This week the #puzzle is: How Long Is the All-Star Streak? #math #probabilities #InfiniteSums #coding #MonteCarlo

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.)

And for extra credit:

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.)

How Long Is the All-Star Streak?

Highlight to reveal (possibly incorrect) solution:

Math StackExchange Program

Method 1:

• 1 win in a row has probability 1/2
• This is actually the probability of 1 team first losing, then winning, and there are 2 teams, 2 * 1/4 = 1/2
• 2 wins has probability 1/4
• n wins has probability 1/2ⁿ
• This checks out, as the sum of all these probabilities is 1.
• The average is 1 * 1/2 + 2 * 1/4 + 3 * 1/8 + … + n * 1/2ⁿ + …
• According to Math StackExchange (and probably a lot of other sources), this is 2.

Method 2:

• Monte Carlo. Same result.

And for extra credit:

Spreadsheet

Method 1:

• This time the probability of n wins is 3 *1/2 * 1/2ⁿ
• The spreadsheet shows, that the probability of a streak of 1 win is 1/4, as a special case
• The average is 1 * 1/4 + 2 * 3/8 + 3 * 3/16 + … + n * 3/2ⁿ⁺¹ + …
• The sum of n/2ⁿ from n = 1 is 2, as seen above; from n = 2, it’s 2 – 1/2 = 3/2
• Most of the above sum (from n = 2) can be rewritten as the sum of 3 * 1/2 * n/2ⁿ = 3 * 1/2 * 3/2 = 2 1/4
• The first part of the sum is 1 * 1/4 = 1/4
• The whole sum, the average, is 1/4 + 2 1/4 = 2 1/2 = 2.5.

Method 2:

• Monte Carlo. Same result.