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Calculation

Posted By: Timothy Chow
Date: Sunday, 5 March 2017, at 12:08 a.m.

In Response To: SMITH and JONES: The amazing "FARE" position (Timothy Chow)

To eliminate reliance on articles outside the bgonline.org domain, let me spell out the calculation that shows that the assumption that D/T is the "perfect play" leads to the conclusion that the equity is undefined. The calculation is made a little simpler in my position because my position is symmetrical.

Let e denote Blue's average equity assuming he enters immediately. (At this stage we are still innocently operating under the assumption that equities always exist.) Let p denote the dancing probability; of course p = 25/36, but I write p for brevity.

Let Blue's equity be E. To compute E, we need to split into cases. The first case is that he enters immediately. This happens with probability 1 – p and the resulting equity is, by definition, e, so the contribution to E from this case is

(1 – p) e.

The second case is that, after D/T, Blue dances and White enters. The equity then is –e by symmetry, and the probability of a dance and then an entry is p(1 – p). So the total so far is

(1 – p) e – 2p(1 – p) e.

Next, we could have D/T, dance, D/T, dance, D/T, enter. The pattern is clear now; we get a factor of 2p for each "D/T, dance" sequence and then we always terminate with (1 – p) e, but with alternating signs. Thus we get the infinite series

(1 – p) e – 2p(1 – p) e + (2p)2(1 – p) e – (2p)3(1 – p) e + …

This is a geometric series with geometric ratio –2p, but since p = 25/36 > 1/2, the series diverges. You might have fun computing the first few terms of this series to see what happens.

There is only one loophole in this argument, which is that the series does in fact converge if e = 0. So it's not quite mathematically watertight. But nobody would argue that if Blue enters then the game is dead even.

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