The Wild Horse

Probability Level 4

Most baseball players round the bases in a conservative fashion. If they hit the ball out of the infield they stop at first, if they hit the ball to the wall they stop at second. This is a safe way to play precisely because you're not testing your luck. Once you hit the ball, everyone knows where you'll end up and it usually isn't challenged.

Another way to play is like Yasiel Puig of the Los Angeles Dodgers, the most exciting player in baseball. Although he is a great hitter who can bat for power, Puig has a very high risk tolerance and tries to make the most out of every single ball he hits.

Whereas $\sim99\%$ of players who hit a ball on the ground to the outfield jog to first base and take a single, Puig views this approach as simply unacceptable. As soon as Puig is out of the batters box, he's constantly judging the defense and the likelihood that he can take an extra base without being tagged out. He's trying to force the defense to make a sloppy move and take advantage of it . Such behavior has led long time LA Dodgers announcer Vin Scully to nickname Puig "The Wild Horse".

Let's use a very simple model for baserunning:

At each base Puig runs to, there is some probability $p$ that the defense makes the play without error and gets him out.
Likewise, there is a chance $1-p$ that the defense messes up, and Puig make it to the base safely.
The opposing team's defensive performance is characterized solely by the parameter $p$ .

Consider Puig's fortune when, on the way to first base, he decides to try for a triple. No matter what happens on the field, he's just going to keep running until he gets to third. At each base along the way (first base, second base, and third base) the defense has a chance to get him out.

Question : At what level of defensive performance, $p$ , does it become more likely for Puig to arrive at third safely than it is for him to be tagged out at some point along the way?

Note

An equivalent question is, when does it become more likely that a cricket batsmen makes three runs than it is for him to be run out along the way?

The answer is 0.206299.

3 solutions

Danny Whittaker
Apr 17, 2014

Pretty simple question from a very complicated situation. I just need (1-p)^3=.5. That is the point when I have a 50% chance of arriving safely.

So, 1-p=(1/2)^(1/3) and p=.2063

On a side note, I really don't like the question. Too much trying to make it relate to baseball when the probabilities are NOT the same for each base in reality.

I'm really confused why this is in Calvin's picks for Level 5 problems. I spent so long to see what I was missing before finally crossing my fingers and typing 0.206, and it turned out to be correct.

Daniel Liu - 7 years, 1 month ago

Jonathan Stieglitz
Apr 13, 2014

The probability of getting out is the same at each base so (1-p) x (1-p) x (1-p) needs to be above 0.5. I wrote a python script that incremented p by .001 while the solution remained above 0.5 with the final result being p=.206.

p = .0000
perc = 1
while perc > .5:
 perc = (1-p)*(1-p)*(1-p)
 print "chance of out: "+str(perc)+", p: "+str(p)
 p += .0005

Nicholas Tomlin
Apr 18, 2014

Since there is a $1-p$ chance of reaching any base, let $(1-p)^3=\frac{1}{2}$ . Hence, $p=\boxed{.2063}$ .

I have a fundamental point to raise here.. the probability that the batsman makes three runs is p p p =p^3. Multiplication is employed here as it is mandatory for him to not be out at each run; only then can he make 3 runs. However the probability of the opposing team taking him out in any of the 3 runs is described as follows: he gets out on the first run -> (1-p) OR he gets out on the second run -> p (1-p) OR he gets out on the third run -> p p (1-p): (1-p) + p (1-p) +p^2(1-p). Thus the final equation to solve becomes this-

p^3>= (1-p) + p*(1-p) +p^2(1-p)

If this is the logic employed, then the solution obtained is different. Where has my logic failed?

Amruta Vasudevan - 7 years, 1 month ago

The chance he gets out at the third base is p (1-p)^2, not p^2 (1-p)

Mehul Gajwani - 7 years, 1 month ago

The Wild Horse

The answer is 0.206299.

3 solutions

0 pending reports