I had a discussion with Claude.AI today. I asked the following question:

Can you explain Bayesian updating. If I have two teams that I believe are evenly matched, with a prior probability of 0.5, how do I update that prior if one team wins three games in a row against the other?

Here is Claude’s response:

Let me walk you through Bayesian updating in this context.

Let’s break this down step by step:

1. Initial Prior:

• P(Team A is better) = 0.5
• P(Team A is not better) = 0.5

1. We need to consider:

• The likelihood of seeing 3 consecutive wins IF Team A is truly better
• The likelihood of seeing 3 consecutive wins IF teams are actually equal

Let’s make some reasonable assumptions:

• If Team A is truly better, they might win any given game with 65% probability
• If teams are equal, each has 50% probability of winning any game

1. Calculating likelihoods:

• P(3 wins | Team A is better) = 0.65 × 0.65 × 0.65 = 0.274
• P(3 wins | Teams equal) = 0.5 × 0.5 × 0.5 = 0.125

1. Using Bayes’ Theorem:
   P(A is better | 3 wins) = P(3 wins | A is better) × P(A is better) / P(3 wins)

Where P(3 wins) = P(3 wins | A is better) × P(A is better) + P(3 wins | Teams equal) × P(Teams equal)
= 0.274 × 0.5 + 0.125 × 0.5 = 0.200

Therefore:
P(A is better | 3 wins) = 0.274 × 0.5 / 0.200 = 0.685

So after three consecutive wins, we would update our belief from 50% to about 68.5% that Team A is the better team.

Would you like me to explain how the probabilities would change with different assumptions about how much better one team might be?

I thought the 0.65 probability was too high. So I asked for the analysis based on a .55 probablility:

I’ll recalculate using 55% as the probability of Team A winning each game if they’re truly better.

1. Initial Prior remains:

• P(Team A is better) = 0.5
• P(Team A is not better) = 0.5

1. New probabilities:

• If Team A is better: 55% chance to win each game
• If teams are equal: 50% chance to win each game

1. New likelihood calculations:

• P(3 wins | Team A is better) = 0.55 × 0.55 × 0.55 = 0.166
• P(3 wins | Teams equal) = 0.5 × 0.5 × 0.5 = 0.125

1. Using Bayes’ Theorem:
   P(3 wins) = 0.166 × 0.5 + 0.125 × 0.5 = 0.146

Therefore:
P(A is better | 3 wins) = 0.166 × 0.5 / 0.146 = 0.569

With this more conservative assumption, three consecutive wins would only move our belief from 50% to about 57% that Team A is better. This makes intuitive sense – if we believe the skill gap between teams is smaller (55% vs 65% win probability), then we need more evidence (more games) to be confident that one team is actually superior.

This illustrates an important principle of Bayesian updating: the strength of our update depends not just on the evidence (3 wins) but also on how different we expect the world to look under our competing hypotheses. When the hypotheses make very different predictions (65% vs 50%), evidence can shift our beliefs more dramatically than when the hypotheses make similar predictions (55% vs 50%).

I like that a lot better. A probability of .569 indicates that a team that goes up 3-0 in a series would win the series 96.5% of the time. That’s very close to the .975 actual series win probability for teams going up 3-0 in MLB history. If teams are evenly matched we would expect them to win 93.75% of the time. So there is some evidence that teams that go up 3-0 in a seven game series are a bit better than their opponents, but not my much.