Talmud for Ketubot 10:3
בראשונה גזרו שמד ביהודה שכן מסורת להם מאבותם שיהודה הרג את עשו דכתיב (בראשית מ״ט:ח׳) ידך בעורף אויביך והיו הולכין ומשעבדין בהן ואונסין את בנותיהן וגזרו שיהא איסטרטיוס בועל תחילה התקינו שיהא בעלה בא עליה עודה בבית אביה שמתוך שהיא יודעת שאימת בעלה עליה עוד היא נגררת מכל מקום אין סופה להיבעל מאיסטרטיוס אנוסה היא ואנוסה מותרת לביתה כהנות מה היו עושות מטמינות היו ויטמינו אף בנות ישראל קול יוצא ומלכותא שמעה ואילין ואילין מתערבבין מה סימן היה להן קול מגרוס בעיר משתה שם משתה שם אור הנר בברור חיל שבוע בן שבוע בן אע"פ שבטל השמד המנהג לא בטל כלתו של רבי הושעיה נכנסה מעוברת תני ולאלמנה מאתים תני אחד אלמנת ישראל ואחד אלמנת כהנים
בראשונה גזרו שמד ביהודה שכן מסורת להם מאבותם שיהודה הרג את עשו דכתיב (בראשית מ״ט:ח׳) ידך בעורף אויביך והיו הולכין ומשעבדין בהן ואונסין את בנותיהן וגזרו שיהא איסטרטיוס בועל תחילה התקינו שיהא בעלה בא עליה עודה בבית אביה שמתוך שהיא יודעת שאימת בעלה עליה עוד היא נגררת מכל מקום אין סופה להיבעל מאיסטרטיוס אנוסה היא ואנוסה מותרת לביתה כהנות מה היו עושות מטמינות היו ויטמינו אף בנות ישראל קול יוצא ומלכותא שמעה ואילין ואילין מתערבבין מה סימן היה להן קול מגרוס בעיר משתה שם משתה שם אור הנר בברור חיל שבוע בן שבוע בן אע"פ שבטל השמד המנהג לא בטל כלתו של רבי הושעיה נכנסה מעוברת תני ולאלמנה מאתים תני אחד אלמנת ישראל ואחד אלמנת כהנים
Jerusalem Talmud Ketubot
MISHNAH: If somebody was married to three women when he died, the ketubah of one was a mina59100 zuz., of the other one 200, and of the other one 300. If the estate is only 100, they divide equally. If it was 200, the one who claims a mina takes 50, and those who claim 200 and 300 each take three gold denars601 gold denar was worth 25 silver denars. 3 gold denars equal 75 zuz, 6 gold denars 150 zuz.. If it was 300, the one who claims a mina takes 50, the one who claims 200 takes a mina59100 zuz., and the one who claims 300 takes six gold denars601 gold denar was worth 25 silver denars. 3 gold denars equal 75 zuz, 6 gold denars 150 zuz.. Similarly, if three who invested together lost or gained they would split in this manner61This refers to the proportional distribution of the last case. Loss and gain of a stock company are distributed per share..62This Mishnah poses difficult problems. First, it deals with a scenario that cannot arise since Mishnah 1 stated that a ketubah which precedes in time has to be satisfied from the estate before the later ketubah can be claimed. Therefore, the case of the Mishnah presupposes that the husband married three woman simultaneously. While this is possible in theory, it is excluded in practice. The problem is a practical one if applied to the case that the amount available in bankruptcy proceedings is not sufficient to cover all claims of equal rank.
At first glance, it seems that the Mishnah is inconsistent, applying three different rules to three different cases. Since the Mishnah must guide the judge to correctly apportion payment of claims c1, c2,c3 if the available amount a < c1+ c2+ c3, for all a between 0 and c1+ c2+ c3, it is obvious that the Mishnah has to be explained by one single consistent formula or algorithm. In addition, the algorithm must be applicable to any number n of claims.
In the tenth Century, Saadya Gaon suceeded in reducing the number of different rules employed in the Mishnah to two (Oṣar Hageonim Ketubot p. 310). A complete discussion of the Geonic treatment of the Mishnah was given by I. Francus, שיטת הגאונים בפירוש משנה וברייתא, סיני פכו-פכז (תשס-תשסא) קצה-ריב (M. H. Katzenellenbogen Memorial Volume).
The first complete solution of the problem of finding a uniform rule for all three cases was given by R. J. Aumann and M. Maschler, Game Theoretic Analysis of a Bankruptcy Problem from the Talmud, J. Econom. Theory 36(1985) 195–213. In the framework of the theory of cooperative games, the authors show that the distribution of the Mishnah corresponds to the nucleolus strategy in game theory. Using only elementary mathematics, a single distribution formula covering all cases was given by M. Balinski in: Quelle équité, Pour la science 311 (September 2003) 82–87; What is Just? Am. Math. MONTHLY 112(2005), 502–511. While these authors succeed in explaining the Mishnah in itself, they do not explain the interpretation given to the Mishnah by the Talmudim. Nor do they explain why in both Talmudim the Mishnah is rejected out of hand. The explanation of the Babli is in language close to the Yerushalmi but translated into mathematical formulas the results are quite different; in both Talmudim the proposed explanation results in a procedure which is not monotone; i. e., it is possible that if the available amount a is increased, the payouts for some claimants are decreased. The solutions both of Aumann/Maschler and Balinski are monotone, increasing with a.
Balinski shows that the Mishnah can be understood in terms of a different distribution problem in Mishnah Baba meṣiʻa 1:1, as indicated by Alfasi (Ketubot 10, 51b in the Wilna ed.). Since this problem is also the basis of the arguments in both Talmudim, Balinski’s method is explained first. In Baba meṣiʻa there are two claimants for an amount of 1. The first claims it all, the second claims ½. Both claims are of the same status. Then the first is awarded 3/4, the second 1/4. While the claims are in ratio 2:1, the awards are in ratio 3:1. The reason given is the rule ascribed to Symmachos, a student of R. Meïr, that “money in doubt is split evenly” [Babli Baba meṣiʻa 2b, Baba qama 35b; Yerushalmi Baba meṣiʻa 8:5 (11d 1. 18); cf. Chapter 2, Notes 9 ff.]. Since the second claimant asks only for ½, the other half is awarded to the first claimant. The first half is in doubt; therefore it is split evenly between the two parties, resulting in the awards mentioned.
In order to arrive at a mathematical formulation, assume that n claims ci are advanced,
c1< c2 < … < cn.
If several claims were identical they would be consolidated into a single claim and the award split evenly between the participants. For any amount a available for distribution, the payout for claimant i is pi(a),
p1(a)+ p2(a)+ … + pn(a)= a.
We also introduce the sum of the claims, S =c1 + c2+ … + cn and M= S/2. For n = 2, Symmachos’s argument determines the payouts uniquely:
If a ≤ c1, the entire sum is in dispute,
p1(a)= p2(a)= a/2.
Therefore, p1(c1)= p2(c1)= c1/2.
If c1 < a ≤ c2, the amount a - c1 belongs to claimant 2 and
p1(a)= c1/2,
p2(a)= a-c1+c1/2= a-c1/2,
and p1(c2) = c1/2,p2(c2)= c2-c1/2.
If c2< a, both claimants have equal rights to the excess over c2,
p1(a)= c1/2+(a - c2)/2
p2(a) = c2-c1/2+(a-c2)/2=(a+c2-c1)/2.
If a = c2+c1 = S, all claims are satisfied in full. Using the function min(x,y) = smaller of x or y, Balinski did express the formulas given above by
(1) pi(a) = min(ci/2, λ) where λ is chosen so that p1(a)+ p2(a)= a;i = 1,2.
An inspection of the formulas shows that the distribution is symmetric about the middle value of the claims:
(2) pi(M-x)+ pi(M+x)= ci.
This means that the distribution for M ≤ a ≤ S is determined if it is determined for 0 ≤ a≤M. In the Mishnah here, it should be sufficient to indicate the distribution up to an estate of 300 if the sum of claims is 600. But the generalization of the procedure for n = 2 to n > 2 is not trivial. As Balinski points out, an arbitrary distribution function which yields the given values for a = ci and satisfies (2) is compatible with the Mishnah. His own solution is to postulate (1) for i = 1, … n. Then (2) is satisfied and the distribution is monotone. For example,
{p1(400),p2(400),p3(400)} = {100-p1(200),200-p2(200),300-p3(200)} = {50,125,225}.
The problem with this solution, as with most other solutions proposed in the last 1000 years, is that the Talmudim take another view. The solution of the Babli is easily described as a recursive computational procedure. We assume that p1(a),…,pi-1(a) have been determined; leaving an amount ai to be distributed. We determine pi(ai). If ci ≥ ai, pj(ai) = ai/(n-i) for j = i, … n. Otherwise, all claims larger than ci are consolidated into one claim Ci = ci+1 + … + cn and pi(ai) is the payout assigned to claimant #i in the two-person distribution problem for claims ci, Ci and amount ai. The fictitious second claimant then receives Pi(ai) = ai+1 and the process continues.
In the example of the Mishnah, c1 = 100, c2 = 200, c3 = 300. For a = 100, by the first alternative pj(100) = 100/3 for all j. For a = 200, c1 = 100, C1 = 400. Therefore, 100 is assigned to the second party and 100 is split evenly. This yields p1(200) = 50, P1(200) = 150. Since 200 > 150, the latter amount is split evenly, resulting in p2(200) = p3(200) = 75. For a = 300, one sees that p1(300) = 50, P1(300) = 250; p2(300) = 100 = 200/2, p3(300) = 50 + 200/2 = 150. This explains the numbers stated in the Mishnah, but what about a = 101? In that case, p1(101) = 50, P1(101) = 51; p2(101) = p3(101) = 51/2 = 25.5. This clearly is unacceptable; the procedure of the Mishnah has to be rejected.
The explanation of the Yerushalmi (cf. Note 69) is not so clearly stated but it seems to be the following: The procedure also is recursive. If ci ≥ ai,pj(ai) = ai/(n-i) for j = i, … n. Otherwise, one considers all two-person problems between claimants i and i+1,…, n. In praxi, this means that one has only to split between ci and cn. If ci < ai < cn then pi(ai) = ci/2. If cn < ai then write ai = cn + δ. Claims n and i overlap to the amount ci - δ. This means that
pi(ai) = δ + (ci - δ)/2 = (ci + δ)/2.
For a ≤ M, the numbers obtained for Babli and Yerushalmi coincide; the Yerushalmi’s method also is not monotone. The results for a > M disagree. For example, for the data of Mishnah 4, the Yerushalmi gives
p1(400)=100, p2(400)=100, p3(400)=200 while the Babli gives
p1(400)=50, P1(400)=350; p2(400)=100, p3(400)=250.
At first glance, it seems that the Mishnah is inconsistent, applying three different rules to three different cases. Since the Mishnah must guide the judge to correctly apportion payment of claims c1, c2,c3 if the available amount a < c1+ c2+ c3, for all a between 0 and c1+ c2+ c3, it is obvious that the Mishnah has to be explained by one single consistent formula or algorithm. In addition, the algorithm must be applicable to any number n of claims.
In the tenth Century, Saadya Gaon suceeded in reducing the number of different rules employed in the Mishnah to two (Oṣar Hageonim Ketubot p. 310). A complete discussion of the Geonic treatment of the Mishnah was given by I. Francus, שיטת הגאונים בפירוש משנה וברייתא, סיני פכו-פכז (תשס-תשסא) קצה-ריב (M. H. Katzenellenbogen Memorial Volume).
The first complete solution of the problem of finding a uniform rule for all three cases was given by R. J. Aumann and M. Maschler, Game Theoretic Analysis of a Bankruptcy Problem from the Talmud, J. Econom. Theory 36(1985) 195–213. In the framework of the theory of cooperative games, the authors show that the distribution of the Mishnah corresponds to the nucleolus strategy in game theory. Using only elementary mathematics, a single distribution formula covering all cases was given by M. Balinski in: Quelle équité, Pour la science 311 (September 2003) 82–87; What is Just? Am. Math. MONTHLY 112(2005), 502–511. While these authors succeed in explaining the Mishnah in itself, they do not explain the interpretation given to the Mishnah by the Talmudim. Nor do they explain why in both Talmudim the Mishnah is rejected out of hand. The explanation of the Babli is in language close to the Yerushalmi but translated into mathematical formulas the results are quite different; in both Talmudim the proposed explanation results in a procedure which is not monotone; i. e., it is possible that if the available amount a is increased, the payouts for some claimants are decreased. The solutions both of Aumann/Maschler and Balinski are monotone, increasing with a.
Balinski shows that the Mishnah can be understood in terms of a different distribution problem in Mishnah Baba meṣiʻa 1:1, as indicated by Alfasi (Ketubot 10, 51b in the Wilna ed.). Since this problem is also the basis of the arguments in both Talmudim, Balinski’s method is explained first. In Baba meṣiʻa there are two claimants for an amount of 1. The first claims it all, the second claims ½. Both claims are of the same status. Then the first is awarded 3/4, the second 1/4. While the claims are in ratio 2:1, the awards are in ratio 3:1. The reason given is the rule ascribed to Symmachos, a student of R. Meïr, that “money in doubt is split evenly” [Babli Baba meṣiʻa 2b, Baba qama 35b; Yerushalmi Baba meṣiʻa 8:5 (11d 1. 18); cf. Chapter 2, Notes 9 ff.]. Since the second claimant asks only for ½, the other half is awarded to the first claimant. The first half is in doubt; therefore it is split evenly between the two parties, resulting in the awards mentioned.
In order to arrive at a mathematical formulation, assume that n claims ci are advanced,
c1< c2 < … < cn.
If several claims were identical they would be consolidated into a single claim and the award split evenly between the participants. For any amount a available for distribution, the payout for claimant i is pi(a),
p1(a)+ p2(a)+ … + pn(a)= a.
We also introduce the sum of the claims, S =c1 + c2+ … + cn and M= S/2. For n = 2, Symmachos’s argument determines the payouts uniquely:
If a ≤ c1, the entire sum is in dispute,
p1(a)= p2(a)= a/2.
Therefore, p1(c1)= p2(c1)= c1/2.
If c1 < a ≤ c2, the amount a - c1 belongs to claimant 2 and
p1(a)= c1/2,
p2(a)= a-c1+c1/2= a-c1/2,
and p1(c2) = c1/2,p2(c2)= c2-c1/2.
If c2< a, both claimants have equal rights to the excess over c2,
p1(a)= c1/2+(a - c2)/2
p2(a) = c2-c1/2+(a-c2)/2=(a+c2-c1)/2.
If a = c2+c1 = S, all claims are satisfied in full. Using the function min(x,y) = smaller of x or y, Balinski did express the formulas given above by
(1) pi(a) = min(ci/2, λ) where λ is chosen so that p1(a)+ p2(a)= a;i = 1,2.
An inspection of the formulas shows that the distribution is symmetric about the middle value of the claims:
(2) pi(M-x)+ pi(M+x)= ci.
This means that the distribution for M ≤ a ≤ S is determined if it is determined for 0 ≤ a≤M. In the Mishnah here, it should be sufficient to indicate the distribution up to an estate of 300 if the sum of claims is 600. But the generalization of the procedure for n = 2 to n > 2 is not trivial. As Balinski points out, an arbitrary distribution function which yields the given values for a = ci and satisfies (2) is compatible with the Mishnah. His own solution is to postulate (1) for i = 1, … n. Then (2) is satisfied and the distribution is monotone. For example,
{p1(400),p2(400),p3(400)} = {100-p1(200),200-p2(200),300-p3(200)} = {50,125,225}.
The problem with this solution, as with most other solutions proposed in the last 1000 years, is that the Talmudim take another view. The solution of the Babli is easily described as a recursive computational procedure. We assume that p1(a),…,pi-1(a) have been determined; leaving an amount ai to be distributed. We determine pi(ai). If ci ≥ ai, pj(ai) = ai/(n-i) for j = i, … n. Otherwise, all claims larger than ci are consolidated into one claim Ci = ci+1 + … + cn and pi(ai) is the payout assigned to claimant #i in the two-person distribution problem for claims ci, Ci and amount ai. The fictitious second claimant then receives Pi(ai) = ai+1 and the process continues.
In the example of the Mishnah, c1 = 100, c2 = 200, c3 = 300. For a = 100, by the first alternative pj(100) = 100/3 for all j. For a = 200, c1 = 100, C1 = 400. Therefore, 100 is assigned to the second party and 100 is split evenly. This yields p1(200) = 50, P1(200) = 150. Since 200 > 150, the latter amount is split evenly, resulting in p2(200) = p3(200) = 75. For a = 300, one sees that p1(300) = 50, P1(300) = 250; p2(300) = 100 = 200/2, p3(300) = 50 + 200/2 = 150. This explains the numbers stated in the Mishnah, but what about a = 101? In that case, p1(101) = 50, P1(101) = 51; p2(101) = p3(101) = 51/2 = 25.5. This clearly is unacceptable; the procedure of the Mishnah has to be rejected.
The explanation of the Yerushalmi (cf. Note 69) is not so clearly stated but it seems to be the following: The procedure also is recursive. If ci ≥ ai,pj(ai) = ai/(n-i) for j = i, … n. Otherwise, one considers all two-person problems between claimants i and i+1,…, n. In praxi, this means that one has only to split between ci and cn. If ci < ai < cn then pi(ai) = ci/2. If cn < ai then write ai = cn + δ. Claims n and i overlap to the amount ci - δ. This means that
pi(ai) = δ + (ci - δ)/2 = (ci + δ)/2.
For a ≤ M, the numbers obtained for Babli and Yerushalmi coincide; the Yerushalmi’s method also is not monotone. The results for a > M disagree. For example, for the data of Mishnah 4, the Yerushalmi gives
p1(400)=100, p2(400)=100, p3(400)=200 while the Babli gives
p1(400)=50, P1(400)=350; p2(400)=100, p3(400)=250.
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