The no-hitter is one of the most impressive feats in baseball. It’s no easy task to face more than 27 batters without letting one of them get a hit. So naturally, no-hitters don’t occur very often. In fact, since 1900 there has been an average of only about 2 no-hitters per year.

But what if you had the opportunity to bet that one *wouldn’t *occur?

That’s exactly what happened to sportswriter C. Trent Rosecrans. He had a friend who kept insisting the Reds would be no-hit his season. And with 24 games left in the season, the friend put his money where his mouth is, betting Mr. Rosecrans $5 that the Reds would be no-hit by the end of the year.

Even if the Reds *do *have one of the worst hitting percentages in baseball, would you take the bet that in 24 games there won’t be an event that occurs only twice in an entire year?

Sounds like a no-brainer.

## Calculating the odds

Back in 2012, I calculated that the odds of throwing a no-hitter were approximately 1 in 1,548. If you update that number to include all the games and no-hitters that have occurred since 2012, the odds become 1 in 1,562. The numbers are very similar, but we’ll use the latter since it incorporates more data.

So there is a 99.936% chance that a no-hitter does not occur in any single game. But the bet was that it wouldn’t occur in 24 games. What are Mr. Rosencrans' chances of winning the bet?

**24 games without a no-hitter** = .99936^24 = .98475 = approximately **98.475%**

I wish *I* could make bets with a winning percentage that was that high! For Mr. Rosecrans, 98.475% of the time he’ll win $5, and 1.525% of the time he’ll lose $5. For his friend, the opposite is true. We can use these numbers to calculate the expected value for each side of the bet.

Reds don’t get no-hit: (0.98475*5) – (0.01525*5) = **$4.85**

Reds get no-hit: (.01525*5) – (0.98475*5) = **-$4.85**

## Making it a fair bet

Obviously this was just a friendly wager and was not meant to be taken too seriously. If Mr. Rosecrans regularly made bets with expected values close to $5 with all of his friends, he probably wouldn’t have many left. But what if he wanted to be a *nice *friend? How much money should he have offered in return to make it a fair bet? We’ll simply set the expected value to 0 and solve for the amount of money he’d lose the 1.525% of the time the Reds were no-hit.

0 = (0.98475*5) – (0.01525*X)

0.01525*X = 4.92375

X = $322.87

To make the bet fair, Mr. Rosecrans should offer to pay his friend $322.87 if the Reds get no-hit. And earlier this week the Reds didn’t get their first hit until the 8^{th} inning. Imagine sweating out *that *game if you had over $300 on the line!

## Adjusting for the Reds

One of the reasons the friend bet on the Reds to be no-hit was that they are one of the worst-hitting teams in their league. Their batting average of 0.238 is ranked 28^{th} in baseball. That means, on average, a Reds batter *won’t* hit the ball 76.2% of the time. So if a pitcher wanted to no-hit the Reds, they would need to face at least 27 batters who didn’t get a hit.

**Probability of having 27 straight batters not have a hit** = 0.762^27 = 0.00065 = **approx. 1 in 1,539**

But remember, just because a batter doesn’t get a hit does not mean they’re out. They can get walked, hit by a pitch, or reach on an error. Unless they pitch a perfect game, the pitcher will face more than 27 batters. Let’s look how the probability changes as we increase the number of Reds batters that the pitcher must face without allowing a hit.

**Probability of having 28 straight batters not have a hit** = 0.762^28 = **approx. 1 in 2,020**

**Probability of having 29 straight batters not have a hit** = 0.762^29 = **approx. 1 in 2,650**

**Probability of having 30 straight batters not have a hit** = 0.762^30 = **approx. 1 in 3,478**

**Probability of having 31 straight batters not have a hit** = 0.762^31 = **approx. 1 in 4,565**

This was *supposed* to show that because they are a poor-hitting team, the Reds have a better chance of being no-hit than the average used above. But as you can see, that’s not the case at all. Despite being one of the worst-hitting teams in the league, it appears that it’s *harder* to no-hit the Reds than the historical average.

Things get even odder when you consider that the average batting average (according to Baseball-Reference.com) is 0.263. Using that number, the odds of having 27 straight batters not have a hit is 1 in 3,788. And those odds drop as you increase the number of batters the pitcher has to face. Applying this probability to the number of games played since 1900, we would expect there to be fewer than 100 no-hitters. And how many have there been? *241*!

This is the same conundrum I encountered when finding the odds of throwing a perfect game. The number of perfect games and no-hitters that have occurred is *much higher* than what we would expect based on historical batting statistics. One explanation could be pitching from the wind-up vs. the stretch. With no runners on base (which is always the case in a perfect game and often the case in a no-hitter), the pitcher can always throw from the wind-up. Assuming pitchers are better when pitching from the wind-up, this would result in a lower batting average than normal, thus explaining the higher number of perfect games and no hitters. This would make for a great analysis using Minitab Statistical Software, but since we can’t separate the data on hand into at bats facing pitchers throwing from the stretch vs. the wind-up, we can't test the theory.

Since the Reds have a batting average .025 points lower than the historical average, it’s probably safe to assume they do in fact have a greater chance of being no-hit. The problem is that it’s nearly impossible to quantify how much greater!

## Looking ahead to next year

With the season almost over, it’s unlikely the Reds will be no-hit this year. But what if the two friends decided to do their bet again next year, only this time, they do it at the start of the season. Let’s use our original probability of throwing a no hitter (the one we’ve observed) and determine what the odds are that the Reds go 162 games getting at least one hit per game.

**162 games without a no-hitter** = .99936^162 = .9015 = approximately **90.15%**

The probability of the Reds getting no-hit is still pretty low, but it’s a lot better than the current bet. I just hope next year the friend gets some better odds than even money!