Baseball Probability

How do I figure this out?

A pitchers throws 60% strikes. A batter does not swing. What is the probability that a batter will get 4 balls before he gets 3 strikes. In other words, what is the probability of gettign a base on balls when a batter doesn't swing and a pticher throws 60% strikes. Thank you

larryperry said:

A pitchers throws 60% strikes. A batter does not swing. What is the probability that a batter will get 4 balls before he gets 3 strikes.

Click to expand...

Please show some work.
What is the probability of four balls straight?

What is the probability of fifth pitch is the fourth ball?

afrazer721 said:

Splitting Events
In our baseball game, the only statistics available say that 60% of pitches are strikes and from this I can say that 40% of pitches are balls. I also know that a batter has a 50% chance of batting right handed and a 50% chance of batting left handed. How can I combine these to get probability I want?
Call y the event that the selected batter is right handed. The event I'm really wanting is x, that this batter will walk first. I split this into two smaller events: x&y and x&not y. These two events are mutually exclusive(both can't occur at the same time). So, P(x)=P(x&y)+P(x&not y). Expressing P(x&y) in terms of conditional probability, I get P(x&y)=P(x|y)P(y) and similarly
P(x&not y)=P(x|not y)P(not y). Put these into the conditional probability formula P(x|y)=P(x&y)/P(y).
I get P(x)=P(x|y)P(y)+P(x|not y)P(not y).
At this point I can use the data I have(x|y)=40%(50%)+60%(50%)=.2+.3=.5=50%.
Thus, the probability of getting a base on balls when a batter doesn't swing and a pitcher throws 60% strikes is 50%.

Click to expand...

I answered this question here.
BTW. This question has nothing to do with conditioning.

A pitchers throws 60% strikes. A batter does not swing. What is the probability that a batter will get 4 balls before he gets 3 strikes. In other words, what is the probability of gettign a base on balls when a batter doesn't swing and a pticher throws 60% strikes.

Click to expand...

If you want a visual representation of the problem, I suggest you try a probability tree diagram. This will allow you to see all the possibilities laid out in a simple format. Though running through all the branches may be a bit tedious, it is a very simple approach. See examples and explanations here:

http://www.mathsisfun.com/data/probability-tree-diagrams.html
http://www.onemathematicalcat.org/Math/Algebra_II_obj/prob_tree_diagrams.htm
http://www.onlinemathlearning.com/probability-tree-diagrams.html

You only have two possible events (a ball or a strike), so you will only have two branches growing off the end of each other branch. You also start with just two branches, which represent the results of the first pitch – either a ball or strike.

Label each branch as either B (for ball) or S (for strike). On each branch labeled B, write “.4”, and on each branch labeled S, write “.6”.

Let each series of branches keep growing until there are three S’s or four B’s along the path from beginning to end. Stop branching whenever one of these events occurs.

Multiply the numbers along each path and write the product at the end of each path. These numbers are the probability of each path/series of balls and strikes occurring.

You can add all the paths that have four B’s in them to find the probability of a walk (base on balls).

Note: This is neither the quickest nor simplest approach, but it may prove instructive if you wish to “get a feel” for this sort of problem.

Well after reply #5, I will give you a back-of-the-book answer. Now you task is to justify the answer.
\(\displaystyle \displaystyle\sum\limits_{k = 0}^2 {\binom{3+k}{k}\left( {0.4} \right)^4 \left( {0.6} \right)^k } \)

pka said:

I answered this question here.
BTW. This question has nothing to do with conditioning.

Click to expand...

Thank you for the constructive feedback. I looked over the problem again and decided to try and map out the events and probabilities with a tree diagram. For instance on the third pitch, P3, I state that 60% chance that pitch is a strike and 40% chance that pitch is a ball. Branching off of strike for the third pitch, I assume there's a 50% chance of a strikeout and 50% chance no strikeout. Thus,
P(strike^strikeout)=.30 and P(strike^strikeout*)=.30. I'll take a look at the tree diagram websites listed above. They look satisfying.
^ means intersection
* means complement

A pitcher throws 60% strikes. A batter does not wing. What is the probability that a batter will get 4 balls before he gets 3 strikes. In other words, what is the probability of the batter walking before they strike out, while the pitcher throws 60 percent strikes?

^ means intersection

I map out events and the probabilities with a tree diagram: The first pitch, P1, has a 60% chance of being a strike. 40% chance of being a ball. The complement of .6 is 1-.6=.4. So far my tree diagram looks like this, <. The top line (branch) is strike, the bottom line(branch) ball. Branching off strike are strikeout, no-strikeout. Conditional probability P1(strikeout|strike)=.6 raised to 1=.6. I raised it to one because it's the first pitch. Thus,
P1(strike^strikeout)=.6*.6=.36. Branching off ball are walk, no-walk . P1(walk|ball)=.4 raised to 1=.4 Hence,
P1(ball^walk)=.4*.4=.16. P1(no-strike|strike)=.4. P1(strike^no-strike)=.24

The second pitch also has a 60% chance of being a strike. 40% chance of being a ball. Branching off of P2 strike are its strikeout, no-strikeout. P2(strikeout|strike)=.6 raised to 2=.36. Thus, P2(strike^strikeout)=.36*.6=.22. Again branching off ball are walk, no-walk. P2(walk|ball)=.4 raised to 2=.16. P2(ball^walk)=.16*.4=.064.

I repeat this for six pitches. I then add up the six (ball^walk) probabilities for the probability of getting a base on balls when a batter doesn't swing and a pitcher throws 60% strikes as my answer. I don't know if this is correct?

afrazer721 said:

Thank you for the constructive feedback. I looked over the problem again and decided to try and map out the events and probabilities with a tree diagram. For instance on the third pitch, P3, I state that 60% chance that pitch is a strike and 40% chance that pitch is a ball. Branching off of strike for the third pitch, I assume there's a 50% chance of a strikeout and 50% chance no strikeout. Thus,
P(strike^strikeout)=.30 and P(strike^strikeout*)=.10. I'll take a look at the tree diagram websites listed below. They look satisfying.
^ means intersection
* means complement

Click to expand...

The batter does not swing the bat at all.
The umpire calls a ball or a strike.
Assuming the ump is actuate \(\displaystyle \mathcal{P}(B)=0.4~~\mathcal{P}(S)=0.6\).
These trials are independent.
Thus \(\displaystyle \mathcal{P}(BSBBSB)=(0.4)^4(0.6)^2\), so the batter walks on the sixth pitch.
But there are \(\displaystyle \displaystyle\binom{5}{2}=10\) ways that can be done.