Posted on Feb 6, 2026
The Fiddler Basketball Association's All-Star Game consists of two teams: "East" and "West". Every year these two teams play a game, each with a 50 percent chance of winning that's independent of the outcomes of previous years.
Many, many years into the future, you look at the most recent results of the All-Star Game. On average, what is the longest current winning streak that one of the teams is on? (Here, having won only the most recent game still counts as a "streak" of one game.)
Run the simulation to estimate the average current streak length. The theoretical value is 2.
Result will appear here.
To spice up the All-Star Game, the commissioner of the FBA has decided that there will now be three teams competing in All-Star Games: "Stars", "Stripes", and "International". Each year, two of the three teams play each other. If one year has Stars vs. Stripes, the next year has Stripes vs. International, the year after that has International vs. Stars, and then the cycle repeats with Stars vs. Stripes.
Many, many years after this new format has been adopted, you look at the most recent results of the All-Star game. On average, what is the longest current winning streak that one of the teams is on? (As before, having won one game counts as a "streak". Also, note that the team with the longest winning streak might not be one of the two teams that played in the most recent All-Star Game.)
Run the simulation to estimate the average longest current streak across the three teams.
Result will appear here.
Let \(M\) be the longest current streak among the three teams. For any team, the event "\( \text{streak} \ge k \)" means it won its last \(k\) games, which happens with probability \(2^{-k}\). For \(k \ge 2\), no two teams can both have streak \(\ge k\) because their last \(k\) games overlap in a head-to-head matchup (only one can win). Therefore:
\[ P(M \ge 1)=1,\quad \text{and for } k \ge 2,\quad P(M \ge k)=3\cdot 2^{-k}. \]
So the expected value is:
\[ E[M]=\sum_{k\ge 1}P(M\ge k)=1+\sum_{k\ge 2}3\cdot 2^{-k}=1+\frac{3}{2}=2.5. \]
The full distribution is \(P(M=1)=\frac14\) and \(P(M=k)=3\cdot 2^{-(k+1)}\) for \(k \ge 2\).
The "current streak" is the consecutive run of the most recent winner. The simulation approximates that by generating many long sequences and averaging the final run length.
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