A Gambler's Fallacy

Thursday, August 7th, 2003

You're walking through a casino and you spot a roulette table. Next to the table is a list of past results that indicate that the past 9 outcomes were all black. You have a preference for betting on red, so you pause for a moment to think. If you decide to make the bet, then you've just committed the gambler's fallacy.

Each spin on a roulette wheel is an independent event, so the past history does not provide any indication of future outcomes. If you rely on the past results, then you are incorrectly assuming that the next spin is a dependent event.

Probability is the study and measurement of uncertainty. This concept is crucial to the study of many areas of computer science, including cryptography, compression, and artificial intelligence (to name a few). The following article provides a brief discussion of independent and dependent events that illustrates the nature of the gambler's logical fallacy.

Independent Events

A event is an independent event if its outcome is not dependent upon, or influenced by, any other event. Two independent events are unrelated and rely on independent variables. We often describe events in terms of their outcomes.

Example 1:

Roll a six sided die twice. What are the odds of getting two sixes?


Each roll is an independent event because the outcome of one roll will not influence the outcome of the other.

Probability of rolling a six: 1/6
Probability of rolling two sixes in a row: 1/6 * 1/6 = 1/36

Using the function P(o) to denote the probability of outcome o, we say:

P(two sixes) = P(6) * P(6) = 1/36

In this case, each P(6) term is an independent term.

Dependent Events

A event is a dependent event if its outcome is dependent upon, or influenced by, another event or set of events.

Example 2:

Select a card at random from a deck. Next, without replacing the first card, select another card at random. What are the odds of drawing two aces?


Our first draw is an independent event, as it is not influenced by any prior occurrence. The odds of drawing an ace with our first card is equal to the number of aces in the deck, divided by the total number of cards in the deck.

Odds of drawing first ace (an independent event) = P(ace) = 4/52 = 1/13

Our second draw is where things get interesting. Since we've already removed one card from the deck, our next draw is influenced by the previous card draw, and has fewer cards to choose from.

Dependent terms (also known as conditional terms) are written as: P(B|A), which specifies the probability of outcome B, given the prior occurrence of outcome A. In the case of our current example, our full equation would be:

P(draw two consecutive aces) = P(ace) * P(ace|ace)

Since we've already determined that P(ace) = 4/52, we simply need to determine the odds for drawing an ace after having already drawn one:

P(ace|ace) = 3/51

We have three remaining aces to choose from, out of a total of 51 cards left in the deck.

Our final result is equal to the odds of drawing a first ace, multiplied by the odds of drawing a second ace immediately afterward:

P(draw two consecutive aces) = 4/52 * 3/51

Notice that in each example, our outcome ranges between 0.0 and 1.0. This must always be the case because an outcome of less than zero would indicate odds that are less than impossible, and an outcome greater than 1.0 would imply an outcome that is more probable than all possible outcomes. Thus, 0.0, or 0%, implies the event will never occur, and a value of 1.0, or 100%, implies that it will always occur.

Combining Outcomes

We can take things quite a bit further by examining cases where we're interested in several different outcomes.

Our previous example was only interested in the case where two aces are drawn. But what about the probability of one or two aces being drawn from the same two card draws? In this case, we're interested in:

P(ace OR ace) = P(ace) + P(ace) - P(ace AND ace)

This equation tells us that our answer is equal to the odds of drawing an independent ace, plus the odds of drawing an independent ace, minus the odds of drawing two aces. This may seem a bit confusing at first, so let's enumerate the result in a slightly different manner.

We begin by listing each of the possible scenarios that would fall under our desired outcome (of at least one ace being drawn):

  1. We draw an ace for the first card, no ace for the second = 4/52 * 48/51

  2. We draw no ace for the first card, an ace for the second = 48/52 * 4/51

  3. We draw two aces = 4/52 * 3/51

If we add all of these scenarios together, we arrive at a roughly 14.8% chance of drawing at least one ace. Our equation for P(ace OR ace) will produce the same result but in a slightly different logical manner, as shown in the diagram below.

P(ace OR ace) adds the results of two independent P(ace) events (shown as P(A) and P(B)). This causes us to accumulate twice the correct amount of P(ace AND ace) (the shaded intersection area above). To compensate, we subtract P(ace AND ace) to arrive at the correct result.

Quick Recap

From the examples we've seen, we are able to define the following:



The probability of independent outcome A.



Equals P(A AND B) / P(A) and defines the probability of dependent outcome B, given that independent outcome A has already occurred. B is a dependent event and is reliant on the outcome of A. We can also refactor this into Bayes' theorem, to produce P(B AND A) / P(A).

P(A) * P(B)


The probability of two independent outcomes occurring. Neither event influences the other.



Equals P(A) * P(B|A) and defines the probability of an independent event followed by a dependent outcome.



Equals P(A) + P(B) - P(A AND B) and defines the probability of either of two independent outcomes.



The probability of either one event OR another happening, but not both. XOR indicates an exclusive OR operation. This result is computed by combining the odds of A happening and NOT B, plus the odds of B happening and NOT A.

Further Examples

Try some of the following examples to test your understanding. These are a bit more challenging than the examples covered so far, and they're designed to make you think!

Example A:

Given a standard deck of cards, what is the probability of picking either a 10 or a black card with a single draw?

Example B:

Given a standard deck of cards, what is the probability of picking a 10 AND THEN a black card with a single draw?

Example C:

Flip a coin. If it lands on tails, re-flip one more time. What are the odds of getting AT LEAST one heads?

Example D:

Given a standard six sided die, what are the odds of rolling it four times and getting AT LEAST one six?

Answer Key

A: 53.8%
B: 3.76%
C: 75%
D: 52%