In 1982, a brilliant young economist named Ariel Rubinstein came up with the answer for how two people should divide a pie in what is perhaps, the most basic form of a bargaining problem. Namely, an alternating series of offers and counteroffers. When we first talked about dividing the pie, we didn't specify any rules or format for a negotiation. Here in Rubinstein's bargaining game, the two parties take turns making offers to each other. To make things concrete, we'll return to Abe and Bea as our negotiators and say that they can make just one offer per day. Abe gets to go first. So on Monday, he makes an offer. If Bea isn't satisfied, she can make a counteroffer on Tuesday. On Wednesday, Abe can make another offer. And on Thursday, Bea can make a second counter and so on. This is a single negotiation. So as soon as one person says yes, the deal is done and the negotiation is over. Thus, the counteroffer on Tuesday is only there as a backup if Abe and Bea don't strike a deal on Monday. And the offer on Wednesday is only there, if the counter isn't accepted on Tuesday and so on. Well, it's possible for the offers to go back and forth forever. That wouldn't be a good idea for either Abe or Bea. It's better to do a deal sooner than later. That doesn't mean either side should accept a bad deal today, but a deal today is always better than the same deal tomorrow. Time is money. A dollar today is worth more than the same dollar tomorrow, because it can be invested and earn interest or dividends. In addition, a delay could in itself make the pie smaller. For example, in the case of a UPS strike, each day an agreement is delayed is a risk loyal customers will develop relationships with other shipping agents. And in the case of the NHL's collective bargaining, hockey games that are missed during a lockout can never be replayed. Let's assume an additional day of bargaining hurts Abe and Bea equally. We'll capture this cost of delay by shrinking the pie. We'll say that the pie shrinks by half each day that they don't reach an agreement. Yes, that's an unrealistically high cost of delay, but it makes the numbers a lot easier. Later, I'll show you how the two parties should divide the pie for any cost of delay and even when they have different costs. As we said, Abe gets to make the first offer. How much then should Bea hold out for? For starters, Bea should never accept zero. The reason isn't due to spite or because she's essential to doing any deal. No, the reason is that Bea can guarantee herself far more than zero. To see how much more, let's say that Abe offers Bea zero on Monday and she says, no. The pie will then shrink from 100 to 50. On Tuesday, Bea can counter 25 and be totally confident Abe will accept. Why? Because if Abe turns down her offer of a 25, 25 split on Tuesday, then come Wednesday, the pie will shrink to 25. What's the best Abe could do then? He couldn't get more than 25 and that's all there is to go around. Thus, Abe should take 25 on Tuesday rather than hope, he'll get all of the 25 pie on Wednesday and here is the key point. When Bea counters with 25 on Tuesday, that means she'll get to keep 25 for herself. Remember, the pie is 50 and she's only giving away 25. So if Bea knows she can be sure to get 25 on Tuesday, there's no reason for her to take anything less than 25, let alone zero on Monday. What we've really shown is that Bea should never accept anything less than the difference between tomorrow's pie and the day after tomorrow's pie. On Monday, tomorrow's pie is 50 and the next day's is 25. So Bea can be sure to get 25. If we think in terms of percentages, Bea should never take less than 25% of the pie. The same argument we used on Monday can be used to show that, if the negotiation gets to Wednesday, Bea shouldn't take less than 25% of the 25 or 6.25. The reason is that Bea could counter on Thursday with an offer of 6.25, she knows Abe would accept. So we've shown that Bea should never take less than 25% of what's available. Bea knows this and Abe should be smart enough to figure it out too or Bea can explain it to him. Therefore, when I argued Bea can be sure to get 25 on Tuesday, I was being too conservative. Come Wednesday, Abe can't get all of the 25 of the remaining pie, because he must concede at least 25% or 6.25 to get Bea to agree. Abe's very best Wednesday scenario isn't 25, it's 75% of 25 or 18.75. That means Bea only has to counter with 18.75 on Tuesday to entice Abe to say, yes right then and there. As Bea gets to keep 31.25 on Tuesday, Abe has to offer her at least that much on Monday. So now we've moved up how much Bea will get on Monday from 25% of the pie to 31.25%. Of course, this argument for Monday also works for Wednesday. And if Bea can get more on Wednesday, then she can give less on Tuesday. And in turn, get more still on Monday. At this point, I hope your head isn't spinning too fast. What Ariel Rubinstein figured out is where this all ends up. If we keep doing this argument, we would end up with a share of the pie or percentage that is a time-consistent lower bound for Bea. When both parties understand this percentage is the least Bea will ever accept in the future, this leads her to a position where it becomes the least she should accept today. Let's give a name to Bea's lower bound. We'll call L the smallest percentage of the pie Bea would ever accept in the future. I'm not saying, Bea should accept it. I'm saying, she should never accept less than L percent. First, we showed that L is at least 25% and then we used Bea and Abe both knowing that fact to show that L is at least 31.25%. To figure out what L must be, let’s consider what is the least Bea can get if she turns down Monday’s offer? What counteroffer can she make that will surely be accepted? She can anticipate that come Wednesday, Abe will still have to offer her at least 25 times L to have any hope she'd say, yes. Leaving him with no more than 25 times 1 minus L. Since that the best Abe can do on Wednesday, he will surely accept 25(1-L) on Tuesday, if Bea makes that offer to him. Bea can guarantee herself 50-25(1-L) on Tuesday. So that's the minimum she should accept on Monday. So when Abe and Bea both understand that Bea will hold out for at least an L share in the future, that means today Bea can be sure to get 50-25(1-L) of the initial pie of 100 or 0.50-0.25(1-L) in share terms. And if she can get that share today, the same argument says, she shouldn't take anything less in the future. Thus, her minimum percentage will be consistent over time if and only if L=0.50-0.25(1-L) or L is equal a third or 33.3%. Let's confirm how all this works. If Bea will never take less than a third of the pie, then the best Abe can do on Wednesday is keep two-thirds of the remaining 25. If Bea offers two-thirds of 25 or 16.67 on Tuesday, Abe is sure to accept. Bea is then left with 33.33 on Tuesday, which is why Abe has to offer her at least 33.33 or a third of the pie on Monday. Pretty neat. We've shown that Bea can quite credibly say, she will never take less than a third of the pie, but perhaps she can do better. What's the most she can ever get? Sadly for Bea, Abe can demonstrate that the most you'll ever get is also one-third of the pie. Abe points out that the most Bea could ever get on Tuesday is the whole pie. Therefore, on Monday, Abe doesn't need to offer Bea anything more than half. He knows she will always accept 50 as that's the very best she could do on the next day. To make matters worse for Bea, she won't even be able to get the whole pie on Tuesday. That's because Abe knows come Wednesday, Bea will be willing to take half of what's still left. Thus, Abe can guarantee himself at least 12.5 on Wednesday. So Bea will need to offer him at least that much on Tuesday, leaving her with 37.5. And since that's the most Bea can get on Tuesday, Abe has no reason to offer her anything more on Monday. Therefore, the best Bea can hope for is down to 37.5% of the pie. But we're not done, as we should now move Bea's best case scenario on Wednesday down from 50% to 37.5%. You can probably see where this is all headed. If we look for a time-consistent upper bound, the most Bea can ever get in this game is a third of the pie. If she will always say, yes to a third on Wednesday, then ABE can guarantee himself two-thirds of the 25 or 16.67 on Wednesday. Thus, Bea can't offer anything less on Tuesday and expect a yes. The most she can get on Tuesday is 50 minus 16.67 or 33.3. And since that's the best Bea can do on Tuesday, she will always say yes when that amount is offered on Monday. So now, we've shown that there's only one internally consistent strategy for Bea. She should reject any offer below a third and she should accept any offer above a third. And armed with this understanding, Abe will offer exactly one-third and she will say yes. What this really means is the person making the offer gets two-thirds of the pie and the person receiving the offer gets one-third. On Monday that's Abe, but if things go wrong on Monday and it's Bea's turn to make an offer on Tuesday, then she'll be able to get two-thirds of the remaining 50, which is 33.3. The same as what she would've received on Monday. How much more the person who makes the offer gets compared to the person who receives the offer depends on how fast the pie shrinks? With our pie shrinking at the Indy 500 speed of 50% per day, the imbalance is two-thirds for the proposer, one-third for the receiver. If the pie only shrinks at 1% per day, then the imbalance is much smaller. The proper gets 100 over 199 or 50.25% and the receiver gets 99 over 199 or 49.75%. The power in the game is limited to one's ability to cause a day's worth of delay. If a day's delay doesn't cost that much, then the two sides aren't pretty equal footing. In our example where the pie shrinks by half each day, Abe has the power to get all that is lost on Monday. Namely, 50%. But on Tuesday, he is in a weak position as now Bea can cause a delay. That leaves Abe with one-third of the remaining half. Therefore, Abe as the person going first can get all of the first half plus a third of the second half, which is two-thirds in total. As the shrinkage rates get lower and lower, the ability to move first becomes less important. And thus, the solution to the bargaining game converges to 50, 50. If the pie shrinks from one to delta each day, then the solution to the game is 1 over 1 plus delta for the proposer and delta over 1 plus delta for the receiver. Thus, when delta is a half, you get 1 over three-halves or two-thirds to the proposer. When delta is 99%, then you get 1 over 1.99, which is 50.25% to the proposer. And if delta is 99.9%, then you get 1 over 1.999, which is 50.025% to the proposer. This makes perfect sense. As the pie shrinks less between rounds, there's less an advantage from going first and so the division gets closer to 50, 50. We've been changing the amount by which the pie shrinks from day to day, but it probably makes more sense to think in terms of offers being made more frequently. If offers and counteroffers can be made an hour apart, then the pie will shrink a lot less between offers than if it takes a month to go back and forth. What matters is the shrinkage, not the time. If the offers can be made with little loss between them, we expect the pie will be divided very close to 50, 50. That was hard. So big congratulations for making it through this. From here on out, everything will be smooth sailing.