Is TG/T possible at an even match score?
Posted By: Timothy Chow
Date: Wednesday, 3 May 2017, at 2:46 a.m.
In a recent post, Sean Garber said:
If XG says this is Too Good/Take at 0-0 to 11, there must be a real, live position that fits the criteria somewhere and really is Too Good/Take at this score.
I think that this is probably not true, although I don't have all the details of the argument worked out.
To the best of my knowledge, the only person who has argued carefully that TG/T is impossible for money, is Doug Zare, in a December 2002 Gammon Village article entitled Giving Gifts (paywall, sorry). I've been aware of the existence of this article for some time, but did not take the time to try to digest his argument until just now. I'm still not sure I completely grasp all the details, but here is the gist as I currently understand it. One way to state Zare's main result is this:
The value of holding a 2-cube in a money game is at most 1.
In other words, if your opponent were to offer you 1 point in exchange for owning the cube (still on 2, not on 4), then you should always agree to the deal, no matter what the position on the board is.
From Zare's result it follows that:
If you hold a 2-cube in a money game, and if you know for sure that your opponent will take or pass with 50% probability, then redoubling to 4 is always correct, no matter what the position is.
Here's why: Suppose your D/T equity is X (and by this I mean your unnormalized equity, which will be twice what the bot says, since the bot divides by 2 to normalize to the cube value). If your opponent is going to flip a coin to decide whether to take or pass, then your equity from doubling is going to be (X+2)/2 since D/P nets you 2 points. But (X+2)/2 = X/2 + 1, and X/2 + 1 is your equity if you sell cube ownership for 1 point. By Zare's result, this is always profitable.
It also follows that:
TG/T is impossible in money play (without Jacoby, at least).
For TG/T means, in particular, that doubling against a coin-flipping taker would be a mistake—you lose equity regardless of which way the coin falls.
How does Zare prove his main result? He starts by considering the value of the cube at the moment of a correct redouble to 4. There are two cases.
Case 1 is a correct RD/T. Let X ≤ 1 denote your normalized RD/T equity (i.e., what the bot reports). Your unnormalized equity is then 2X. How much was cube ownership worth to you at that moment? Had your opponent already owned the cube on 2, then your equity would have been X, since this is precisely the same situation as after the RD/T except with the cube on 2 instead of 4. The fact that in reality you own the cube has given you an equity of 2X instead of X, so cube ownership is worth 2X – X = X ≤ 1 to you.
Case 2 is a correct RD/P. Again let X ≥ 1 denote your normalized RD/T equity. Your actual equity is 2 since you're just cashing a 2-cube. If your opponent owned the cube on 2, then your equity would be X. So the value of cube ownership is 2 – X ≤ 1.
This shows that the value of cube ownership is at most 1 at the moment of a correct redouble, but what about if your current position is a ND? Here's where Zare's argument gets more technical, and I won't try to give all the details, but the basic idea is that the value of cube ownership is a weighted average of all your future potential redoubles, and since you're taking a weighted average of numbers that are all individually at most 1, the resulting average must also be at most 1. The argument gets particularly subtle because correct checker play can depend on cube location, but as I said, I'm going to gloss over these details.
Now the question is, can this argument be adapted to match play at an even score? Suppose you own the cube (not necessarily on 2; call its value c). Let P be your MWC if you win a single game at the current cube level. Because it's an even score, your MWC if you lose a single game at the current cube level is 1 – P. Suppose, as in Case 1, that you currently have a correct RD/T, and that your MWC after the RD/T is Q ≤ P. Let Q' be your MWC if your opponent owned the cube on c (instead of 2c). Then I think what we need to show is that Q – Q' ≤ P – 0.5 since "P – 0.5" is the equivalent of "value of 1" for money. This claim is pretty intuitive because it amounts to saying that at an even score, if you have a correct redouble, then you must be a match favorite after the redouble, and also a match favorite if your opponent owns the cube at half that value. Not quite sure how to prove this rigorously though.
For Case 2, I think what needs to be shown is that if you have a correct RD/P, then you would still be a match favorite if your opponent instead owned the cube at its current value. Also intuitively plausible, but I'm not sure how to prove it.
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