Tuesday, June 01, 2010

choosing lots to find 70 zekeinim -- unfair probability?

One other Rashi from last week's parsha provided much food for thought in my home. Hashem told Moshe to select 70 zekeinim to help him. Problem: there is no way to divide 70 into 12 equal parts -- we need some way to determine which shevatim got only 5 representitives and which got 6. The solution was to draw lots. 72 lots were created, 70 of which said "zakein" and 2 which were blank. But, asks the Taz in Divrei David, is this method really fair? Let's say you are the first one to draw -- the odds of your getting one of the "losing" lots is 2/72. But let's say you are the last one to draw -- if no one else got a blank lot, you are certain to draw a losing ticket! The fewer lots in the mix, the greater the odds of drawing a losing ticket. The Yerushalmi provides some help and says there were 74 lots created, so that even the final two people to draw would have a chance to win. But even so, their odds would be 50/50, far higher than the initial person's odds of 1/36! (Perhaps it was my BIL's mention of the Monty Hall problem in a comment to the previous post that attracted my attention to this issue, as both problems deal with the effect of choice on probability.)

By rights I should have discussed this earlier in Parshas BaMidbar where the same issue comes up in regards to pidyon. Each Levi served as a substitute for a bechor. Problem: there were 22,000 Levi'im, but 22,273 bechorim. The 273 overflow had to pay pidyon, but how do you determine who the 273 "losers" who have to pay are? Again, the answer was to draw lots, and again the question is whether the method is really fair.

One might answer simply that the lots were not truly random but were guided by hashgacha, but if that's the case, why not dispense with the pretense of lots entirely? Another interesting answer offered by the Taz in BaMidbar is that since the lots were drawn from a fixed jumble, we apply the principle of kol kavua treat everyone has having 50/50 odds. This seems hard to swallow. Kol kavua is not a metziyus; it's a legal fiction. You can't use a halachic reality to answer a mathematical question about real metziyus. Kol kavua doesn't mean the odds are really 50/50, but rather despite the tilt in odds, for purposes of law we treat kol kavua as 50/50.

My son and I both thought this question doesn't get off the ground. No matter who chooses first and who chooses last, the odds remain the same. Imagine dealing a deck of cards to 52 people. Whether you stop after 10 cards or 20 or 50, the odds of any one person holding the ace of spades is still 1/52. The Taz cites a Rashba"sh who writes that the more lots chosen, the greater the odds the losing lot is among those taken already (simple principle of rov). I'm not sure I follow the logic here completely, and for either of these answers to work I think (though I may be wrong) you have to assume that no one revealed their choice in advance. To tell you the truth, the more I think about this, the more confused I get, probably (no pun intended) because I am a lost cause when it comes to math. If anyone has another solution or cares to shed some light, please comment away.


  1. Anonymous11:29 PM

    interesting post. i assume the taz is coming from the following standpoint. he agrees to you and your son that e/o has a 1/36 chance. but if the first drawers reveal what they had then the subsequent drawers odds either have changed or seem to have changed in their eyes.
    another option is even though in metizus e/o has 1/36 chance, nonetheless the later drawers have less options to choose from than the first. lehavdil, when playing scrabble, i would feel more comf. getting the q or x in the beginning of the game as opposed to the last drawing even if i havent seen any of the other letters drawn yet. the feeling is its more available. hu hadin hacha.

  2. Mike S.6:56 AM

    The kicker is in the phrase "if no one else drew a blank lot" The likelihood of that keeps dropping as more pick. Thus the first person to pick has a 1/36 chance of being excluded, and a 35/36 chance of being included. The second one's chances are computed as follows:

    if the first person was excluded (1/36 of the time) he has a 1/71 chance of being excluded for a total of 1/2556 (1/36 times 1/71) chance of exclusion through this route.

    If the first person was included (a 35/36 chance) he has a 2/71 chance of being excluded so a probability of being excluded by this route of 70/2556 (35/36 * 2/71)

    His total probability of exclusion is the sum of those 2 or 71/2556 = 1/36.

    The calculation for the others is similar but more tedious. Thus, if no one else had picked the blanks, yes the last two would get them, but there is only a 1/36 chance of no one picking a blank before them.

    This is not similar to the Monte hall problem, which turns on the fact that Mr. Hall has a choice of which unpicked door to show you, and can show you a goat no matter which door you pick.

  3. Mike S -- sounds like a good explanation of the Rashbash answer. Thanx!

    Following up on Anonymous' point, if the first one's reveal their choice and show that they did not get the blank, wouldn't that change the odds for the last people drawing? I thought that was the case and am not sure why the Taz assumes that no one shows their cards until the end.

  4. Mike S.10:07 AM

    Depends by what you mean by change the odds for the last people drawing. if you mean, the chance that they would be selected overall, no. They all have a 35/36 chance of being selected; however, at the time they draw, if they know which lots have already been drawn, the odds may be different. Consider the last person to draw--he is going to get the lot that remains, so at the time he draws, assuming the other lots have been revealed, he knows what the will get. But there is a 35/36 chance that both blank lots will already have been drawn before he picks, and a 1/36 chance that only one blank lot has already been drawn. In mathematical terms the probabilities are equal but not independent.

    The Monte Hall problem is completely different because Monte hall can choose whether to offer you a switch and which door he shows you. Therefore, what your odds are depends on what you assume Monte Hall does. If you assume he is malicious and will only offer you the chance to switch if you have picked the car, of course you should keep your choice; if you assume he only offers a switch if you have picked a goat, of course you should switch.

    The implied assumption of the problem (and I think what he did on the show) is that he will show you a goat no matter what door you pick. In that case your odds of getting the car are 1/3 if you stay, and 2/3 if you switch which you can understand as follows. when you initially picked you had a 1/3 chance of getting the car. Since Monte can show you a goat whether you picked the car or a goat, when he shows you a goat you learn nothing about you picked door; it still has a 1/3 chance of having the car. However, in the other 2/3 of the time, Monte has told you which door has the car, so you no longer have to guess.

    If you assume Monte reveals one of the remaining doors at random (the closest the problem can come to our case) it doesn't matter what you do. 1/3 of the time you will have picked the car, and you will lose if you switch. of the remaining 2/3, half the time (1/3 of the total) the door will reveal the car and you will get a goat whether you switch or not, and the other half (1/3 of the total) you will get the car by switching. Thus there is a 2/3 chance you will see a goat when the door is revealed and it is 50-50 which other door has the car, and a 1/3 chance you will see a car revealed, and then you know you get a goat. Your chance of getting a car is 1/3 no matter which strategy you choose under these assumptions.

  5. Interestingly, the other famous lottery -- the dividing up of the land, was done in a different fashion.

    Instead of having the leaders of the sh'vatim "pick" their lot, there were two sets of lots -- one with the names of the sh'vatim and the other with the names of the portions. For each portion, a lot would be drawn from *each* set of lots.

    This, incidentally, is how the United States handled the draft in the Veitnam war. There were two sets of balls -- one with all the calendar dates of the year and the other with balls numbered 1 to 366.

    All that this does, of course, is to randomize the draw order. But since the draw order is the main objection here, one wonders why the other lotteries described (levi'im vs. bechorim, z'keinim) weren't done that way.

    The Wolf

  6. To present the same answer as Mike's in a less math-y way:

    Until the moment of the first selection, the odds were 1/72 for each. Yes, the odds change after that point. But how they change is also a matter of odds.

    Let's put it another way... What if everyone grabbed a slip bevas achas? What if they also opened it bevas achas? Now, what if they opened their slips one after the other?

    Even if you come up with a fully fair system, there is a point in time where 70 people had a 100% chance of wining, and two people had a 0% chance -- after all the lots are known. Probability changes as we gain information.

    Here we just have a chance to gain information incrementally.

    Still, if you divided up the candidates millions of times, each one would be selected to be "zaqein" roughly 70/72 of those millions.

    (Philosophically inclined people might notice that I'm contrasting the Bayesian explanation of statistics with frequentist statistics.)


  7. As for the pretense of the lot, you know that the Gemara in Yesh Nochlin says that the chalukas ha'aretz, which involved a lot, as Wolf said, was preceded by an announcement based on the Urim ve'Tumim that this pick will be shevet X; after the shevet was picked out, thus validating the Ruach Hakodesh, the Urim ve'Tumim would indicate that the portion picked would be Y, and so it was. That's not what happened here, though, but at least you can't say that a lottery is a waste of time when it is the handmaiden of hashgacha.

  8. The Yerushalmi (Sanhedrin 1:4, IIRC) has this concern, as do many of the Tosafists in their commentary on Chumash. (Many other commentaries discuss this too, with numerous approaches, including having an equal number of blank slips as slips with "zakain" witten on it.)

    Personally, I don't see the problem, especially since G-d was involved in the process and the odds are the same for all going in (only if you believe in the "magic" of rov is there a problem). The Bavli (Sanhedrin 17a) doesn't seem to have an issue either.

    The Sifray avoids the numbers game by not telling us how many said "zakein," although the Gra's version has a number. What is clear from the yerushalmi, however, is that if there is a concern about the odds changing, the "blank" slips were not the last two, so that whomever picked one still had a better than 50% chance of getting "zakain."

  9. I forgot to mention (there are many approaches about this lottery) that one suggestion made by a meforush is that each slip was put back in the box after it was drawn, so the odds were exactly the same for everybody.

  10. RDK,

    If you keep putting the chits back in, it would be possible to have fewer or more than 70 zeqeinim. (Likely to happen over 13% of the time.)

    Of course HQBH could simply insure that exactly 70 of the men chose "zaqein" chits. But the whole point of the question was the pretense that this is random.


  11. Micha,

    That was the whole point- see Rashi, Sifray and Talmud's wording of what Moshe said to them after they drew a "chit." If it said "zakain" he said "G-d has sanctified you;" if it was blank he said "G-d doesn't want you, what can I do for you?"

    It is pretty clear from the discussion of the sources that HKBH guided the outcome; the only issue is whether there could be any complaints about how it was done that would undermine this. Having 140 slips (or 74, or any number larger than 72) would make it even clearer that it was HKBH's doing.

  12. And my point was that once is assuming that statistics and having a 70/72 chance of being selected is irrelevent, the question you are answering doesn't exist either. The words in the title -- "unfair probability" -- aren't relevant, as probability is only for random variables.

    The goral could have had just two chits -- "zaqein" and blank -- and Hashem could have insured that exactly 70 men selected it.