Jordan Eludes Defenders, and Three-Game Losing Streaks

By Ben Blatt

See the updated math below the original post.

There’s no shortage of ways to describe Michael Jordan’s greatness. But here’s a particularly astonishing one: MJ was never part of a three-game losing streak from November 1990 until he retired from the Bulls (for the second time) in 1998. This streak, brought to my attention in Bill Simmons’ recent “Trade Value List” column, includes 500 regular season games and 126 playoff games, for a total span of 626 games. Just how unlikely was this streak? Let’s start by determining the chances for a typical .500 team.

We can accomplish this through the use of a statistical tool known as binomial distribution. Statisticians use this technique to figure out the probability of a success-failure event occurring x times in n number of opportunities, provided that we know the chance of the event occurring any one time. In order for us to use it here, we have to determine n (the number of chances the Bulls had at losing three in a row), p (the probability of a single three game losing streak) and x (the number of single three game losing observed in n trials). With the Bulls, we observed 0 three game losing streaks in n trials, so x equals zero. If we are looking at .500 team, the probability (p) of a team losing three games in a set three game span is simply .500 cubed, or .125. The number of chances the Bulls had at losing three games in a row was 624. This is because there were 624 separate chances for the Bulls to lose three consecutive games (games 1-3, 2-4, 3-5…all the way to 624-626).

Using x=0, p=.125 and n=624, we get a probability of a team going on such a streak at 0.00000000000000000000000000000000006502%, about one out of every two-undecillion tries. If you haven’t heard of undecillion before, don’t worry. Neither have I, or Microsoft Word. Just realize how unlikely of an event that is.

But of course, Michael Jordan and the Bulls were no .500 team. After the losing streak in November 1990 until Jordan retired in 1993, and from when he rejoined the team in 1994 to when he retired again in 1998, the Bulls had a winning percentage of .786. This means the chance of them losing any three games in a row in a set three game span was .214, their losing percentage, cubed: .009800. Using p=.009800, x=0, and n=624, we get a probability of the Bulls going on such a streak at .2114%, or about one out of five hundred.

This is still quite amazing. Even for a team that had the ability to maintain such a high record for such a long time, we should have expected them to have at least one three game losing streak 99.8% of the time.

There are many possible reasons that could explain how Jordan and the Bulls were able to accomplish this. Most of them have to do with the problem of interdependence: each trial in a binomial distribution is supposed to be independent, but each 3-game period shares two games with the periods both preceding and following it. Jordan and the team could have felt a sense of urgency after losing a couple games. Also, the .2114% figure only refers to the chances for those particular 626 games – there’s a much better chance of going on this streak at any point of his career as opposed to this specific set of games.  Or, it could be just that the schedule rarely had them lined up against three challenging teams.

However, they were able to maintain this streak even in over 100 playoffs games when consistently matched against good teams. Additionally, the study assumed that the Bulls had a constant winning percentage of .786. It’s probable that there were times when it was lower, making the streak much more likely to have broken. There are mitigating factors either way. Whatever the reasons, Jordan’s streak is impressive.

Unsurprisingly, the only two teams this season without three consecutive losses are the Lakers and the Cavaliers, who have streaks dating back to January 2008 and March 2008 respectively. Even the Lakers in their three-peat from 2000-2002 were not immune to three game losing streaks. Jordan’s streak, even among other elite teams in the NBA, is rare and amazing.

Ed. note: A couple of astute commenter below outlined a method of calculating the probability that fixed the problem of interdependence. Their calculations put the probability at about one in 150, a larger but still unlikely figure.

Update: As a few of you have pointed out, the methods used to calculate the odds were a bit skewed because all events were treated as completely independent events. While it is true the chance of every three game span has the same chance of being lose-lose-lose, we must remove all prior  lose-lose-lose streaks from the calculations when calculating future streaks. For instance,  when calculating the chances games 3-5 break the run of no streaks it is not strictly the chances of getting LLL, but the chances of getting LLL assuming there was no prior LLL. This means games 1-3 could not have been LLL which means the chances game three was L is a bit lower than 1/2. We are left with a slightly higher probability of 3-5 not being LLL than simply 7/8.

To fix this error we can use a recursive equation which takes into account these past runs changing the probability of the next events. We ended up using an equation associated with the Bernoulli process with the formula g(n)=g(n-1)+(1-(g(n-4))((p^1)*(1-p)^3). In this equation g(n) is the probability of having at least one three game winning streak over n games given the probability of winning one game is p. This equation works because it takes into account the probability of avoiding the streak up to game n-1  (g(n-1)) and then adds the probability of the next game causing the first three game losing streak (1-(g(n-4))((p^1)*(1-p)^3). We know if the nth game causes an end to the streak, then the three games before it must be WLL. If the games one game and two games before game n were not  LL then the next game could not cause a three game losing streak. If the game three games before game n was a L then the sequence would have been LLL and a three game losing streak would have been recorded already. Therefore we want to know the probability of WLLL which is ((p^1)*(1-p)^3). We must multiply this by the probability that there was no three game losing sequence in the games before this which is the (1-(g(n-4)).

When using this formula to calculate the odds we get .000000000000166% chance of avoiding such a streak with a .500 team which is about one in one quadrillion. With Jordan’s percentage of .786 there is a .71% chance of avoiding such a streak which is about 1/140. While these numbers are higher than the probabilities initially calculated, they are still stunningly low and show how improbable of a streak it was even for a team with as high a winning percentage as the Bulls.

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13 Responses to Jordan Eludes Defenders, and Three-Game Losing Streaks

  1. yanzhang says:

    The math here is a bit flawed, because the 624 events are not independent. For example, suppose we knew that the 2nd event (no 3-game losing streak between games 2-4) happened. Given only this, there are 7 equally possible situations for games 2-4, which are WWW,WWL, WLW,WLL, LWL, LLW, LWW. This makes the probability of having a win between the 2nd and 3rd game slightly higher than before (since the LLL possibility is removed), which slightly (only a little bit, since the LLL possibility was so small anyway) reduces the probability of having a 3-game losing streak in games 1-3, which is your first event.

    Each of these “slightly”s is very small, but we have 624 events here, so maybe they’ll add up. I don’t see offhand how much it will change the result, so the point of your post probably remains.

  2. yanzhang says:

    A friend and I just did a computerized calculation, and we get 0.0061, which is about one in 160, not that far from your answer. This is a good thing, because if I just wanted to do an approximation I’d have done what you did too. Besides, like the article points out, there are bigger gray areas like the actual win %, the selection bias of the interval, and a bunch of other factors. One thing we can all agree on – MJ was amazing.

  3. andrew says:

    The hypothesis is correct, but I believe the logic to be flawed.

    Based on the theory at http://mathforum.org/library/drmath/view/56637.html

    The probability of 0 3-loss streaks given the probability of any one loss is 0.214 in 626 trials is 0.00718 or approximately 1 in 140

    Essentially you can’t use your method because the trials are not independent. i.e. if you get WWL in your first 3 games, you have failed your first chance to get a 3-loss streak, but over the next 3 games, you are more likely to get a 3 run streak as LLW or LLL both would count (given the previous game was already a loss).

    I know it’s warped. Probability is hard. :D

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  11. Daniel B says:

    Here’s what I think the analysis misses (not a bad thing, this is tedious stuff and the estimates are still very useful) in terms of interdependence: not only does there need to be no prior LLL, but there also can’t be an immediately prior WWW or LWW because it overlaps.

  12. Daniel B says:

    Here’s, I think, the best way to calculate it. Get a really good pRNG and have it throw out 626 three-digit numbers. If a number from 781-999 manages to appear 3 consecutive times, then an LLL happened, otherwise it didn’t. Then do this a few thousand times and report the results.

    • Daniel B says:

      Doh, I keep hitting enter too early. It’d actually be possible to do this in a reasonable amount of time with a C program or maybe even in Excel.

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