AlphaGo and Laplace

On Friday AlphaGo had won its first two matches against Go champion Lee Sedol and I was wondering what was the probability it would win the third (out of five).

1. If we consider AlphaGo and Lee Sedol to be pretty evenly matched, then the chances of two wins to AlphaGo are 25% - hardly unusual - so the chances of a further AI win might still plausibly be 50:50.

But .. two wins in a row?
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2. If we adhere strictly to a frequentist dogma, then with two wins in two trials we must assume that AlphaGo would win its next match with probability 1. But that's crazy.
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3. Laplace had a rule for this kind of thing: the Rule of Succession. Here's how Wikipedia describes it.

"If we repeat an experiment that we know can result in a success or failure, n times independently, and get s successes, then what is the probability that the next repetition will succeed?"

We count a success as scoring 1 and a failure as scoring zero, then Laplace tells us the probability of a success the next - (n+1)th - trial is:

(s + 1)
(n + 2).

Read the article for the counterintuitive derivation.

So we plug in the numbers: two games (n = 2) and two successes (s = 2) so the probability of AlphaGo winning the third game is 3/4. Sounds about right, I thought.

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On Saturday, AlphaGo duly won the third match in a row and the series, but today (Sunday) it lost its fourth match.

The last match is on Tuesday, so what are AlphaGo's chances?  (3 + 1)/(4 + 2) = 2/3.

In case you were thinking of placing a bet.

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Update Tuesday: AlphaGo wins the fifth match, and the series 4:1.