[Avodah] Is it better for one person to do a vadai mitzva...

[Chana Luntz]

RSM writes:

> I wrote
> >>
> Asei doche lo taaseh even if one could somehow both do a 
> mitzva and avoid the lo taaseh. Tzitzit is doche k'laim, and 
> wollen tzitzit can be put on a linen garment, even if linen 
> tzitzit (without t'chelet) could possibly be used, both 
> performing a mitzva (in perhaps a lesser way) and avoiding 
> the lav. Despite the wording of the principle, I think it's 
> fair to say that the lo taaseh is hutra, not merely d'chuya, 
> by the asei.
> >>
> I neglected to qualify the above, as I should have, with 
> "mikkar hadin, min haTorah". We do have the principle of 
> Reish Lakish, cited in connection with both yibum Yevamot 
> 20a) and tzitzit (Menachot 39b), that "efshar l'kayem et 
> sheneihem, m'kayem". Tosfot in Menachot 40a d"h kevan says 
> that this is m'dirabban, and the Rambam  Tzitzit 3:6 says 
> "din hu" that it is permissible to put wool tzitzit on a 
> linen garment, but then says we do not do so because of the 
> above cited principle. 

I may be wrong about this, but I understood that while this may be true
of tzitzit, it is not true of yibum.

As I understand it, the sugya which concludes on Yevamos 20a which you
cite is discussing a case where the brother of a cohen gadol dies
childless, so that his wife falls to the cohen gadol in yibum.  Of
course the cohen gadol is forbidden to a widow by means of a lav.  The
question is how can the cohen gadol perform yibum (or can he perform
yibum) given that there is an issur lav - and the conclusion appears to
be that he can, because aseh doche lo ta'aseh.  They say that this is OK
for those who disagree with Resh Lekish, but given that Reish Lekish
holds that efshar l'kayem et shneihem, mikayem, how can this be true
according to Reish Lekish, given that he can perform chalitza and still
keep the lav, and the answer that is given, and which seems to be the
conclusion (and certainly the Encyclopedia Talmudit understands it as
such in the paragraph I quoted) is that chalitza b'makom yibum aino
mitzvah.  That is, while the principle of Reish Lakish is applicable
elsewhere, eg with regard to tzitzit, it is not true of yibum because
chalitza cannot be considered an appropriate substitute for yibum,
allowing one to say that one is mekayim shneihem.  That means, as I
understand it, that while there may be a whole debate about Reish
Lakish's principle and whether it is d'orisa or d'rabbanan, elsewhere,
that debate does not start to apply to yibum/chalitza, because chalitza
is already discounted as a mitzvah when one is discussing yibum, so the
whole Resh Lakish principle does not apply if the option is chalitza as
opposed to yibum.  And it was from that statement that I was drawing the
inference that if we discount chalitza as a mitzvah to the extent that
we don't even get into the whole Resh Lakish discussion about being
mekayem shnehem, because in the face of yibum, chalitza is considered
not to be a mitzvah, then surely we would apply the same principle in
our case.

Earlier you wrote:

> As far as the probability of getting 5 yibumim goes, RCL has 
> gotten the math wrong. The possible combinations of the 
> yevamim and yevamot is,  5 factorial (5*4*3*2*1) = 120, not 
> five to the fifth power (=3125).Thus the chance that there 
> will be 5 yebumim is, as REReich correctly points out, 1/120, 
> an exponentially :) larger chance than RCL's  1/3125  This is 
> because the five yibumim are not independent of one another. 
> If , for example the first yibum is a "match", the chance 
> that the second one is also a match is 1/4, not 1/5, since 
> there are only 4 yevamot the second yavam may marry, one of 
> whom is "his" yevama. In the probability examples we are 
> probably familiar with, this illustrates the idea of "with 
> replacement" and "without replacement". If five people 
> blindly choose a ball from a box which contains one white 
> ball and 4 black balls, there is a fundamental difference 
> between if the ball chosen is replaced in the box after each 
> choice or not. In the former case the number of possible 
> choices is 5 to the fifth power, since each choice has no 
> influence on any other; in the latter 5 factorial, since they 
> do. In our case,obviously the yevamot are married "without 
> replacement"

While I would seem to have gotten the maths wrong, because as you say,
the yevamot choices to have an influence on the next choice, I am
struggling a bit to see how this case matches the white/black ball case
either.  If five people choose a ball from a box with one white ball and
five black balls, then in the end *somebody* gets the white ball.
However in our case, it is quite possible that nobody gets *his* yevama.
The reason for the difference is that the white ball is the same for
each of the five, but in our case, the yevama is a different one for
each.  In fact, if the first person chooses someone who is not his
yevama to marry (and hence does chalitza to his real yevama) then he has
made sure that somebody else of the five will also do chalitza, because
that person's yevama has been taken by him.  It is only if the first one
actually gets his yevama that he leaves everybody else with their yevama
still in the pool.  That would seem to still work with five factorial
for five yevamos, but that is not the figure I am really interested in.
What I really want to know is what is the likelihood of there being one
or more yevamos - ie how likely is it that you will beat the certain
result if you had one man marry all five women, and hence definitely did
one yibum.   

> Saul Mashbaum

Regards

Chana
>