OK, so is this guy magic, or what?
September 7, 2006 6:58 PM   Subscribe

I need a little help figuring out the odds of a baseball pool scenario actually occurring. (And correctly constructing this question, it looks like).

My co-workers run a baseball pool (this is, of course, hypothetical). As I don't do any betting myself and pay no attention to sports, I have an at-best cursory understanding of the many ways people can wager money on these sorts of things. That being said, here is the deal:

-Each week, each bettor is assigned 2 new teams, drawn randomly from a hat. Teams are picked until all are assigned.
-The first bettor to have either of his/her teams who scores a total of 16 runs in that week (just one team, not both combined) wins the pool.
-If no team achieves this goal, the money is thrown back in and the whole process starts again the following week.

What is the chance that one bettor would win the pool on 5 of the 6 weeks that the pool existed? And what effect would consecutivity have on that result (ie 5 weeks in a row vs 2)?

Extra credit for making it understandable to the many laymen I will be showing your correct answer to.

Scuttlebutt around the building has that this is approaching statistical impossibility, and they have stopped the pool as a result. I have my doubts, but it seemed like an interesting enough question to ask.
posted by nevercalm to Sports, Hobbies, & Recreation (18 answers total)
Urm. How many bettors are there?
posted by Kwantsar at 7:28 PM on September 7, 2006

We're missing a key piece of information -- how many people participate in the pool? Given that there are 30 MLB teams, I'm inclined to believe 15 people, but you don't really make that clear.
posted by Doofus Magoo at 7:32 PM on September 7, 2006

Response by poster: Yes, sorry....there are 15 people in the pool....(left that out bc the wife was itchin to walk the dog and I didn't have time to google how many teams there actually were....I thought it was 32, actually.)
posted by nevercalm at 7:36 PM on September 7, 2006

Here's how I'd figure it out, were I not so lazy:

Take the stats from last year (or if you're ambitious, the last decade) and find out how many times any team has scored 16 runs in a week. The probability of any one team doing it in a given week, based on past performance, then looks like:

P(A) = (# of total 16-run weeks by any team) / [(# of weeks sampled) * (# of teams)]

You could get technical and start trying to do team-by team analysis, or figure out offensive production based on team rosters, but that gets pretty complicated. Doing this should let you... errrr... ballpark it.
posted by chrisamiller at 7:38 PM on September 7, 2006

Roll a 15-sided die. What are the chances of rolling the same number 5 out of 6 times?

1 in 16.8 million that the same number will come up all 6 times.
1 in 186,000 that the same number will come up 5 times out of 6.
posted by LobsterMitten at 7:40 PM on September 7, 2006

Der; those previous numbers are if it's a 16 sided die.

Here are the numbers for a 15-sided die:
1 in 11.4 million that it will be 6 out of 6 the same
1 in 136,000 that it will be 5 out of 6 the same
posted by LobsterMitten at 7:43 PM on September 7, 2006

It's been (too) many years since I had practice with statistics, but disregarding the possibility of nobody's teams scoring 16 runs in a week (which seems like it would be pretty rare; that's less than three runs per game -- for all 30 teams), it sounds like each bettor has a 1/15 chance of winning, assuming a truly random distribution of teams.

From there I think it's a pretty short route to figure out the chance of that 1/15 hitting five out of six times, but I'll be ding-danged if I can remember how. The chance of them winning five consecutive times, however, is easy -- 1/(15^5), or about 1/760k. I think.

However, if the chance of nobody winning in a given week is in fact statistically significant, then you'd need to do something like what chrisamiller describes.
posted by Doofus Magoo at 7:46 PM on September 7, 2006

The die analogy only works if it is guaranteed that any 2 given teams will score a total of 16 or more runs in week, which isn't actually true. So the odds are somewhat less that 1 in 136,000. Which is to say that guy is a bloomin' cheat.
posted by TonyRobots at 7:48 PM on September 7, 2006

Rolling a die is not the proper way look at it. For one, it's possible that none of the teams could get 16 runs, or that all of them could. You can't roll a die once and get every number.
posted by chrisamiller at 7:49 PM on September 7, 2006

Lobstermitten has it.

The total number of possible outcomes of winners is 15^6 = 11.4 million.

The total number of ways that 5 can be won i worked out by saying there are 6 different draws he could have lost, and 14 other potential winners for each draw. This gives a chance of 1 in 11.4 million/(14*6) = 1/135602.

For winning all of them in a row, the idea is the same, except there are only 2 ways it can be done, not 6, so the answer comes out at 1/406808.

These are very very small odds. If I was you, I'd kick the winner out of the draw.
posted by scodger at 7:55 PM on September 7, 2006

Those calculations are from my math consultant, not me.

They are the odds for a particular person (call him Jim) to win.

This assumes that every week some team will score 16 runs; i.e., assuming a winner every week.

Let's say we want to know the odds, in advance, of Jim winning 5 out of 6 times. That's like wanting to know the odds of rolling a 1; so we'll use that as the example...

The odds that a 1 will come up on the 15-sided die are 1/15.
The odds that a 1 won't come up are 14/15.

So if we want to know the chances that we will roll a 1 five times, we take the odds that we will roll a 1 any given time, and raise that to the fifth power.

So 14/(15^6) is the odds that you roll five 1s and some other sixth roll that is not 1, in a specified order.

To find the odds of this happening in any order, not a specified order, we multiply by the number of times we're going to roll. In this case, six.

14/(15^6) x 6

Then, to get the odds that anybody at all in the pool will be this lucky guy (or, in our example, that SOME number, not just 1, will come up five times), we multiply the odds for Jim by the number of people in the pool -- here, 15. (Might be easier to think of it as dividing the big number by 15)

So the actual odds that you were after are:
1 in 9067 that someone in the pool will win 5 out of 6 times.
posted by LobsterMitten at 7:59 PM on September 7, 2006

This guy says that there were an average of 4.82 runs scored per game in the years 2000-2004. Given that teams play 4-6 games per week, it seems like the odds of any team scoring 16 runs in a week are very good.

Thus, the odds of winning the pool are mostly dependent on how early in the week their team comes through. There's no good way to predict when a team will have a blowout game or two and meet their mark.

Thus, without crunching all kinds of crazy baseball stats, I would model it as the roll of a die, as has been explained.
posted by chrisamiller at 8:01 PM on September 7, 2006

To justify the assumption mentioned in my previous comment about there always being a winner, the math consultant has just found some numbers. Last year, the lowest-scoring team in the majors -- Washington -- averaged 23 runs per week.
posted by LobsterMitten at 8:01 PM on September 7, 2006

And math consultant, now hot on the trail of obscure facts, says you have to go back to 1992 to find a team that didn't average 20 runs a week. Maybe whoever's running the pool should increase the number of runs.
posted by LobsterMitten at 8:05 PM on September 7, 2006

Yeah, sorry about that, LobsterMitten. I was working under the assumption that scoring 16 runs was a rare occurrence, because the poster mentioned the pool rolling over.
posted by chrisamiller at 8:05 PM on September 7, 2006

A more important question might be how on earth this guy would have the opportunity to cheat. The only thing that it seems he could do to rig the results is to fix the initial draw.

If you're reasonably sure the draw is fair (e.g., if Mr. Improbable Victory isn't the one creating the ballots, or supplying the deck of cards, or whatever), then you may have to accept Occam's Razor (or at least my understanding of it) and accept the fact that while his run of victories may have been highly implausible, that doesn't mean he deserves to be branded a cheat.
posted by Doofus Magoo at 8:06 PM on September 7, 2006

chrisamiller, nothing to be sorry about! As you can see, team Lobster was working it out as we went too. Now we've got out the big books of old stats and are into serious nerd territory. :)

And as to what to do with Mr. Lucky, I'm actually quite surprised at how *likely* it is that someone will win 5/6. I don't think there needs to be foul play.
posted by LobsterMitten at 8:11 PM on September 7, 2006

Response by poster: Wow, this is a little mind-reeling at 8.20am.

There is no opportunity to cheat. Team names are each written on the back of a playing card while two or three people are in the room. Then another person totally independant of the first group then chooses the teams for everyone. To my knowledge, the gent who won 5 times has never drawn. (Or bought beer, pizza, lunch or a bag of chips for all the other people in the pool after essentially taking their money for over a month, which I guess qualifies him as something other than a "gent.") Not to mention that the person running the poll is someone who everyone seems to trust pretty highly.

At any rate, I think the possibility of cheating is so remote that I haven't even heard mention of it or accusations of him as a cheat. It's more of a "Aw, c'mon, what are the odds, fer crissake? I'm sick of giving this guy my money," which lead to some discussions around the coffee urn and me posting here.

I second being surprised at how likely it is. In all the forehead slapping, I was thinking it would be pretty far north of one in a million, if not more. This has me thinking that I've gotta start getting in on the action.
posted by nevercalm at 5:33 AM on September 8, 2006

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