Let's say there are 10 teams: A-J. Each team always participate in each of the game. Only 1 team wins, others lose. Probability of any team to win is unknown (different for each team) and to be calculated.

There were 3 games: Game1, Game2, Game3.

Team A lost Game1 and Game2 and won Game3. What is the probability of the team A to win Game4?

Now, more complex second question: let's say probability to win is unknown, but known to change over time. How would this affect the answer?

  • $\begingroup$ The "best" estimate is $1/3$. And it is essentially useless. $\endgroup$ Aug 10, 2014 at 19:47
  • $\begingroup$ @AndréNicolas Why 1/3 and why it is useless? :) $\endgroup$
    – Artem
    Aug 10, 2014 at 19:51
  • $\begingroup$ The estimate is $1/3$ because the team had won $1$ out of $3$. It is useless because the sample size is ridiculously small, and because we know nothing about the strengths of the teams. $\endgroup$ Aug 10, 2014 at 19:56
  • $\begingroup$ "Probability of any team to win is unknown (different for each team) and to be calculated." How is this supposed to be calculated? Without knowing this, as André precised, estimates are useless. $\endgroup$
    – JohnWO
    Aug 10, 2014 at 19:57

1 Answer 1


If the probability is unknown in advance then the estimate after $3^{rd}$ game is:

$P(A) = \frac13$; and same for teams who won $1^{st}$ and $2^{nd}$ games (if not a single team won both, in which case it would be $\frac23$ for that)

For the rest $P(x) = 0$.

The $2^{nd}$ does not seem like a question, but is almost obvious, probabilities calculated will change initially, at least till we get a large sample of experiments (games in this case). In the beginning, As Andre said, estimates are almost useless because of low number of trials. These will get more meaning after every experiment.


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