# Is Babe Ruth's statement meaningful?

Quote from Babe Ruth:

Every strike brings me closer to the next home run.

As I understand memorylessness, this is meaningless. For every at-bat, there is a certain probability that he will strike, and there is a certain probability that he will hit a home run, and that's that. The likelihood of a home run at any particular point in time does not increase as strikes accrue.

However, I have an intuitive understanding of what he means. Is there some statistically-rigorous way to express it or make sense of it?

Maybe it makes sense for someone looking back on Babe Ruth's career with the benefit of hindsight. Or, maybe if we imagine an omniscient deity who can see the entire timeline of the universe at once. The deity can indeed see that, from any particular moment, there are N strikes remaining before Ruth hits the next home run. Another strike reduces that number to N-1. So, indeed, every strike brings him closer to the next home run.

Epilogue

If I could go back in time and rewrite this question, I would have omitted all the baseball references and simply described a guy rolling dice, hoping for a seven. He says, "I'm hoping for a seven, but I'm not bothered when I get something else, because every roll of the dice brings me closer to that seven!" Assuming he eventually rolls a seven, is his assertion the gambler's fallacy? Why or why not?

Thanks to @Ben for articulating that this is not the gambler's fallacy. It would have been the gambler's fallacy if he had instead said, "Every roll of the dice which does not result in a seven makes it more likely that the next roll results in a seven."

The guy didn't make any such statement, and he didn't make any statement at all about probability, merely about the passage of time.

By assuming that there is a seven in his future, we have made it undeniably true that every roll of the dice brings him closer to the seven. In fact, it is trivially true. Every second that ticks by, even when he is sleeping, brings him closer to that seven.

• Just to help with googling, what you're describing at the end is Eternalism en.wikipedia.org/wiki/Eternalism_(philosophy_of_time) and I agree that this is the framework in which Ruth's statement makes sense. Nov 16, 2021 at 15:55
• Nov 16, 2021 at 16:37
• I am with @StephanKolassa that this is Gambler's fallacy. "Uggg, bad cards again! I guess that brings me closer to my next straight-flush, though!"
– Dave
Nov 16, 2021 at 16:52
• I think treating this as the Gambler's Fallacy might not give Ruth's quotation due credit. It's not necessarily even about probability. Wouldn't most people understand it as an exhortation to make the effort? In other words, being able to observe an event in a stochastic point process depends on continuing the process.
– whuber
Nov 16, 2021 at 17:25
• I actually think the last paragraph is more accurate than it might seem. Babe Ruth himself is implicitly conditioning on there being a next home run (and implicitly that it is a finite number of swings away). Given that he will hit a home run some time in the future, as far off as it might be, it is true that each swing brings him one step closer to it. Babe Ruth himself is acting as the deity, but allowing N to be unknown but finite, and therefore not having any omniscience beyond the knowledge that he will hit a next home run.
– Noah
Nov 16, 2021 at 18:37

## 4 Answers

#### It is both meaningful and (usually) correct

You are overcomplicating this by bringing probability into a simple non-probabilistic assertion. You need not invoke an omniscient deity in order to accept that there is a reality that exists independently of knowledge of it. (You seem to be operating under the assumption that reality is only admissible to discussion if there is an omniscient being with total knowledge of it; this is a reasonably common misconception of probability, which is examined in this related question.)

The simplest rigorous examination of this statement is a non-statistical analysis based on looking at the underlying population of values pertaining to all the balls Babe Ruth ever hit. Let $$X_1,...,X_N$$ be the ordered career outcomes of all balls faced by Babe Ruth, with $$X_i = \bullet$$ denoting a strike and $$X_i = \diamond$$ denoting a home-run (we need not specify the notation for other possible outcomes). At the end of ball $$n$$ the number of balls until the next home-run is:

$$B_n \equiv \min \{ k \in \mathbb{N} | X_{n+k} = \diamond \}.$$

Now, we know that a strike and a home-run are mutually exclusive --- i.e., no single ball can be both. Consequently, if ball $$n+1$$ is a strike (i.e., if $$X_{n+1} = \bullet$$) and if $$B_n<\infty$$ (i.e., if Babe has at least one home-run left in his career) then we can easily show that $$B_{n+1} = B_n-1$$. This confirms Babe's statement that his strike brings him (one ball) closer to his next home-run.

The only exception to this is when Babe gets to the point where he has already hit his last home-run, so that there are no more home-runs left to come in his career. At this point with have $$B_n = \infty$$ and getting a strike on ball $$n+1$$ still gives $$B_{n+1} = \infty$$. In this latter case Babe is no closer to the next home-run, because there is no next home-run.

Of course, at the time of Babe's last home-run, he probably didn't know that would be his last. (According to this historical account, Babe's last home-run was on 25 March 1935. He went on to play five more times without another home-run.) At that point his saying would be wrong, and looking back in hindsight we now know this.

Ultimately, this statement by Babe Ruth is no more controversial than if he asserted, "The elapsing of time spent not getting a home-run brings me closer to my next home-run". That is of course also true, setting aside the situation where he has no future home-runs to get closer to.

Finally, I do not agree with other comments/answers here that assert that this is the gambler's fallacy. It could (but might not) be a manifestation of the gambler's fallacy if he instead said, "Every strike makes it more likely that I will get a home-run in the future". That could be an example of the gambler's fallacy because it would assert that a bad outcome now makes a good outcome in the future more likely. (On the other hand, if strikes are not independent then it might not be.) In any case, merely asserting that the elapsing of time required for a bad outcome to occur now makes a subsequent good outcome closer in time is not the gambler's fallacy, and is not a fallacy at all.

• Arguably, even if Ruth had claimed that each strike makes a home run more likely, it still isn’t necessarily the gambler’s fallacy—he might very well learn something with each attempt that doesn’t go quite right, so his at-bats aren’t independent. Nov 17, 2021 at 2:18
• @KRyan I am not a baseball expert but the opposing pitcher is also learning things about Ruth which makes Ruth's next home run more or less likely. If they think he can consistently hit it out of the park, then maybe they will purposely walk him. This all goes to show that his next at-bat is not independent. Nov 17, 2021 at 3:00
• @emory Yes, absolutely true—in which case Babe Ruth’s statement could be taken as an assertion that he’s getting more out of it than his opponents were. Considering his record, this may be plausible. Certainly isn’t fallacious, even if someone could demonstrate it doesn’t happen to be true. Nov 17, 2021 at 3:29
• This is a great answer, but viewed through that lense the statement is also sort of ... trivial. Yeah, every strike brings him "closer" to the next home-run, but so would actually hitting a home-run (or even sitting out a round and eating a Snickers instead). It's questionable if he wasn't actually more thinking along the lines of the gambler's fallacy. Nov 17, 2021 at 11:58
• @Joshua it seems you know a lot more about baseball than I do, but my analysis was limited to disproving that each at-bat was independent - which I think I did. A baseball expert such as yourself could construct the exact dependency. Nov 17, 2021 at 15:26

Suppose that at at-bat $$t$$ Babe has access to the information in the filtration $$\mathcal{F}_t$$. Write that Ruth's next home run with be at at-bat $$N$$. Further suppose that each at-bat has probability $$p$$ of being a homerun.

Based on the currently available information at at-bat $$t$$, our best guess of $$N$$ is $$\mathbb{E}[N \mid \mathcal{F}_t] = t + \frac{1-p}{p}$$. At time $$t+1$$,our best guess is $$\mathbb{E}[N \mid \mathcal{F}_{t+1}] = t + 1 + \frac{1-p}{p}$$. Notice that the expected home run time is always a constant $$\frac{1-p}{p}$$ at-bats in the future. Ruth is forgetting that the filtration updates.

As I see it, there are two interpretations to this statement.

The Non-gambler's fallcy interpretation is merely a statement about the deterministic course of events that will occur in the future. In that case, you can index all hit hits that are home runs (i.e. the set $$\{k_1, k_2, \dots k_m\}$$ where $$k_i$$ is the index corresponds to his $$i$$'th homerun (total of $$m$$ homeruns) out of the set of all his hits. After swinging for his $$j$$th career ball, for $$k_i < j < k_{i+1}$$, he is one hit closer to his $$k_{i+1}$$'th hit, which is a homerun. This is not a probabilistic statement, but a statement that is true if you take a deterministic viewpoint of events.

But, in this interpretation, you could say it's meaningless because the fact that he hit the $$j$$'th ball has nothing to do with getting closer to his next home run. All that mattered here was the passage of time. You could equally say that sitting at the dinner table, or falling asleep, or sitting in a room and counting to ten also brings him closer to his next homerun.

The gambler's fallacy interpretation would be this:

"After each strike, the expected number of subsequent hits before my next homerun is lower."

This is the Gambler's fallacy, assuming that each hit in isolation has an equal probability and is independent of other hits. (But this is a questionable assumption.) Under this assumption, if the probability of each homerun is $$p$$, you can calculate that the expected number of swings before the next homerun is $$E[\text{number of swings before homerun}] = \sum_{i=1}^\infty i (1-p)^{i-1} p = \frac{1}{p}.$$ This number is independent of the history of hits, making the statement the gambler's fallacy.

Now, my suspicion is that, if you presented both the interpretations to Babe Ruth, he would suggest that the first interpretation is closer to what he means than the second.

• I don't follow your first argument, because getting up to bat and swinging is quite different from falling asleep (that is, doing nothing). I think your model might not be capturing some important aspects of the situation.
– whuber
Nov 19, 2021 at 21:56

This is very fun to think about. I will answer in terms of probabilities since that is how you framed your original question. Ben has chosen to provide a non-probabilistic answer by retrospectively looking at the sequence of Babe's observed outcomes.

Let's consider a simplified Bernoulli probability model for whether Babe hits a home run. If we let $$X$$ be the result of an at-bat where $$X=1$$ "with probability $$p$$" for a homerun and $$X=0$$ otherwise, then $$P(X=1|p)=p:=\underset{n\rightarrow \infty}{lim}\frac{1}{n}\sum_{i=1}^n X_i$$ is a statement about the proportion of homeruns over many, many (infinite in fact) at-bats. $$p$$ is an unknown fixed constant. When we use probability we are describing the emergent pattern of events over many, many samples. In order for a pattern to emerge that contains homeruns, an eventual home run is inevitable with enough at-bats. It might be more clear to simply say that each at-bat brings him closer to the next home run. This, then, is a no-brainer, a statement of the obvious − Babe cannot strike out indefinitely. Using our simplified Bernoulli model, does this mean that the long-run probability of Babe hitting a home run depends on his past performance?

Where we sometimes run into trouble is if we try assigning probability statements to single events. We can be confident in a single event occurring based on our knowledge of the long-run performance of the process, but a single event does not have a probability. Probability is a proportion so always ask yourself, "A proportion of what?" A proportion of many samples.

The memoryless property is stating that the emergent pattern over many, many samples is the same no matter how the sequence begins, i.e $$P(X=1|p, x_1,...,x_n)=P(X=1|p)$$ where we are considering that each at-bat is independent of the others. I don't think these ideas are at odds with Babe's statement. Dave's answer (now since deleted) is correct in terms of expected values, but he is incorrect to apply these probability statements to a single at-bat. In some sense Babe is owed a homerun (in repeated trials) because he cannot strike out indefinitely and still have an expected value of $$p$$ (or $$\frac{1}{p}$$ for the geometric distribution of the number of at-bats) where $$0. Some discuss this idea using the phrase "regression to the mean." Perhaps Babe's statement and your intuition are leading us to the CDF of the geometric distribution, $$P(K\le k)=1-(1-p)^{k}$$ where $$K$$ is the number of at-bats until Babe's next homerun. $$P(K\le k)$$ is also a limiting proportion of many samples or trials, where each homerun constitutes the conclusion of a single sample or trial. Such statements about repeated trials give us confidence in what we should expect to see in our reality since our reality can be viewed as one such trial.

Suppose we had many measurements allowing us to estimate Babe's long-run homerun rate, as well as knowing his observed sequence. We could use this information to calculate a predictive p-value testing a hypothesis about the number of at-bats until Babe's next home run. This predictive p-value is also a long-run probability, but it gives us confidence in our hypothesis regarding the next observed result while incorporating information on Babe's long-run performance AND his most recent performance. This is not a p-value testing a hypothesis about $$p$$, it is a p-value testing a hypothesis about the next observed result, $$H_0$$: $$K\ge k$$ or $$H_0$$: $$K\le k$$, unconditional on the unknown fixed true $$p$$. This is the sort of analysis used to construct prediction intervals for, say, a time series or a Poisson process. Here is an answer discussing the prediction of a single coin toss, analogous to predicting the result of the next at-bat without considering the recent sequence leading up to the next at-bat.

As a separate but related thought experiment, think of flipping a coin with $$p=0.4$$ for the probability of heads and getting straight tails in 10 flips, where $$X=1$$ denotes a flip landing on heads and $$K=k$$ is the number of flips until a heads appears. If we are confident $$p$$ is indeed close to $$0.4$$ from earlier experiments based on 1,000 flips (and nothing about the coin or the flipper has changed) then we must be witnessing a rare event and it would only be natural to bet on $$K=11$$ since the likelihood of such a streak continuing is exceedingly rare. The predictive p-value testing the hypothesis $$H_0$$: $$K\ge 12$$ $$[X=0$$ (tails) on the next flip$$]$$ is the probability of the discrepancy between the observed result and the hypothesized result or something more extreme, $$1-\Phi\bigg(\frac{\frac{400}{1000}-\frac{0}{11}}{\sqrt{0.4(1-0.4)/1000 + 0.4(1-0.4)/11}}\bigg)=0.004,$$ approximated using a Wald-type test. We are therefore $$100(1-0.004)\%=99.6\%$$ confident the next flip will result in a heads. Additionally the p-value testing the hypothesis $$H_0$$: $$K=11$$ $$[X=1$$ (heads) on the next flip$$]$$ is $$\Phi\bigg(\frac{\frac{400}{1000}-\frac{1}{11}}{\sqrt{0.4(1-0.4)/1000 + 0.4(1-0.4)/11}}\bigg)=0.981$$. I am not suggesting the coin becomes more likely to land heads or tails in any given flip (see my discussion above regarding probability). I am suggesting that if we were to repeat this experiment many thousands of times, the proportion of times where the coin produces a $$\hat{p}=0.4$$ in 1,000 flips and then lands on tails after a string of 10 consecutive tails (or something more extreme) is incredibly small.

The key to succeeding in the long run with this predictive p-value strategy is to bet only on the length of a run before any data are observed (Neyman-Pearson's error rate). Otherwise, if you have historical data, are in the middle of a run, and interested not only in the ultimate length of the run but also the result of the next flip, one can view the predictive p-value as a weight of the evidence without error rate guarantees (Fisher's evidential p-value).

If we take it as known that $$p=0.4$$ and reference the CDF of a geometric distribution we see that the probability of the number of flips being at least 12 until a head is observed is $$(1-0.4)^{11}=0.004$$. This probability from the geometric CDF could be viewed as a p-value testing the hypothesis $$H_0$$: $$p=0.4$$. We could instead retain this hypothesis and call into question whether the next flip will land heads. One might also examine the conditional probability $$P(K>k|K\ge 11)$$.

Knowing the limiting proportion of heads, if we want a safe bet on the outcome of each coin toss that will pay off in the long run then we should just always bet tails. This is like investing in the S&P500. If we want a safe bet on the length of a run then the geometric distribution would indicate we should bet on shorter-length runs.

Above is a time series of 50 flips from a coin with $$p=0.4$$. Observing runs of a few heads or tails is not uncommon, but if a run were to continue long enough we would naturally anticipate it to end, e.g. consider the string of tails leading up to flip 21. In order for $$p$$ to remain a constant $$0.4$$ the coin cannot land on heads or tails indefinitely. This is the probabilistic way of interpreting Babe's statement without invoking the gambler's fallacy, i.e. that the limiting proportion $$p$$ changes as a result of what we observe.

Here is a wikipedia page on prediction intervals and predictive p-values. Here is a paper on the topic.

• Comments are not for extended discussion; this conversation has been moved to chat.
– whuber
Nov 18, 2021 at 18:51
• Just a comment of opinion - this answer in places is too lengthy without good reason. In particular that picture with two bars of proportions seems like a filler. Maybe it would be received better if concentrated on a concrete point. Nov 20, 2021 at 14:43
• @Karolis, I appreciate your comment and initially opted for a brief answer as you suggested. However, whuber was critical of my solution, even questioning how I was interpreting probability (see the comments moved to chat). For this reason I have lengthened my solution and included the bar chart to make it unequivocal. Nov 20, 2021 at 15:00