They split the prize money if they draw. I believe the match ends immediately if one person wins six games.

From the standpoint of intuitive probability Sedol has basically secured his non-defeat. It's impossible to envision Sedol losing 4 games out of the remaining 4 games. I'd even go so far as to say that Li won't win 3 out of 4, so Sedol will win. You heard it here fist .

however, just prior to his last win, he had in fact lost 4 in a row against Gu Li... just not all in the same event. So it is indeed possible.

A.k.a bulls**t

Well, Lee Sedol has currently won 22 out of 44 games vs. Gu Li, exactly 50%. If we leave out the one triple ko game then it's 51%. If each player has a 50% chance of winning each upcoming game then the odds are now:

Lee Sedol Wins: 68.75%
Draw: 25%
Gu Li Wins: 6.25%

True.. under the added assumption that the outcomes of the games are uncorrelated. Which probably isn't true. But with only 44 games between them, there's no way there's enough data to extract solid information about the correlations, so emeraldemon's calculation is probably about as good as you are going to get.

It does make me wonder if there have been enough n-game matches out there to build a better model, though.

Tonight,

http://www.usgo.org/news/2014/08/myungw ... day-night/

7th game,

Lee Sedol defeated Gu Li by resign.

Lee Sedol vs Gu Li 5:2

Now that Lee Sedol has won game seven, it looks like this:

Lee Sedol Wins: 93.75%
Draw: 6.25%
Gu Li Wins: 0%

No, because now Sedol has won 23 out of 45.
BTW could one of the players improve his odds in this jubango given the score by changing his style? I obviously don't mean by playing better go but something like playing risky or unexplored moves/fuseki.

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I think I am so I think I am.

Even assuming 50% chance each of winning a game, isn't the odds on a draw 12.5% rather than 6.25%? Gu Li needs to win 3/3, which is presumably 1/8 chance (at 50% probability per game.)

Also, clyclops - are you suggesting a purely frequentist approach to updating the winning probablities? In the abstract, I prefer a Bayesian method, since a frequentist approach produces silly answers if the lifetim score were, say 2-0 to one player or the other.
It's slightly off topic, but it seems to me that Bayesian probability theory is on far sounder theoretical basis than the frequentist approach.

The problem is more fundamental than the difference between "frequentist" and Bayesian theory. Related to whether or not belief in the independence of the outcomes of each game.

Thus the belief that Gu Li has zero chance of winning is based on the belief that result of the last few games means that the probability of the next game is no longer 50% (the 50% being based on their game history over long time).

So if we have been flipping a coin for a while, and the result prior to the current match had been 53 heads to 47 tails (no reason to suppose not 50-50 out of 100 trials) but the results so far in a current set of ten trials was two heads followed by seven tails expecting the tenth try to be more likely a tail (because of the previous seven tries).

That is disbelief in "independence of the events" (and would affect Bayesian thinking as well as frequentist).

Thank you for the edifying and informative lecture Mike, but is not true, it's that to win the Jubango he needs to win 6 games, has won 2 so far, and there are 3 games left: 6 - 2 = 4. 4 > 3. QED

Mike - I didn't deny the chances of Gu Li winning were 0! My post was on the odds of a draw.

A time varying binomial model has a lot more complexity than a simple binomial model. It's not immediately apparent to me how one would go about working out the time weighting.

Furthermore, the model could also become arbitrarily complex quite quickly. For example, it is intuitively pluasible to weight towards more recent games. This acknowledges that P(winnning) is actually changes over time. However, it could also be affected by other things such as:

Time limits

Location

Prize money

Colour played

Amount of komi, and other changes in rule-sets

Amount of beer drunk the night before a game

The issue is that the data set is relatively small, and insufficient to give any confidence in a very granular model.

According to go4go, since the jubango started, the score in all games is 6-4 to Lee Sedol. (Gu Li won a couple not in the jubango, and Lee Seedol won 1 not in it.) This may or may not be sufficient evidence to abandon the 50% hypothesis. You could argue that the jubango games are the best evidence for P(winnning) in the jubango - that acknowledges that some of the other factors on my list could be in play.

On a side note, a friend (who is a go player) has set up a Markov Chain Monte Carlo prediction model for the English football leagues. This features time-varying parameters and a (presumably) Bayesian updating approach. I believe he assigns attack and defence scores to each team, and the scores are updated after each result is received. Maybe something along those lines could be set up for Go.
http://www.archduke.org/footballpreds/

Edited to add: The MCMC approach would presumably take all known games as input, and assume games against other players gives useful information for games between the two.

While most of the modeling you all are talking about sounds too complex for me to make any comment on, I feel confident in saying this: Lee Sedol winning 6/10 games cannot possibly invalidate a 50% null hypothesis. That's only 1 more than expected!

In fact, the chance of getting exactly 5/10 wins (assuming a null hypothesis that each game is a coin toss) is only 24.6%. So there's over a 75% chance of getting a skew of 6/10 or more towards one or the other of them.

Applying the same reasoning towards just the Jubango games: Even with the null hypothesis that every game is a coin toss, the chance of one player being ahead at 5/7 or more is a whopping 45.3%! So this result really isn't too surprising even if you believe Gu Li and Lee Sedol are exactly evenly matched.

But you have to divide by two. It is not 'one player' that is ahead. It is one particular player. So the odds of this happening randomly are only about 23%.

Are you sure we wouldn't be having this same conversation if Gu Li was 5:2 up?

You have to love Feynman.

So if it goes 5-5, that's just it, right? There's no tie breaker?

_________________
I think I am so I think I am.