The simplest solution is just that P1 cuts the pie in thirds. The other two pick first, in whatever order. Since P1 gets the last pick, he has to ensure that the last piece of pie is as big as each of the other two, because that’s the piece he gets!
But the more important question is that you better make sure that P1 knows what he is doing and knows how to handle a knife. So if you’ve got a rapist, a blackmailer, or a bankrobber or whatever in the cell, make sure it’s the guy who is convicted of killing with a knife or assault with a knife who gets to cut.
if p1 cuts one big slice and 2 slivers, then p2 picks the big slice, then p3 is screwed thru no fault of his own.
just thought of yet another variation though. if p2 and p3 contest the big slice, then they could just do a cut/choose on it, as well as the 2nd choice.
Based on the assumption that all 3 likes pie, why would P1 cut one big slice and 2 slivers? In your scenario so P3 would get screwed, but so would P1. So why would P1 do it? Your comment makes no sense.
And if P2 and P3 contest the big slice and do a cut choose, then there would be 4 slices. Your 2nd comment makes even less sense than the first one.
P1 would make that cut because he had a deal with P2 - they would split the 99% of the pie that they got by ganging up. If you read the OP carefully, you’ll understand that the problem requires that every prisoner be able to ensure that he gets what he thinks is at least 1/3, no matter what.
Sure if you put it that way. P1 still would lose out because P3 and P2’s mother-in-laws are actually first cousins and they made a deal that their son-in-laws would split the pie if P2’s sister baked P3’s third cousin an apple pie on P1’s grandparents’ 35th anniversary.
And P1 would still lose out because P2 was in jail for fraud, he was a pathological liar to boot.
Why would a prisoner make a big cut, even though he doesn’t get first chance to get that big piece? Because he doesn’t understand the problem fully. There really shouldn’t be a penalty for any prisoner getting “screwed”, just because someone else didn’t know how to fairly cut a pie; that’s not a fair division scheme. Some of these ideas, while very simple, allow for two prisoners to gang up on the third, leaving him no recourse. Some allow every prisoner a chance to get what he thinks is a fair slice, even if someone made a gross error.
Each person cuts a piece what ever size they want. Then they get to choose which piece of the three they want in order of who cut the smallest piece. Arguments in piece size is mediated by the third person in each decision.
I’m trying. I’m sorry - I don’t mean to sound like I’m attacking your solution. A couple of the others posted so far have this same flaw. This is not actually going to be used to arbitrate any pies - it’s a math problem. As such, you can’t assume certain things you’re not told, for instance, that each prisoner will work to be awarded a large slice. But if they want 1/3 of the pie, and they can reason logically, they have to be able to get it.
I like the solutions so far that have been posted by 23skidoo and areider, because they ensure that the pie will be cut into no more than four pieces. I wonder if there are any solutions that ensure that the pie will be cut into exactly three pieces.
It may work as a math problem in which we asume that each person has perfect control in cutting and perfect powers of observation, and greed is the only mitigating factor. IRL, however, it always pays to let the other person cut. There is always the chance that the cutter will make a mistake. The chooses always gets at least half the pie. The cutter always gets half a pie at most.
Suppose it was, say, a bar of butter rather than pie. Cutter divides the bar in half, width-wise. Now, choosing involves making an imaginary line where one thinks the halfway point is. So really, both sides are making a guestimate of the halfway point. Assuming other things equal, it would seem that both participants have the same chance to be mistaken on where the half line really lies.
OTOH, is cutters job is more difficult? He has to create the halfway line where Chooser only has to determine the correctness of one existing line. But again, Chooser must draw his own imaginary line.
But, I think he does have an edge based on the possibility of a gross error by Cutter simply by virtue of going second. Cutter has no such option. Also, Chooser has a better means of comparing, he could line the half-bars up.
With pie, Chooser could possibly stack the slices and see if one doesnt occlude the other.