Bookies & Statisticians: How are odds assigned?

The following baseball sucker bet generally pays 8 to 1:

“On any specific day, you can pick any three hitters whom you believe will collectively have six or more hits among them. You must specifiy the game for double-headers. If one of your guys doesn’t play, or if a game is rained out, too bad. Your other two guys better have great days. Doubles and triples are considered one hit.”

Here’s my question: By what process are the “8 to 1” odds determined to be optimum (balancing the bookie’s advantage against the odds required to maximize bettors’ participation)?

Please address the above sucker bet in your explanation. Let’s also discuss a couple of assumptions that must be included in the calculations: If each batter gets five at-bats in his game, and if each batter’s average is .300, then our expectation is that they will collectively get 4.5 hits (3x5x.3 = 4.5).

Does the co-log of the hyperbolic log of 2 (.693) eneter into this? This is the factor that determines the number of trials required before an event with “x to 1” odds becomes a 50-50 proposition.

Thanks in advance,

Sam

I think you’re asking at least two questions: 1. How does one compute the probability of winning a bet like the “sucker bet” that you describe? 2. How do bookies arrive at odds, given, perhaps, knowledge of winning probabilities.

For the bet you describe, if we assume that each hitter gets exactly five at-bats, and that all the fifteen at-bats are mutually independent and lead to a hit with the same probability, then the distribution of the total number of hits is a binomial distribution. If we assume that the probability of a hit is 0.3 (which is not the same as batting 300 because outcomes such as walks don’t enter into the batting average), I calculate a 27.8% chance of at least six hits. So that wouldn’t be a sucker bet at all if those assumptions were correct, but who knows what the actual numbers are? Also, it won’t really be binomial because the number of at-bats is not really fixed, the hit probability is not the same in all at-bats, and at-bats are probably not independent (I imagine that hitters have good days and bad days, but I don’t really know).

So, how are odds set? It was my understanding that bookies attempt to get equal amounts of action on both sides of a bet. They (ideally) have no interest in the outcome. They’ve essentially arranged for one customer to bet against another, in exchange for a fee (if one side gets 1:2 odds, the other gets slightly worse than 2:1). The odds are set to be what they need to be to arrive at this, and this may change over time. Essentially, the marketplace of betters sets the odds.

Perhaps the bet that you describe doesn’t work that way. But then it would be like a casino table game. Should a casino pay out 98% of the bet on average, or go down to 95% and lose some customers? Marketing considerations will determine that.

Skeptic42,

Thanks for responding. Let me clarify: I am asking both questions, but mostly the 2nd one. Your analysis of the 1st question is incisive, and so let’s assume the 5 at bats and the .300 average for all three batters. (There are many players with higher avgs, but the higher the avg, the more likely he is to be walked. This argues that the bettor should pick three players whose avgs are low enough that they will actually be pitched to.) The bet is a sucker bet because the 4.5 hit expectation is lower than the 6 hit requirement. What I want to know is: How do we convert that 1.5 hit difference into the 8 to 1 odds offered by bookies?

Bookies do attempt to balance action on “Line” and on “Over/Under” bets. Line bets are bets on which team will win, and they incorporate a point spread (e.g., “San Francisco to beat Los Angeles by 2 runs.” If you bet on SF, their victory margin has to be in excess of 2 pts for you to collect. If you bet on LA, they either have to win outright or lose by no more than 1 run for you to collect. Ties are a push). Over/Under bets are on the total runs for both teams (e.g., if bookies calculate that SF should get 5 runs and LA should have 3 runs, the O/U is 8).

The sucker bet I described, however, is known as a proposition (“prop”) bet, also sometimes as a “gimmick” bet. There is no way for a bookie to balance his action against a bettor who, for example, picks Jeter, Bonds, and Ramirez as his three hitters. The bet is structured so that the bettor has to pick three guys who will get six hit among them; there’s no option for picking three guys who won’t get those six hits. Balancing the action is not possible in bets like this.

So, again, my question is: How do we convert a 1.5 hit difference between the expectation and the bet requirement into an 8 to 1 payoff?

Thanks again,

Sam

There is only one thing that you can say for sure about any casino proposition bet (assuming a competent sports book manager which is not always the case, especially in start-up books that are trying to create action) and that is that the prop bet will not pay anything close to true odds. If I wanted to hazard a guess at the likelihood of any 3 batters getting 6 hits in one day and every casino offering the bet is paying 8 to 1, my guess is that the actual likelihood is at least 15 to 1. If I wanted to hazard a bet I would take Ichiro and 2 other batters and figure out how many times those 3 accomplished the feat over the past 100 games. That would give me a better idea than any formula, I think. Without any idea of who you are picking my guess is that your picks did not do this more than 7 times in those 100 games.

How important is this? My experience is that it is just this sort of little overlooked possibility that gambling institutions can make money on. I don’t know jack about baseball: how often is a game going to get rained out or a player not going to play, making the odds drastically worse for the person betting.

No one batting .300 would average 1.5 hits a game though:

Derek Jeter .318 2267/1764 1.29

Alex Rodriquez .306 2168/1831 1.18

Rod Carew .328 3053/2469 1.24

Wade Boggs .328 3010/2440 1.23

Even Ty Cobb .366 4189/3035 1.38

what is your point?

This can be modeled as a standard Poisson distribution problem. You sum the expected number of hits by the three players in the game, and subtract the probability of them getting 0 through 5 hits in the game from 1.

E.g., if their expected number of hits is 3.5, then the probability of them getting 6 or more hits is

1 - (e[sup]-3.5[/sup] + 3.5 * e[sup]-3.5[/sup] + 3.5[sup]2[/sup] * e[sup]-3.5[/sup]/2 + 3.5[sup]3[/sup] * e[sup]-3.5[/sup]/6 + 3.5[sup]4[/sup] * e[sup]-3.5[/sup]/24 + 3.5[sup]5[/sup] * e[sup]-3.5[/sup]/120) = 14.3%, about 1 chance in 7.

Then I guess we are all on the road to wealth. All we have to do is head for vegas and pick our 3 hitters and collect our 8 to1 on a 7 to 1 poroposition. I’ll tell you what though, you guys go without me, I have enough tax problems.

I like your approach, and I think we’re getting close to the solution, but denquixote is right: an expectation of 1 in 7 isn’t going to have bookies offering 8 to 1 odds. Something has to be wrong with the math, but I still think you might be on the right track.

Thanks,

S_

That assumes that the number hit by any player has a Poisson distribution. See no reason to assume that. Of course the binomial I mentioned above is just a crude approximation too, for the reasons I mentioned. A real answer would come from looking at historical data.

I wouldn’t say that makes it a “sucker bet” without reference to the odds being offered (and, of course, the probability of a win). Expectation < requirement doesn’t even guarantee that your chance of winning is less than a half (the expectation is not (necessarily) the same as the median). But that’s not really my point.

We don’t. We use theory or historical data to get a probability that the bet will be won (4.5 vs. 6 is not enough information; the distribution around 4.5 could be wide or narrow). Odds that would be fair follow trivially from this. A bookie will give you worse odds than this. In effect, the bookie will mark up the price of the bet. Asking how the size of this is determined is akin to asking how retailers determine their markup.

It seemed clear to me: that the mean 1.5 hits per game in the OP is too high, not only for the hypothetical 0.300 hitters mentioned, but also for hitters with even higher averages.

It was a 6 to 1 proposition according to the math in question. But we don’t actually know the distribution of number of hits; that was just a crude approximation.

MY guess is that the book that offers this bet, or the statistician that proposes it to the various books in town has all of the historical data on a program for probably at least 10 years and can tell you how many times any particular group of 3 hitters accomplished this feat in one season, not how many times 3 guys did this - which probably happens every time there is a full slate of games. My guess is still that it happens no more than once in 15 days for the most accomplished 3 batters.

Den,

How about this: What is the formula by which we can establish the probability that any three hitters scheduled to play in a game will collectively get six hits? Let’s set aside, for the time being, the hitters’ batting averages. Wouldn’t we need to know the total number of hitters among all teams (as the denominator) and every combination of hitters as the numerator? That’s do-able, but how do we factor in the six hit requirement? And, once we’ve come up with a factor that applies to every possible combination of three hitters, how do we incorporate the likelihood that there are going to be groups of three hitters who are more apt to get the six hits than other groups of three hitters are going to be?

Thanks,

Sam

I am not a mathemetician so I really can’t even begin to comment on your ideas, but I have been to las vegas and I can guarantee you that advanced mathematics degrees are not a prerequisite for running a sports book. Those guys may do some kind of computer simulation based on battting averages, official at bats, home field advantage, weather etc. but i doubt it. You wanted initially to know how they arrive at the odds and I think they calculate the historical frequency and then probably double or triple it to remain on the safe side or in other words an 8-1 prop would probably pay 15-1 to 25 -1 in order for it to be paying true odds. I think you are giving the oddsmakers way to much
credit.

In my opinion, if you want to determine the likelihood of success that a player will have today, you have to see how he performed in a similar situation in the past. You cannot fail to factor in batting average (particularly against the prospective opposing pitcher if it is available, or at least against left and right handed pitchers who are similar), whether the batter is at home (in order to get the maximum at bats he should be on the road, but some players perform significantly poorer on the road), how cold is it, are they playing in Denver (where more hits and runs seem to occur) and a variety of other factors. A scientific methodology may help, but in my mind there are too many factors

I cannot seem to find any statistical site with multi-hit game records for batters although i am sure they are out there. I did find a record of Ichiro’s hits on a game by game basis for this season. In his first 83 games he was hitless 14 times, 1 hit 27 times, 2 hits 32 times, 3 hits 7 times, 4 hits twice and 5 hits once. Therefore there was better than a 50 % chance of Ichiro getting 2 or more hits. Unfortunately he is unique, but there may be other players with high multi-hit game totals.

Assuming there were two others like him it would seem that betting on those 3 at 8 to 1 would be a good thing. Basically there is a 50% chance of 2 or more hits or 1 in 2 times 1 in 2 times 1in 2 = 1 in 8 plus those 3,4,5 hit games where the other 2 have the needed support should cause a favorable outcome, IF there were 2 others like Ichiro (at least this year anyway.)

You’ve also got to ensure that they’re all playing on the same day. Not too hard – for example, tomorrow (Friday) there are 34 teams playing, which means you have a pool of at least 306 batters to choose from. Why not check Friday’s box scores, and count up the number of players with 1,2,3,4… hits? As long as we assume Friday is a representative sample, we’d be able to calculate how many 3-batter combinations would pay out, and how many would not. Assuming 306 batters, there are 4,728,720* unique unordered lists to submit to the bookie, if we also assume that you know who’s playing and who’s not playing. So if the number of ways to get to six hits on Friday is on the close order of 600,000 then 8:1 is decent odds.

    • I used “306 choose 3” to get this number to avoid duplicate lists.

1.5 hits per game is way too high. The 200 most prolific hitters (most number of hits, not best batting average) are averaging 1.04 hits per game. Of those 200, only Suzuki has cracked 1.50, and only 5 (including Suzuki) are doing better than 1.40 hits per game. It’s possible that there are players sitting in the middle of the league who have better hits-per-game statistics, but the top ten most prolific hitters are all in the top twenty hits-per-game (in my sample) so it looks like the two are strongly correlated.

I think the key to winning the bet is to look for a night when at least two of your three favorites are facing some of the weakest pitchers in the league, and cross your fingers.