Probability of 20 straight

We go through the probabilities associated with streaks in a variety of settings.  While we use 20 as a benchmark for the calculations below, the discussion is meant to more broadly capture the spirit of a streak of approximately this value.

Sports
The 2002 Oakland A's won 20 straight games, an American League record since 1901.  If a team has a 50% probability of winning any one game, then the probability associated with a 20-game streak is 50%^20, which is nearly 0%.

Before the streak, what probability of winning would need to be associated with the 2002 Oakland A's, so that the winning streak has a 50% chance of occurring?  

p^20 = 50%

ln(p^20) = ln(50%)

ln(p) = ln(50%)/20

p = e^(-.035)

p = 97%

Keep in mind that this record was set relatively recently so it is more difficult to empirically estimate how rare such a record exists.  Streaks occur in both directions however, and the American League losing streak was 21 games, set by the Baltimore Orioles.  In 1988, this too is a relatively recent occurrence.  The winning and losing streak records in the parallel National League, on the other hand, both occurred nearly a century ago.

Financial markets
Let’s turn our attention to the 2013 Tuesday winning streak in the Dow Jones Industrial average index.  The streak is currently at 17, while the previous record was for a higher number of straight Wednesdays.  Once again, we’ll do this analysis on 20 straight wins, which happens to be the average of this Tuesdays' streak and the prior Wednesdays' streak.

We know from the calculation above that the probability associated with 20 straight wins, from a fair random walk process, is nearly 0%.  What daily gain probability would be associated with a 20% probability of seeing at least one weekday having a streak of 20 wins (e.g., 20 Tuesdays, or 20 Wednesdays, etc.)?

[1-(p^20)]^5 = 100%-20%

[1-(p^20)] = 96%

p^20 = 4%

p = 86%

Say that the daily chance for an up-move on the Dow Jones Industrials is, in fact, nearly 86%.  What is the precise probability associated with one weekday having this streak of 20?  This is a more precise calculation versus the calculation above of “at least one” weekday having this streak of 20.  This is a binomial distribution where r=1 weekday, out of n=5 weekdays.  The binomial distributions solves for the joint probability of events multiplied by the permutation of events.  One may notice that ordering matters in the equation, but it is on both sides of the product so the ordering factor cancels.

permutation * joint probability
= n!/[r!*(n-r)!] * p^r * q^(n-r)

= 5!/[1!*(5-1)!] * [86%^20]^1 * [1-86%^20]^4

= 18%

We see that this 18% chance for one weekday is slightly less than the 20% for the chance for “at least one weekday” having this streak of 20 (note that when the probability of an up-move is 50%, the higher probability still quickly converges to 0%).  Now assuming nothing else changes, what is the probability of seeing both one weekday having a streak of 20 gains, while simultaneously another weekday having a streak of 20 losses?

There are two ways to solve this, one with a nested binomial formula and the other with a trinomial formula.  We show both below.

For the nested binomial we solve first for the complement in three workdays:

l = probability of a down-move, which is 100%-86% (or 14%)

l^20 = nearly 0%

complement = 100% - 86% - 0%

= 14%

permutation * joint probability

= n!/[r!*(n-r)!] * p^r * q^(n-r)

= 5!/[2!*(5-2)!] * [86%^20]^2 * [1-86%^20]^3

= 2%

We then solve for the two remaining workdays being a streak of 20 wins and 20 losses:

p = 86%^20 / (86%^20 + 0%^20)

= 100%

permutation * joint probability

= n!/[r!*(n-r)!] * p^r * q^(n-r)

= 2!/[1!*(2-1)!] * 100%^1 * 0%^(1)

= 0%

So 2%*0% implies a 0% chance using nested binomial.  Let’s now see this is through a trinomial formula:

s = number of weekdays with a streak of 20 losses

permutation * joint probability

= n!/[r!*s!*(n-r-s)!] * p^r * (l^20)^s * q^(n-r-s)

= 5!/[1!*1!*(5-1-1)!] * [86%^20]^1 * [14%^20]^1 * q^(n-r-s) 

= 20 * 4%^1 * 0%^1 * [(100%-4%-0%)^20]^(5-1-1)

= 0%

So both the nested binomial and the trinomial approaches show a near 0% probability of having both a 20 winning streak and 20 losing streak, at the same time.  

Conclusion

A better explanation for the streaks in the financial market data, as opposed to the sports data above, is that the probability associated with an up-move is not stationary and fixed at 50%.  Ex-post we can see regimes that have randomly oscillated about, even as the historical de-trended average is balanced about 0%.  A market regime that is only part way biased in one direction, say a 60% daily chance for a performance gain, could see a streak in the mid-single digits with the same probability that we see the current financial market streak of 20.

And recall that streaks can and do occur, in both directions.  Another difference with plain-vanilla financial markets streaks versus sports streaks, is that sports probabilities are paired within zero sum games.  A streak of 20 wins by Team X reconciles with 20 losses for all the teams that had played against Team X.  It is rare however, as we have eluded above in the financial markets example, to have record losing streaks in sports that occur in the same season(s) as record winning streaks.  While negative correlation does appear in plain-vanilla financial markets, and the art of multivariate regression can asses the combination of other asset class performance(s) in one direction versus that counterbalance of an isolated asset class (e.g., X) performance in the opposite direction, it is never inherently perfectly (negative).  Still rare, though more likely for financial markets, are large streaks in one direction are then serially mean-reverting (i.e., followed by again somewhat large streaks in the opposite direction).

A side note: the May Barron's article was meant to cite this link, instead of the article above.