**A SINGLE EVENT DOESN’T HAVE A PROBABILITY**

A believer in single-event probabilities takes the view that a single flip of a coin or roll of a dice has a probability. I do not. A probability represents the frequency with which an outcome occurs over the very long run, and it is only an average that conceals random variations.

The outcome of a single coin flip can’t be reduced to a percentage or probability. It can only be described in terms of its discrete, mutually exclusive possibilities: heads (H) or tails (T). The outcome of a single roll of a die or pair of dice can only be described in terms of the number of points that may come up, 1 through 6 or 2 through 12.

Yes, the *expected* frequencies of H, T, and and various point totals can be computed by simple mathematical operations. But those are only *expected* frequencies. They say nothing about the next coin flip or dice roll, nor do they more than approximate the actual frequencies that will occur over the next 100, 1,000, or 10,000 such events.

Of what value is it to know that the probability of H is 0.5 when H fails to occur in 11 consecutive flips of a fair coin? Of what value is it to know that the probability of rolling a 7 is 0.167 — meaning that 7 comes up every 6 rolls, on *average* — when 7 may not appear for 56 consecutive rolls? These examples are drawn from simulations of 10,000 coin flips and 1,000 dice rolls. They are simulations that I ran once, not simulations that I cherry-picked from many runs. (The Excel file is at https://drive.google.com/open?id=1FABVTiB_qOe-WqMQkiGFj2f70gSu6a82. Coin flips are at the first tab, dice rolls are at the second tab.)

Let’s take another example, one that is more interesting and has generated much controversy of the years. It’s the Monty Hall problem,

a brain teaser, in the form of a probability puzzle, loosely based on the American television game show

Let’s Make a Dealand named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to theAmerican Statisticianin 1975…. It became famous as a question from a reader’s letter quoted in Marilyn vos Savant’s “Ask Marilyn” column inParademagazine in 1990 … :Suppose you’re on a game show, and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice

Vos Savant’s response was that the contestant should switch to the other door…. Under the standard assumptions, contestants who switch have a 2/3 chance of winning the car, while contestants who stick to their initial choice have only a 1/3 chance.

Vos Savant’s answer is correct, but *only* if the contestant is allowed to play an unlimited number of games. A player who adopts a strategy of “switch” in every game will, in the long run, win about 2/3 of the time (explanation here). That is, the player has a better chance of winning if he chooses “switch” rather than “stay”.

Read the preceding paragraph carefully and you will spot the logical defect that underlies the belief in single-event probabilities: The long-run winning strategy (“switch”) is transformed into a “better chance” to win a particular game. What does that mean? How does an average frequency of 2/3 improve one’s chances of winning a particular game? It doesn’t. Game results are utterly random; that is, the average frequency of 2/3 has no bearing on the outcome of a single game.

I’ll try to drive the point home by returning to the coin-flip game, with money thrown into the mix. A $1 bet on H means a gain of $1 if H turns up, and a loss of $1 if T turns up. The expected value of the bet — *if repeated over a very large number of trials* — is zero. The bettor expects to win and lose the same number of times, and to walk away no richer or poorer than when he started. And for a very large number of games, the better *will* walk away approximately (but not necessarily exactly) neither richer nor poorer than when he started. How many games? In the simulation of 10,000 games mentioned earlier, H occurred 50.6 percent of the time. A very large number of games is probably at least 100,000.

Let us say, for the sake of argument, that a bettor has played 100,00 coin-flip games at $1 a game and come out exactly even. What does that mean for the play of the next game? Does it have an expected value of zero?

To see why the answer is “no”, let’s make it interesting and say that the bet on the next game — the next coin flip — is $10,000. The size of the bet should wonderfully concentrate the bettor’s mind. He should now see the situation for what it really is: There are two possible outcomes, and only one of them will be realized. An average of the two outcomes is meaningless. The single coin flip doesn’t have a “probability” of 0.5 H and 0.5 T and an “expected payoff” of zero. The coin will come up *either* H or T, and the bettor will either lose $10,000 or win $10,000.

To repeat: The outcome of a single coin flip doesn’t have an expected value for the bettor. It has *two possible values*, and the bettor must decide whether he is willing to lose $10,000 on the single flip of a coin.

By the same token (or coin), the outcome of a single roll of a pair of dice doesn’t have a 1-in-6 probability of coming up 7. It has 36 possible outcomes and 11 possible point totals, and the bettor must decide how much he is willing to lose if he puts his money on the wrong combination or outcome.

In summary, it is a logical fallacy to ascribe a probability to a single event. A probability represents the observed or computed average value of a very large number of like events. A single event cannot possess that average value. A single event has a finite number of discrete and mutually exclusive *possible* outcomes. Those outcomes will not “average out” in that single event. Only one of them will obtain, like Schrödinger’s cat.

To say or suggest that the outcomes will average out — which is what a probability implies — is tantamount to saying that Jack Sprat and his wife were neither skinny nor fat because their body-mass indices averaged to a normal value. It is tantamount to saying that one can’t drown by walking across a pond with an average depth of 1 foot, when that average conceals the existence of a 100-foot-deep hole.

It should go without saying that a specific event that *might* occur — rain tomorrow, for example — doesn’t have a probability.

**WHAT ABOUT THE PROBABILITY OF PRECIPITATION?
**

Weather forecasters (meteorologists) are constantly saying things like “there’s an 80-percent probability of precipitation (PoP) in __________ tomorrow”. What do such statements mean? Not much:

It is not surprising that this issue is difficult for the general public, given that it is debated even within the scientific community. Some propose a “frequentist” interpretation: there will be at least a minimum amount of rain on 80% of days with weather conditions like they are today. Although preferred by many scientists, this explanation may be particularly difficult for the general public to grasp because it requires regarding tomorrow as a class of events, a group of potential tomorrows. From the perspective of the forecast user, however, tomorrow will happen only once. A perhaps less abstract interpretation is that PoP reflects the degree of confidence that the forecaster has that it will rain. In other words, an 80% chance of rain means that the forecaster strongly believes that there will be at least a minimum amount of rain tomorrow. The problem, from the perspective of the general public, is that when PoP is forecasted, none of these interpretations is specified.

There are clearly some interpretations that are not correct. The percentage expressed in PoP neither refers directly to the percent of area over which precipitation will fall nor does it refer directly to the percent of time precipitation will be observed on the forecast day Although both interpretations are clearly wrong, there is evidence that the general public holds them to varying degrees. Such misunderstandings are critical because they may affect the decisions that people make. If people misinterpret the forecast as percent time or percent area, they maybe more inclined to take precautionary action than are those who have the correct probabilistic interpretation, because they think that it will rain somewhere or some time tomorrow. The negative impact of such misunderstandings on decision making, both in terms of unnecessary precautions as well as erosion in user trust, could well eliminate any potential benefit of adding uncertainty information to the forecast. [Susan Joslyn, Nimor Nadav-Greenberg, and Rebecca M. Nichols, “Probability of Precipitations: Assessment and Enhancement of End-User Understanding“,

Journal of the American Meteorological Society, February 2009, citations omitted]

The frequentist interpretation is close to be correct, but it still involves a great deal of guesswork. Rainfall in a particular location is influenced by many variables (e.g., atmospheric pressure, direction and rate of change of atmospheric pressure, ambient temperature, local terrain, presence or absence of bodies of water, vegetation, moisture content of the atmosphere, height of clouds above the terrain, depth of cloud cover). It is nigh unto impossible to say that today’s (or tomorrow’s or next week’s) weather conditions are like (or will be like) those that in the past resulted in rainfall in a particular location 80 percent of the time.

That leaves the Bayesian interpretation, in which the forecaster combines some facts (e.g., the presence or absence of a low-pressure system in or toward the area, the presence or absence of a flow of water vapor in or toward the area) with what he has observed in the past to arrive at a guess about future weather. He then attaches a probability to his guess to indicate the strength of his confidence in it.

Thus:

Bayesian probability represents a level of certainty relating to a potential outcome or idea. This is in contrast to a frequentist probability that represents the frequency with which a particular outcome will occur over any number of trials.

An event with Bayesian probability of .6 (or 60%) should be interpreted as stating “With confidence 60%, this event contains the true outcome”, whereas a frequentist interpretation would view it as stating “Over 100 trials, we should observe event X approximately 60 times.”

The Bayesian approach to learning is based on the subjective interpretation of probability. The value of the proportion p is unknown, and a person expresses his or her opinion about the uncertainty in the proportion by means of a probability distribution placed on a set of possible values of p.

It is impossible to attach a probability — as properly defined in the first part of this article — to something that hasn’t happened, and may not happen. So when you read or hear a statement like “the probability of rain tomorrow is 80 percent”, you should mentally translate it into language like this:

X guesses that Y will (or will not) happen at time Z, and the “probability” that he attaches to his guess indicates his degree of confidence in it.

The guess may be well-informed by systematic observation of relevant events, but it remains a guess. As most Americans have learned and relearned over the years, when rain has failed to materialize or has spoiled an outdoor event that was supposed to be rain-free.

**BUT AREN’T SOME THINGS MORE LIKELY TO HAPPEN THAN OTHERS?**

Of course. But only one thing will happen at a given time and place.

If a person walks across a shooting range where live ammunition is being used, his is more likely to be killed than if he walks across the same patch of ground when no one is shooting. And a clever analyst could concoct a probability of a person’s being shot by writing an equation that includes such variables as his size, the speed with which he walks, the number of shooters, their rate of fire, and the distance across the shooting range.

What would the probability estimate mean? It would mean that if a very large number of persons walked across the shooting range under identical conditions, approximately S percent of them would be shot. But the clever analyst cannot specify which of the walkers would be among the S percent.

Here’s another way to look at it. One person wearing head-to-toe bullet-proof armor could walk across the range a large number of times and expect to be hit by a bullet on S percent of his crossings. But the hardy soul wouldn’t know on which of the crossings he would be hit.

Suppose the hardy soul became a foolhardy one and made a bet that he could cross the range without being hit. Further, suppose that S is estimated to be 0.75; that is, 75 percent of a string of walkers would be hit, or a single (bullet-proof) walker would be hit on 75 percent of his crossings. Knowing the value of S, the foolhardy fellow offers to pay out $1 million dollars if he crosses the range unscathed — one time — and claim $4 million (for himself or his estate) if he is shot. That’s an even-money bet, isn’t it?

No it isn’t. This situation is exactly analogous to the $10,000 bet on a single coin flip, discussed above. But I will dissect this one in a different way, to the same end.

The bet should be understood for what it is, an either-or-proposition. The foolhardy walker will either lose $1 million or win $4 million. The bettor (or bettors) who take the other side of the bet will either win $1 million or lose $4 million.

As anyone with elementary reading and reasoning skills should be able to tell, those possible outcomes are not the same as the outcome that would obtain (approximately) if the foolhardy fellow could walk across the shooting range 1,000 times. If he could, he would come very close to breaking even, as would those who bet against him.

To put it as simply as possible:

When an event has more than one possible outcome, a *single trial* cannot replicate the *average outcome* of a large number of trials (replications of the event).

It follows that the *average outcome* of a large number of trials — the probability of each possible outcome — cannot occur in a single trial.

It is therefore meaningless to ascribe a probability to any possible outcome of a single trial.

**MODELING AND PROBABILITY
**

Sometimes, when things interact, the outcome of the interactions will conform to an expected value — if that value is empirically valid. For example, if a pot of pure water is put over a flame at sea level, the temperature of the water will rise to 212 degrees Fahrenheit and water molecules will then begin to escape into the air in a gaseous form (steam). If the flame is kept hot enough and applied long enough, the water in the pot will continue to vaporize until the pot is empty.

That isn’t a probabilistic description of boiling. It’s just a description of what’s known to happen to water under certain conditions.

But it bears a similarity to a certain kind of probabilistic reasoning. For example, in a paper that I wrote long ago about warfare models, I said this:

Consider a five-parameter model, involving the conditional probabilities of detecting, shooting at, hitting, and killing an opponent — and surviving, in the first place, to do any of these things. Such a model might easily yield a cumulative error of a hemibel [a factor of 3], given a twenty five percent error in each parameter.

Mathematically, 1.25^{5} = 3.05. Which is true enough, but also misleadingly simple.

A mathematical model of that kind rests on the crucial assumption that the component probabilities are based on observations of actual events occurring in similar conditions. It is safe to say that the values assigned to the parameters of warfare models, econometric models, sociological models, and most other models outside the realm of physics, chemistry, and other “hard” sciences fail to satisfy that assumption.

Further, a mathematical model yields only the expected (average) outcome of a large number of events occurring under conditions similar to those from which the component probabilities were derived. (A Monte Carlo model merely yields a quantitative *estimate* of the spread around the average outcome.) Again, this precludes most models outside the “hard” sciences, and even some within that domain.

The moral of the story: Don’t be gulled by a statement about the expected outcome of an event, even when the statement seems to be based on a rigorous mathematical formula. Look behind the formula for an empirical foundation. And not just any empirical foundation, but one that is consistent with the situation to which the formula is being applied.

And when you’ve done that, remember that the formula expresses a point estimate around which there’s a wide — very wide — range of uncertainty. Which was the real point of the passage quoted above. The only sure things in life are death, taxes, and regulation.

**PROBABILITY VS. OPPORTUNITY**

Warfare models, as noted, deal with interactions among large numbers of things. If a large unit of infantry encounters another large unit of enemy infantry, and the units exchange gunfire, it is reasonable to expect the following consequences:

- As the numbers of infantrymen increase, more of them will be shot, for a given rate of gunfire.

- As the rate of gunfire increases, more of the infantrymen will be shot, for a given number of infantrymen.

These consequences don’t represent probabilities, though an inveterate modeler will try to represent them with a probabilistic model. They represent opportunities — opportunities for bullets to hit bodies. It is entirely possible that some bullets won’t hit bodies and some bodies won’t be hit by bullets. But more bullets will hit bodies if there are more bodies in a given space. And a higher proportion of a given number of bodies will be hit as more bullets enter a given space.

That’s all there is to it.

It has nothing to do with probability. The actual outcome of a past encounter is the actual outcome of that encounter, and the number of casualties has everything to do with the minutiae of the encounter and nothing to do with probability. *A fortiori*, the number of casualties resulting from a possible future encounter would have everything to do with the minutiae of that encounter and nothing to do with probability. Given the uniqueness of any given encounter, it would be wrong to characterize its outcome (e.g., number of casualties per infantryman) as a probability.

Related posts:

Understanding the Monty Hall Problem

The Compleat Monty Hall Problem

Some Thoughts about Probability

My War on the Misuse of Probability

Scott Adams Understands Probability

Further Thoughts about Probability