Evidence receipt / uncertainty
Published · transcript-backedEric Jang: uncertainty
15 May 2026 Dwarkesh Podcast Eric Jang – Building AlphaGo from scratch
“I also don’t know if this one is proved to have logarithmically or square-root-bounded regret, but I think the algorithm was derived to look something like this.”
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Everything needed to verify it.
- Speaker
- Eric Jang
- Attribution
- Verified speaker
- Claim type
- uncertainty
- Recorded
- 15 May 2026
- Publisher
- Dwarkesh Podcast
Transcript context
…Just to make sure I’m understanding, let me put it in my own words. Let’s focus on UCB. Conceptually, you can think of it as two different things: the Q and the exploration term. Let’s be clear about what Q is. Q is basically saying, once we do these rollouts—you’re actually running all these simulations—you go down the tree and figure out: if I end up at the terminal value of this tree, do I win this game or not? You average whether I win or not across all the leaves of this tree starting from this node. That average, you put in Q. So Q represents the probability that I’ll win this game starting at this node. That is your sort of exploit. It’s saying: I’ve run these simulations, and I think this is a good move or not. The other term is saying: have I explored this branch enough yet relative to the other actions I could be exploring, or have already explored? If I haven’t explored this branch yet, maybe I think it has a low score, but I just haven’t explored that many leaves down this node in this tree. So I should try this even though Q, the exploit term, is telling me it’s not that valuable. Because ln(n) grows slower than n, over time you will move from the argmax being dominated by the exploration term, which is the second term here, to the argmax being dominated by the Q term, which is when you’ve done enough simulations and are confident that this is the branch to go down. Yes, that’s right. The motivation for UCB was to come up with an algorithm where, if you don’t know the payoff of the different actions you can select, this strategy, given some exploration term here, bounds your regret in terms of how wrong you can possibly be. I don’t know the proof. I also don’t know if this one is proved to have logarithmically or square-root-bounded regret, but I think the algorithm was derived to look something like this. You can tell these terms grow a little differently, and this is to account for the fact that Go has many more actions for any given move compared to your standard bandit problem. One small clarification: you talked about simulations and probabilities. We should remember that Go fundamentally is a deterministic game. Where does the notion of probability come from here? If you had a very powerful computer, there are no probabilities. You can just compute the true average of the mean action value. So where does the probability come in? In computer Go before AlphaGo, we’ve always done some sort of Monte Carlo method where we take the expected Q value averaged over a randomly selected tree. That randomly selected tree is where probabilities come in. The interpretation of Q is: what is the expected action value under the random distribution induced by some random search process? Where does the random search process come in? That’s where Pa, of action, comes in. If we assume a naive algorithm where you have a uniform probability of taking any valid action, then this would just be one over the number of valid moves. You would be taking this average over a very diffuse tree. This is a valid integral, but it’s very slow because you’re going to consider a lot of trees that have very low value. It’s essentially almost like an importance sampling problem. Only a few actions and paths contribute high value, and almost everything else is low value. So that’s a tricky problem here. This is the action selection criterion for how you decide which moves to go down. As you move down in tree search, you will eventually run into a node where it’s quite clear you’ve won or lost. At the very end of the game, when there are no valid moves left to play under Tromp-Taylor scoring, you can decide whether you won or lost. This is the final return of the whole game. We can assign a value, U, to a terminal leaf node of the tree, but how do we assign values to the nodes prior to that, the parents? You take the mean action value, which is essentially your average. Suppose these were all leaf nodes. The mean action value of this node is just the average of whether you won or lost at the leaf nodes. Correspondingly, you can walk up the chain and say the mean action value of this node—let’s call it Qb—is just a weighted average of these ones here. The weighted average could depend on whether you have a different sampling distribution. hain and say the mean action value of this node—let’s call it Qb—is just a weighted average of these ones here. The weighted average could depend on whether you have a different sampling distribution. But the basic intuition is that you want to resolve the game where you have a deterministic win or lose, and then you can go backwards—this is called the backup step—and assign values to these nodes or actions corresponding to the average over the final terminal leaf. If you were to do this without neural networks, it would still be intractable. You would have trouble finding which actions to sample. A lot of the actions would contribute very low value, especially if you’re trying to fight your way out of a losing position. Only a few actions give you high value, so the search in practice is still very expensive. But the idea is that because Go follows a tree structure, you can inform a very good estimate of the value of this node based on the downstream values, assuming they’re all correct and you’ve searched deep enough.…
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