Analysis 7A31 Main |
| by Richard Pavlicek |
Perhaps the most written-up deal in bridge history is the grand slam that decided the 1975 Bermuda Bowl, the tournament of infamy that rocked the bridge world. I wont rekindle the nefarious side issues, which are readily available from many sources, including my May 2006 bidding poll Italy Wins in Bermuda. This look back will focus on bridge, or more specifically, the play of that fateful grand.
On the famous deal Italy bid to an egregious 7 , which was blessed with a lucky lie, allowing the contract to make. Had 7 failed, North America would have won the Bermuda Bowl; and even on the fortuitous layout, a stronger American defense might still have defeated the grand. This was the fateful Board 92:
Q J 8 | West | North | East | South | ||
N-S vul | A J 9 6 5 | Kantar | Garozzo | Eisenberg | Belladonna | |
K 8 2 | Pass | 2 | ||||
A Q | Pass | 2 | Pass | 2 | ||
7 6 5 2 | 4 3 | Pass | 3 | Pass | 3 NT | |
K 4 3 2 | Q 10 8 7 | Pass | 4 | Pass | 4 | |
J 5 3 | Q 10 6 4 | Pass | 4 NT | Pass | 5 | |
K 10 | 7 5 4 | Pass | 5 | Dbl | Rdbl | |
A K 10 9 | Pass | 5 | Pass | 5 NT | ||
| Pass | 7 | All Pass | |||
A 9 7 | ||||||
7 South | J 9 8 6 3 2 |
The 2 opening was natural (Precision) and 2 was an artificial force. The next four bids were natural, then 4 began a control-showing sequence; 4 NT was not Blackwood per Italian tradition, after which some of the actions are mysterious not just to me but apparently to the Italians as well, considering the final bid.
Eisenbergs double of 5 was strange, because his diamond holding suggests that lead would be more effective, perhaps setting up a trick before dummys heart suit could be established. Was he just antsy from passing too much? Conflicting accounts also have Eisenberg doubling 5 , but this is incredulous; Broadway Billy may march to a different drummer, but he never lost his drumsticks.
Kantar dutifully led a heart, which Belladonna ruffed and led a club to the 10, queen
Game over!
Much speculation has been made on the outcome if Kantar had played the king on the first club lead, suggesting a singleton. Belladonna would then have an alternate line to exploit a trump coup against Easts presumed 10-7-5-4. The catch of course is that it would be likely to fail if Kantar had played the king from a doubleton, a risk-free falsecard. Belladonna remarked to the media shortly after that he would have played for the coup (going down on the layout) but this may have been influenced by celebration and too much vino di Italia.
Analysis 7A31 Main | Top Grandiose Nostalgia |
Recently a longtime cyber friend, Charles Blair, asked if I thought Belladonna should have played for the trump coup if the K had appeared. This is subjective of course, as it depends on how often the K is played from a doubleton. Assuming say 50-50, my gut reaction favored the coup line, since it would sometimes work against king-doubleton. Charles felt otherwise, convinced that drawing trumps was right. Considering Charles is a mathematician (retired, U. of Illinois) I didnt like my chances, but it piqued my interest to do a detailed analysis.
For the first analysis I will ignore the double of 5 (which shouldnt happen anyway) and assume a nondescript diamond lead, the only unshown suit.* Declarer would win in hand with the A and lead a club (king, ace) then take one of two lines: (A) Cash the Q; if West shows out, East must be 4=3=2=4, else failure. (B) Cash the A (diamond pitch), ruff a heart, cross to the Q and ruff another heart. If either opponent shows out, win the Q (coup wont work); else cash two spades ending in dummy, ruff a heart, cross to the K, ruff a diamond, then high-crossruff Q-J.
*The reason for changing the lead is to achieve an unbiased state. When a defender chooses to lead dummys suit it portends foul distribution, while choosing to lead the unshown suit is innocuous.
The following table compares Line A and Line B when the K is singleton, hence East holds 10-7-5-4. For each case that allows success, the number of hands is calculated by combinatorics, and the totals show the number of hands on which each line succeeds.
Case | West | East | Line A Success | Line B Success |
---|---|---|---|---|
1 | 1=5=6=1 | 5=3=1=4 | 0 | 0 |
2 | 1=6=5=1 | 5=2=2=4 | 0 | 0 |
3 | 2=4=6=1 | 4=4=1=4 | 0 | 0 |
4 | 2=5=5=1 | 4=3=2=4 | 15·56·21 = 17,640 | 15·56·21 = 17,640 |
5 | 2=6=4=1 | 4=2=3=4 | 0 | 0 |
6 | 3=3=6=1 | 3=5=1=4 | 0 | 0 |
7 | 3=4=5=1 | 3=4=2=4 | 0 | 20·70·21 = 29,400 |
8 | 3=5=4=1 | 3=3=3=4 | 0 | 20·56·35 = 39,200 |
9 | 3=6=3=1 | 3=2=4=4 | 0 | 0 |
10 | 4=2=6=1 | 2=6=1=4 | 0 | 0 |
11 | 4=3=5=1 | 2=5=2=4 | 0 | 0 |
12 | 4=4=4=1 | 2=4=3=4 | 0 | 0 |
13 | 4=5=3=1 | 2=3=4=4 | 0 | 0 |
14 | 4=6=2=1 | 2=2=5=4 | 0 | 0 |
15 | 5=2=5=1 | 1=6=2=4 | 0 | 0 |
16 | 5=3=4=1 | 1=5=3=4 | 0 | 0 |
17 | 5=4=3=1 | 1=4=4=4 | 0 | 0 |
18 | 5=5=2=1 | 1=3=5=4 | 0 | 0 |
19 | 5=6=1=1 | 1=2=6=4 | 0 | 0 |
Total successful hands | 17,640 | 86,240 |
To simplify the analysis I eliminated all East-West patterns with a singleton heart or any void. Besides being rare and of little significance, these cases would often evolve differently; e.g., a bid with 7+ cards in a red suit, or a Lightner double.
The next table compares Line A and Line B when the K is doubleton, which comprises four holdings as West can have K-10, K-7, K-5 or K-4. Line A always works. Line B only works with spades 3-3, or if a 6-2 heart break dictates abandonment.
Case | West | East | Line A Success | Line B Success |
---|---|---|---|---|
20 | 1=4=6=2 | 5=4=1=3 | 6·70·7·4 = 11,760 | 0 |
21 | 1=5=5=2 | 5=3=2=3 | 6·56·21·4 = 28,224 | 0 |
22 | 1=6=4=2 | 5=2=3=3 | 6·28·35·4 = 23,520 | 6·28·35·4 = 23,520 |
23 | 2=3=6=2 | 4=5=1=3 | 15·56·7·4 = 23,520 | 0 |
24 | 2=4=5=2 | 4=4=2=3 | 15·70·21·4 = 88,200 | 0 |
25 | 2=5=4=2 | 4=3=3=3 | 15·56·35·4 = 117,600 | 0 |
26 | 2=6=3=2 | 4=2=4=3 | 15·28·35·4 = 58,800 | 15·28·35·4 = 58,800 |
27 | 3=2=6=2 | 3=6=1=3 | 20·28·7·4 = 15,680 | 20·28·7·2 = 7840 |
28 | 3=3=5=2 | 3=5=2=3 | 20·56·21·4 = 94,080 | 20·56·21·3 = 70,560 |
29 | 3=4=4=2 | 3=4=3=3 | 20·70·35·4 = 196,000 | 20·70·35·4 = 196,000 |
30 | 3=5=3=2 | 3=3=4=3 | 20·56·35·4 = 156,800 | 20·56·35·4 = 156,800 |
31 | 3=6=2=2 | 3=2=5=3 | 20·28·21·4 = 47,040 | 20·28·21·4 = 47,040 |
32 | 4=2=5=2 | 2=6=2=3 | 15·28·21·4 = 35,280 | 15·28·21·2 = 17,640 |
33 | 4=3=4=2 | 2=5=3=3 | 15·56·35·4 = 117,600 | 0 |
34 | 4=4=3=2 | 2=4=4=3 | 15·70·35·4 = 147,000 | 0 |
35 | 4=5=2=2 | 2=3=5=3 | 15·56·21·4 = 70,560 | 0 |
36 | 4=6=1=2 | 2=2=6=3 | 15·28·7·4 = 11,760 | 15·28·7·4 = 11,760 |
37 | 5=2=4=2 | 1=6=3=3 | 6·28·35·4 = 23,520 | 0 |
38 | 5=3=3=2 | 1=5=4=3 | 6·56·35·4 = 47,040 | 0 |
39 | 5=4=2=2 | 1=4=5=3 | 6·70·21·4 = 35,280 | 0 |
40 | 5=5=1=2 | 1=3=6=3 | 6·56·7·4 = 9408 | 0 |
Total successful hands | 1,358,672 | 589,960 |
Line B notes: Cases 27 and 32 fail against K-10 or K-7 (overruff) so the club factor is reduced to 2. Case 28 fails against K-10 (overruff) so the club factor is reduced to 3.
The above tables show that Line A succeeds against 1358672 hands with king-doubleton, and 17640 hands with king-singleton. Line B succeeds against 589960 hands with king-doubleton, and 86240 hands with king-singleton. Let p = the probability of king-doubleton after the king is seen. The proportion of Line A success is then 1358672p + 17640(1-p), and Line B is 589960p + 86240(1-p). Equating the two will discover the value of p for which the lines are equivalent, hence:
1358672p + 17640(1-p) = 589960p + 86240(1-p)
Solution: p = 5/42, or approximately 0.119 or 11.9 percent
This means that if the king (once seen) would be doubleton 11.9 percent of the time, the lines are equal. Anything higher would favor Line A; anything lower would favor Line B.
The above does not indicate how often the king should be played from a doubleton, but this can be calculated. The probability of a specific doubleton (missing five cards) is 39/1150, so the probability of any king-doubleton is 4×39/1150 = 78/575. Now suppose the king is played 1/t of the time, which makes the probability of king from king-doubleton (1/t)×(78/575) = 78/(575t). The probability of a singleton king is 13/460. Therefore, the ratio of king from king-doubleton to the whole (king from king-doubleton + singleton king) must equal the 5/42 probability obtained above, hence:
78/(575t)/(78/(575t)+13/460) = 5/42
Solution: t = 888/25, or 1/t = 25/888
This means that if the king is played 25/888 of the time from king-doubleton, the lines are equal. Anything more frequent (say 1/35 of time) would favor Line A; anything less frequent (say 1/36 of time) would favor Line B. Charles was certainly on the mark, as Line A is better if the falsecard is made as little as once in 35 times. The flaw in my initial assessment was to compare a blank king with the actual K-10 doubleton, rather than all king-doubletons, which quadruples the weight of those cases.
The conclusion is clear. In an expert game the falsecard would be made far more often than 1/35 of the time, probably over half the time, so playing for the coup is a long shot. Basic Bridge 101: Draw trumps!
Analysis 7A31 Main | Top Grandiose Nostalgia |
Now lets consider the impact of Eisenbergs double of 5 . Does it affect the calculations? And if so, which line does it favor? And theres also the real-life possibility that it may have affected Kantar, distracting him from his usual sterling play.
Clearly the double affects the plausible heart distributions, of which East should have at least four. This is confirmed by Wests 2 lead (fourth-best or low from three) which marks East with four or five hearts; hence all other heart layouts will be ignored. Further, Easts double must show strength in hearts, so Ive halved the number of heart combinations to allow for the occasions where East would not be strong enough to double hardly an exact adjustment but a good approximation.
The lines of play are altered slightly after a heart lead, but essentially the same. Declarer begins by ruffing the first trick and leading a club (king, ace) then Line A is the same. Line B only varies by cashing three rounds of spades after ruffing the third round of hearts, and cashing the A at some point along the way.
As before, the first table compares Line A and Line B when the K is singleton. Almost a wipeout, as the heart layout restrictions prevent Line A from ever working, and Line B only works in Case 7.
Case | West | East | Line A Success | Line B Success |
---|---|---|---|---|
3 | 2=4=6=1 | 4=4=1=4 | 0 | 0 |
6 | 3=3=6=1 | 3=5=1=4 | 0 | 0 |
7 | 3=4=5=1 | 3=4=2=4 | 0 | 20·35·21 = 14,700 |
11 | 4=3=5=1 | 2=5=2=4 | 0 | 0 |
12 | 4=4=4=1 | 2=4=3=4 | 0 | 0 |
16 | 5=3=4=1 | 1=5=3=4 | 0 | 0 |
17 | 5=4=3=1 | 1=4=4=4 | 0 | 0 |
Total successful hands | 0 | 14,700 |
Case numbering is kept the same as in previous tables, but patterns with East having 2, 3 or 6 hearts (deemed impossible) are removed.
The next table compares Line A and Line B when the K is doubleton. Line A works against all cases deemed possible, while Line B succeeds only in Case 28 and 29.
Case | West | East | Line A Success | Line B Success |
---|---|---|---|---|
20 | 1=4=6=2 | 5=4=1=3 | 6·35·7·4 = 5880 | 0 |
23 | 2=3=6=2 | 4=5=1=3 | 15·28·7·4 = 11,760 | 0 |
24 | 2=4=5=2 | 4=4=2=3 | 15·35·21·4 = 44,100 | 0 |
28 | 3=3=5=2 | 3=5=2=3 | 20·28·21·4 = 47,040 | 20·28·21·3 = 35,280 |
29 | 3=4=4=2 | 3=4=3=3 | 20·35·35·4 = 98,000 | 20·35·35·4 = 98,000 |
33 | 4=3=4=2 | 2=5=3=3 | 15·28·35·4 = 58,800 | 0 |
34 | 4=4=3=2 | 2=4=4=3 | 15·35·35·4 = 73,500 | 0 |
38 | 5=3=3=2 | 1=5=4=3 | 6·28·35·4 = 23,520 | 0 |
39 | 5=4=2=2 | 1=4=5=3 | 6·35·21·4 = 17,640 | 0 |
Total successful hands | 380,240 | 133,280 |
Line B note: Case 28 fails against K-10 (overruff) so the club factor is reduced to 3.
The revised tables show that Line A succeeds against 380240 hands with king-doubleton, and none with king-singleton. Line B succeeds against 133280 hands with king-doubleton, and 14700 hands with king-singleton. Plugging the new values into the first equation gives:
380240p = 133280p + 14700(1-p)
Solution: p = 5/89, or approximately 0.056 or 5.6 percent
This means that if the king (once seen) would be doubleton 5.6 percent of the time, the lines are equal. Anything higher would favor Line A; anything lower would favor Line B.
Applying the 5/89 probability to the second equation gives:
78/(575t)/(78/(575t)+13/460) = 5/89
Solution: t = 2016/25, or 1/t = 25/2016
Quite a difference. The lines are now equal if the king is played 25/2016 of the time from king-doubleton. Line A is better if the king is played at least once in 80 times; Line B is better if the king is played at most once in 81 times. Evidently Eisenbergs double was even more wayward than it looked, diminishing the chance that Belladonna would have pursued the ill-fated coup if given the opportunity.
As a final note, appropriate for this Bermuda Bowls sideshows, on the first club lead Kantar should have tossed the king in the air exclaiming Rabbis rule! with Eisenberg adding Dont worry, pard, Ive got you covered.
Analysis 7A31 Main | Top Grandiose Nostalgia |
© 2014 Richard Pavlicek