[gumpzilla]

Very interesting question. I haven't read the analysis of anybody else yet. I'll call the numbers of players A and B lowercase a and b, respectively. There are 2 antes in the pot, each player has paid 1.

B should never switch if A has switched and b > .5, since then b will be likely to outperform A's second number. Likewise, if A has switched and b < .5, B should always switch. So when A switches, B's strategy is easy.

It is when A stands pat that things get complicated for B, because this is the only area where A can "bluff." The simple strategy of switch whenever A stands pat fails to situations like a = .65, b = .7 where A benefits by tricking B into switching a winning number.

My guess is that most likely some kind of mixed strategy is going to be necessary here, and I don't have the experience to work out what's optimal in that regard. Let's assume that B adopts a slightly more complex strategy than above: B switches only when b < .5 or when A stands pat and .5 < b < beta, where beta is some threshold value above which B always stands pat. If A knows B will employ this form of strategy, and happens to know beta, how does A proceed? We can break down into three regimes:

b < .5: EV of 0 when A switches, since b will switch too, and EV of 2a - 1 when A stays, since b will switch. So, when b < .5, A should stand pat if and only if a > .5. Averaging over all a, A has an EV of .25 whenever b < .5.

.5 < b < beta: B will stand pat if A switches, so switching nets A an EV of (1 - 2b)for switching and an EV of (2a - 1) for staying. 2a - 1 exceeds 1 - 2b for a > 1 - b, so a should stand pat for all a > 1 - b and switch for a less than that. Averaging this one is more complicated, so it's quite possible that I've made a mistake here, but I get that averaged over all a for a given b, a picks up an EV of (1 - b)^2 here. This analysis is only valid for b > .5, so the best A can do over this range is .25, and the larger b gets the worse A does, which makes sense, so this looks plausible. We'll average over b at the end.

When b > beta, A's choice is simple. B will never switch, so A's strategy becomes stand pat when winning, change when losing. The EV when switching is (1 - 2b), and the EV when standing pat is 1 (because A only stands pat when winning.) Averaged over all a, the EV in this range is (1 - 2b)b + (1-b) = 1 - 2b^2. As a sanity check, this gives EV of -1 for A when b = 1, as it should.

Okay, so now we know how A does on average for any given b. We need to average over all b for a given beta to figure out the value of the game for a particular strategy. I get (on the off chance that anybody has actually been following all of this thus far, this is another integral you probably want to check me on) that the average over all b for a given strategy is 1/6 + (beta^2)*(beta - 1).

At beta = 2/3, this EV gets pushed down to 1/54, which is the minimum. So for a fixed strategy, beta = 2/3 is the best B can do, and he's a slight loser there, just under 2 cents on the dollar by my calculation. I'm actually surprised he's not more of a loser. And mixed strategies are beyond me at the moment.