Baseball Musings
Baseball Musings
April 07, 2008
What's the Probability?

A nice probability quiz at The Numbers Guy. Here are two interesting ones:

4. In baseball, suppose the American League champion is better than the National League champion, such that it has a 55% probability of winning each game against the NL champ. Then the NL champ nonetheless will win a best-of-seven-games series four in 10 times. What is the smallest odd number, X, for which a World Series between these two league champs that is best-of-X will ensure that there's a 95% probability of a just result -- the superior AL champ winning?

5. Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind each of the other two, a cow. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, to reveal a cow. He then says to you, "Do you want to change your choice to door No. 2?" Is it to your advantage to switch your choice, assuming you prefer cars to cows?

I believe the answer to four is 383. I have a Python script that computes 95% confidence intervals, and the low end of 383 is 192. The low end of 381 is 190.

On question 5, the answer is to switch. I love asking this question. The answer is if you switch, you win the car 2/3 of the time. When you make the first choice, you'll be wrong 2/3 of the time. If you switch, you'll be right 2/3 of the time!


Posted by David Pinto at 10:44 PM | Statistics | TrackBack (0)
Comments

... as long as the host will never reveal the car. Essentially, if the host never reveals the car, you are getting your initial choice, or *both* of the other two doors. This is the Monty Haul problem (which can be restated so that it's not advantageous to switch, so one must be careful).

Posted by: Subrata Sircar at April 8, 2008 05:45 AM

To complicate things further, suppose a random stranger comes up after the cow door has been revealed. He has no idea which door the contestant has chosen, and he is given the choice of door 1 or door 2. What are his odds?

Offhand I would say 50-50.

Posted by: rbj at April 8, 2008 09:16 AM

No.

It's easier to think about if you assume 100 choices at the beginning, and then you're down to the final two, but the logic is the same.

On the first selection, you had a 1/3 chance of being right (assuming all is fair). If you switch, you have a 1/2 chance of being correct. Which is why you'll be correct (1/3)/(1/2) or 2/3 of the time.

Posted by: Ken Houghton at April 8, 2008 09:28 AM

No, I'm not talking about a guy who had 3 or 100 choices. I'm talking about someone who comes in after it is whittled down to two choices, and has no knowledge of any of the other choices or who has chosen what.

Posted by: rbj at April 8, 2008 09:37 AM

rbj, it depends on whether you mean: how likely is it given what *we* know, or, how likely is it given what *he* knows. The question without qualification is much more naturally read in the first way. In that case, of course, the answer is indeed unchanged, as Ken Houghton says.

Probability is always relative to some kind of information base, prior probability function, or the like.

Catch you over at BB,
RIY

Posted by: RIYank at April 8, 2008 09:43 AM

RIY, I'm adding a wrinkle. I understand the original contestant should switch to increase his odds to 2/3. But what if someone new comes in without all the information, just a choice of 1 or 2. Are his odds, based upon his limited knowledge, different than the original contestant.

In other words, can you have two people, standing side by side, have differing odds of winning based upon differing levels of knowledge. OC switches because he knew about door 3 which was a dud, but new guy gets no boost because he had no knowledge of door 3.

Posted by: rbj at April 8, 2008 09:50 AM

Ken is right, because the contestant has more information than the naive observer in rbj's example. Without the 100 example (which is helpful), think of it this way: the odds that the contestant selected the right door is always 1/3, and it is always possible to remove one of the "cow" doors and be left with one "car" and one "cow" door. The three possible outcomes are "you chose correctly", and two versions of "you chose poorly", and removing one door doesn't change that, so you switch because the host was nice enough to tell you that if you were wrong initially (which happens 2/3 of the time), the car's behind door #2.

Ken's equation is wrong, as is shown by his 100 example. In the 100 example, your odds of selecting correctly at first is 1/100, and removing 98 doors changes odds of the car being behind the other door to 99/100 (1-1/100), which is not (1/100)/(1/2)

Posted by: Ryne at April 8, 2008 09:54 AM

I've never seen the Monty Hall problem fail to start a lively debate.

The thing I like to point out is that the math only holds if the host will *always* offer the switch. Since he knows where the car is, he could decide to offer the choice only when your initial pick was right.

Posted by: Mecha-Shiva at April 8, 2008 11:29 AM

I really don't understand how you guys are getting to your conclusions. Honestly, I think that you guys are complicating something that is really simple. At the beginning you have a 33% chance of being right. when you eliminate the 3rd door, it ups your chances to 50%. You have 2 doors and one right answer. Changing to door 2 isn't going to change the fact that you still only have a 50% shot at winning the car.

Posted by: patrick at April 8, 2008 12:15 PM

Who cares about Monty Hall. Im looking forward to the best-of-383 series.

By the way, the BINOMDIST function in Excel gives a different answer. But I have to make two assumptions: they play all the games regardless of the cumulative won-lost record, and every game is an independent event. Unrealistic, of course.

Under these assumptions, a team with a .55 edge has a 95% chance of winning 135 or more games out of 269 games played.

Posted by: Casey Abell at April 8, 2008 01:53 PM

Oh, a third assumption in my BINOMDIST treatment is that the .55 probability is exactly the same for every game. This could never apply even roughly in the real world, of course, because winning chances would fluctuate so sharply with the starting pitchers, not to mention other factors.

Posted by: Casey Abell at April 8, 2008 02:10 PM

patrick said, "At the beginning you have a 33% chance of being right. when you eliminate the 3rd door, it ups your chances to 50%."

Incorrect. If you don't change your answer after a door is eliminated, your odds remain at 33%. If you really want to stick to 50%, then let's play a little game:

We go through this exercise 1,000,000 times. Every time you get the correct door, I'll give you a dollar. And every time you get the door wrong, you only have to give me $0.80. Deal?

Posted by: Nor at April 8, 2008 02:53 PM
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