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.