Two Batting Average Riddles
Posted by Rob Herman at July 30th, 2006
1 (Not too hard). On every day of a given season, player A has a better batting average than player B. Does it follow that player A has a better batting average than player B over the whole season? Prove or give a counterexample.
2. (I don’t know the answer to this one.) At some point during the season, a player has a batting average of less than 80%. At some later point, that player has a batting average of greater than 80%. At some point, his average must have been exactly 80%. (If you don’t believe me, try it.) Why is this the case? What other percentages is this true for?
P.S. I don’t know why his batting average is so high. Maybe this is some weird hitter-friendly version of baseball.
ok off the top of my head:
Warning potential answers
1. Counter Example:
Game One: Player A hits 1 for 4 or 0.250, Player B hits 0 for 1 or 0.000
Game Two: Player A hits 1 for 1 or 1.000, Player B hits 3 for 4 or 0.750
Season : Player A hits 1 for 5 or 0.200, Player B hits 3 for 5 or 0.600
Season A=2 for 5 or .400 B= 3 for 5 or .600
(unless on every given day it means you do a cumulative batting average, which doesn’t make much sense because on the last day the premise is that A has a better cumulative than B)
2. No idea, I really lack the formal proof math training needed to prove this.
During the season player A hit 2 for 5 in your example. giving him an average of .400. The example still holds I just like to nittpick
For the second part.
The proof:
(NOTE: [LT] = less than, [GT] = greater then)
At some point immediately after missing x number of times out of y pitches (having an average of 1-x/y[LT].8) the batter will hit the ball and have an average greater than .8 (1-x/(y+1)[GT].8). thus
1-x/y[LT].8[LT]1-x/(y+1)
subtract 1 and multiply by -1 gives you
x/(y+1)[LT].2[LT]x/y
Let z be a positive real number such that x/z = .2 thus
x/(y+1)[LT]x/z[LT]x/y
multiply by x and raise to the power of -1
y[LT]z[LT]y+1
thus z can not be an integer
now let us look at x/z=.2
divide by .2 multiply by z
x/.2=z or x/(2/10)=z or 10*x/2=z or 5x=z
x is an integer (number times the batter has missed) so 5x is an integer thus z is an integer therefore y[LT]z[LT]y+1 is false
1-x/y[LT]1-x/z[LT]1-x/y+1
by the proof above one can conclude that any precentage x where (1-x)^(-1) is an integer can not be “skipped over” when going from a lower average to a higher average.
BTW the last line of the proof was suppose to be deleted.
so .5, 2/3, .75
any percentage of the form x/(x+1) where x is an integer greater than 0 meets this criteria (.8=4/5)
(1-(x/(x+1)))^-1=(((x+1)/(x+1))-(x/(x+1)))^-1=((x+1-x)/(x+1))^-1
=(1/(x+1))^-1=(x+1)/1=x+1 since x is an integer x+1 is an integer
Nuts, I didn’t check this soon enough to chime in. Oh well. Cheers, John.
incidentaly there are only 15 averages which can be expressed in the standard 3 significant digits used for batting averages that fit this criteria they are:
.500, .750, .800, .875, .900, .950, .960, .975, .980, .990, .992, .995, .996, .998, .999