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Skipping over winning percentages?

November 2, 2008 pm30 10:12 pm

(A version of this was once a Putnam problem)

Is it possible to have a winning percentage of .750, and bring it up to a winning percentage of .900, without ever having a winning percentage of exactly .800?

For example:

wins losses per cent
start 3 1 .750
next 4 1 .800


Try instead:

wins losses per cent
start 3 1 .750
next 3 2 .600
next 4 2 .667
next 5 2 .714
next 6 2 .750
next 7 2 .778
next 8 2 .800

Oops again!

Can it be done?

7 Comments leave one →
  1. Brent permalink
    November 3, 2008 am30 12:41 am 12:41 am

    No, it can’t. I have a simple proof but the current margin of time is not wide enough to write it down. =)

    Maybe I’ll come back later if no one else has written a solution…

  2. Brent permalink
    November 3, 2008 pm30 5:28 pm 5:28 pm

    Think of playing games as taking steps in the first-quadrant lattice, with the x-axis being total number of games played, and the y-axis being number of losses. Playing a game corresponds to taking a single step to the right (if you win) or a single step up (if you lose). Records of .800 corresponds to the line 5L = T. It turns out that .750 and .900 are red herrings — it’s impossible to get from *any* percentage less than .800 to any higher without going through .800! Here’s why: note that the line 5L = T intersects an integer lattice point for every single y-coordinate: (1,5), (2,10), (3,15), and so on. Since you can only take one step at a time right or up, if you find yourself to the left of these points, there’s no way to get to the right of them without actually passing through one of them.

    The key point here turns out to be that .800 can be expressed in the form (n-1)/n, which means that for *any* number of losses there *exists* a number of wins which gives a winning percentage of .800. For any percentage with this property (like .750, .900, .8888888…, and so on), it is impossible to go from a lower percentage to a higher one without passing through that exact percentage first. For a percentage like .600 = 3/5 this is not the case: it’s quite possible to go from (say) .550 to .675 without passing through .600 (I leave this as an exercise for the reader).

  3. November 4, 2008 am30 12:56 am 12:56 am

    The crux here is that if your winning percentage is 0.800, the number of wins has to be a multiple of the number of losses: W = 4xL. If you start out with a winning percentage below 0.800, and end up above 0.800, at some critical game you must have gone from below 0.800 to being at or above 0.800. But that can only happen with a win, hence the number of loses says fixed for that game, and there’s no way to go from W 4L if W and L are integers.

    Amusingly, once your winning percentage hits 0.900, it is entirely possible to lose games and end up with a winning percentage below 0.800 without hitting 0.800 on your way down. (e.g. 9/10, 9/11, 9/12 would put you back at 0.750 without ever being at 0.800). We can skip over 0.800 on the way down because the number of losses isn’t an integer multiple of the number of wins.

  4. November 4, 2008 am30 12:58 am 12:58 am

    “…there’s no way to go from W 4L if W and L are integers.”

    (Didn’t like my inequalities…) Read:

    …there’s no way to go from W being less than 4L to W being greater than 4L if W and L are integers.

  5. November 4, 2008 pm30 4:38 pm 4:38 pm


    what other percents can’t be skipped over on the way up? (.500 is the most obvious).

    Better extension:

    what percents can’t be skipped on the way down?

  6. November 11, 2008 am30 4:34 am 4:34 am

    Another way to thinking about skipping over percentages on the way down…

    If your winning percentage is falling, that means your losing percentage is rising. So apply the analysis of rising and skipping winning percentages to the (identical) problem of rising and skipping losing percentages, and then subtract from 100%. Voila! 1/n.


  1. Carnival of Mathematics #43 « The Number Warrior

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