Ways	Fits in All Sides
6-4	13c6 × 39c7 × 7c4 × 32c9	4 × 4	414,574,312,849,272,576,000

For a comparison in all deals, multiply the number of straight flushes by 39!/(13!)³ × 4 (hands in all deals) and the number of 6-4 fits by 26c13 (combinations to split the remaining cards into two hands) and the winner appears:

Comparison	Occurrence in All Deals
Straight flushes	4,250,363,376,377,263,298,186,880,000
6-4 fits	4,311,821,598,220,144,353,945,600,000

So there you have it! Proof that bridge is superior to poker. Why settle for 4.25 octillion straight flushes when you can get 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.

4. Equal number of 7-card and 6-card fits

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 really want to know:

Fit Length	Combinations	Ways	Fits in All Sides
7 cards	13c7 × 39c19 × 26c13	4 × 2	9,840,824,874,877,284,288,000
6 cards	13c6 × 39c20 × 26c13	4 × 2	9,840,824,874,877,284,288,000

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.

To upscale to deals, simply multiply by 26c13 (combinations to split the remaining cards into two hands) though ‘simply’ may be an overbid. Most calculators would resort to scientific notation, but every digit is honored here:

Fit Length	Occurrence in All Deals
7 cards	102,350,483,193,648,682,965,772,800,000
6 cards	102,350,483,193,648,682,965,772,800,000

With over 102 octillion 7-card fits — an average of about two per deal — the challenge of finding an 8-card trump fit is magnified. That’s a heap to avoid!

The equality of fit lengths clearly holds for any two lengths that add to 13, although counts are reduced as the lengths diverge. For example, the number of 8-card fits and 5-card fits are equal, with about 69.5 octillion each.

5. More voids than hands with 3-4 aces

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.

Length	Combinations	Ways	Voids in All Hands
0	13c0 × 39c13	4	32,489,701,776

Counting hands with 3-4 aces in done separately — first three, then four — to ensure no redundancy:

Aces	Combinations	Ways	Hands
3	4c3 × 48c10	1	26,162,863,584
4	4c4 × 48c9	1	1,677,106,640
Hands with 3-4 aces			27,839,970,224

For the final answer, multiply by 39!/(13!)³ × 4 (hands in all deals) to produce:

Comparison	Occurrence in All Deals
Void suits	10,978,672,851,964,438,613,962,905,600
Hands with 3-4 aces	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…

6. More inches than 5-5 fits

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 Type	Combinations	Ways	Fits in All Sides
5-5	13c5 × 39c8 × 8c5 × 31c8	4 × 2	279,837,661,173,258,988,800

Finding the number of inches to Einstein Cross Quasar is straight multiplication: 8,000,000,000 light-years × 9,460,730,472,580,800 meters/light-year × 39.37007874 inches/meter. Yes, and it emphasizes how useful my work can be… No need to call Elon Musk!

Multiplying the fit count by 26c13 (combinations to split the remaining cards into two hands) gives the following:

Comparison	Final Count
5-5 fits in all deals	2,910,479,578,798,597,438,913,280,000
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.

7. More angstroms than 3-card suits

The counting of 3-card suits was already shown in Problem 1 and repeated here:

Length	Combinations	Ways	Suits in All Hands
3	13c3 × 39c10	4	727,292,733,024

Angstroms to Owl Nebula, albeit ridiculous, is another straight multiplication: 2600 light-years × 9,460,730,472,580,800 meters/light-year × 10,000,000,000 angstroms/meter.

Multiplying the number of 3-card suits by 39!/(13!)³ × 4 (hands in all deals) gives the following:

Comparison	Final Count
3-card suits in all deals	245,761,227,312,338,046,214,479,974,400
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: 2.45980E+29 angstroms.

Sorry, Peter, 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!

8. More picometers than touching cards

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).

Length	Holdings	1	2	3	4	5	6	7	8	9	10	11	12
2	78	12	0	0	0	0	0	0	0	0	0	0	0
3	286	110	11	0	0	0	0	0	0	0	0	0	0
4	715	360	135	10	0	0	0	0	0	0	0	0	0
5	1287	504	504	144	9	0	0	0	0	0	0	0	0
6	1716	280	700	560	140	8	0	0	0	0	0	0	0
7	1716	42	315	700	525	126	7	0	0	0	0	0	0
8	1287	0	21	210	525	420	105	6	0	0	0	0	0
9	715	0	0	0	70	280	280	80	5	0	0	0	0
10	286	0	0	0	0	0	84	144	54	4	0	0	0
11	78	0	0	0	0	0	0	0	45	30	3	0	0
12	13	0	0	0	0	0	0	0	0	0	11	2	0
13	1	0	0	0	0	0	0	0	0	0	0	0	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
Touching cards in all hands			1,792,979,462,400

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? Evidently there’s some underlying truth about touching cards that eludes me. It certainly raised some doubts about my method, but verification by other means produced identical counts.

Picometers to Aldebaran, like previous cosmic nonsense, is a straight multiplication: 65 light-years × 9,460,730,472,580,800 meters/light-year × 1,000,000,000,000 picometers/meter. Relax! We’re almost done.

Multiplying the touching-card count by 39!/(13!)³ × 4 (hands in all deals) gives the following:

Comparison	Final Count
Touching cards in all deals	605,869,979,469,049,895,596,093,440,000
Picometers to Aldebaran	614,947,480,717,752,000,000,000,000,000

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.

Octillitis

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.

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© 2021 Richard Pavlicek