Only the second owner to join adds any value, so they get the same slice as the sole owner in the first example, ie, half. Since each owner has a 50% chance of being the second owner, they each get 25%. With 3 owners it’s 1/3*.5=1/6. Etc. the math works out very neatly this way.
Only the second owner to join adds any value, so they get the same slice as the sole owner in the first example, ie, half.
Show your work and/or explain what’s wrong with my reasoning above? I already took that into account, but got a different answer, because there’s more permutations where the second owner is added late in the sequence than where the second owner is added early in the sequence.
(Note that if my reasoning from the previous comment is wrong, it has to be wrong about both the owners and the workers, since they need to sum to the correct total.)
You’re right, this was less clear than I intended it to be; my apologies. The proposal I offer here is that since the first owner to join offers nothing, they aren’t included in the calculation; you just run them as though the second owner was the only owner, and get the same result for the value of the second owner as you did for the sole owner the first time. This has the advantage of being very computationally simple and gets you the same results as the full calculation for this case. It’s a specific case of a broader way you can simplify shapely calculations, where when the payout can be calculated by (A1+A2+A3…AN)*(B1+B2+B3…BM)+…+(X1+X2+…XO), you’ll get the same values for combining “Like Terms” (Not exactly the same concept but I don’t remember what the real word was, been a while) and running the shapely calculation for A, B,…X and then dividing the total of A by which fraction of A each A1, A2, etc terms makes up. For more complicated payout amounts it’s a bit more annoying to do this way but you can still do it if you’re careful to set it up properly. This applies even when one of the terms, eg A2, is 0, which then gets us our specific case here. I don’t remember the exact setup for the proof my professor used, but I don’t think it’s neccesary? It feels like one of those things where it’s obvious once you run it through both ways for an example, but if it isn’t lmk and I’ll see about digging through my old notes.
If I understand you, you’re saying that this simplified calculation should give the same answer as the full calculation, and that this should be obvious to me after I try it both ways.
But you’re also saying that the simplified calculation is equivalent to the earlier problem with only 1 owner, which means I already worked it both ways (one way in the 1-owner problem and the other way in the 2-owner problem) and I got different answers. So, no, the equivalence is not obvious after running it both ways.
My suspicion is that this half-remembered theorem you are trying to apply actually applies to a different situation and you are using it wrong. But if you don’t think that’s the case, then yes, I’d like to see a more precise explanation. (And ideally, an explanation of what you think I did wrong in my own calcs.)
Only the second owner to join adds any value, so they get the same slice as the sole owner in the first example, ie, half. Since each owner has a 50% chance of being the second owner, they each get 25%. With 3 owners it’s 1/3*.5=1/6. Etc. the math works out very neatly this way.
Show your work and/or explain what’s wrong with my reasoning above? I already took that into account, but got a different answer, because there’s more permutations where the second owner is added late in the sequence than where the second owner is added early in the sequence.
(Note that if my reasoning from the previous comment is wrong, it has to be wrong about both the owners and the workers, since they need to sum to the correct total.)
You’re right, this was less clear than I intended it to be; my apologies. The proposal I offer here is that since the first owner to join offers nothing, they aren’t included in the calculation; you just run them as though the second owner was the only owner, and get the same result for the value of the second owner as you did for the sole owner the first time. This has the advantage of being very computationally simple and gets you the same results as the full calculation for this case. It’s a specific case of a broader way you can simplify shapely calculations, where when the payout can be calculated by (A1+A2+A3…AN)*(B1+B2+B3…BM)+…+(X1+X2+…XO), you’ll get the same values for combining “Like Terms” (Not exactly the same concept but I don’t remember what the real word was, been a while) and running the shapely calculation for A, B,…X and then dividing the total of A by which fraction of A each A1, A2, etc terms makes up. For more complicated payout amounts it’s a bit more annoying to do this way but you can still do it if you’re careful to set it up properly. This applies even when one of the terms, eg A2, is 0, which then gets us our specific case here. I don’t remember the exact setup for the proof my professor used, but I don’t think it’s neccesary? It feels like one of those things where it’s obvious once you run it through both ways for an example, but if it isn’t lmk and I’ll see about digging through my old notes.
If I understand you, you’re saying that this simplified calculation should give the same answer as the full calculation, and that this should be obvious to me after I try it both ways.
But you’re also saying that the simplified calculation is equivalent to the earlier problem with only 1 owner, which means I already worked it both ways (one way in the 1-owner problem and the other way in the 2-owner problem) and I got different answers. So, no, the equivalence is not obvious after running it both ways.
My suspicion is that this half-remembered theorem you are trying to apply actually applies to a different situation and you are using it wrong. But if you don’t think that’s the case, then yes, I’d like to see a more precise explanation. (And ideally, an explanation of what you think I did wrong in my own calcs.)