Independent Events and The Gambler's Fallacy

In the play Rosencrantz and Guildenstern are Dead, the play opens with the title characters playing a game where they gamble on the result of coin flips. Rosencrantz wins if the coin comes up heads; Guildenstern wins if it's tails. After they've gotten eighty-nine heads in a row, Guildenstern starts coming up with possible reasons for the seemingly impossible streak. Is he subconsciously flipping heads on purpose to punish himself for something? Are they in a time loop? Is God punishing him? Or is it "a spectacular vindication of the principle that each individual coin spun individually...is as likely to come down heads as tails and therefore should cause no surprise each individual time it does"?

Mathematically speaking, that last theory is exactly what is going on. Guildenstern is describing the principle of independent events in probability. This principle says that two events are independent from each other if the result of one event does not affect the probability of the other event.  In the case of coin tosses, the results of previous tosses don't affect the results of the current toss, so the probability of getting heads on any individual toss is still 50% even if all the previous ones were heads. 

Though the idea of independent events makes sense overall, its application to situations like Rosencrantz and Guildenstern's coin game tends to defy our gut intuition about probability. We know that the chances of getting 89 heads in a row is so incredibly small that it's effectively impossible. As a result, the natural inclination is to think "the next flip must be tails, in order to even out the odds!" This assumption is what's known as the gambler's fallacy - the mistaken belief that because a certain outcome hasn't happened for a while, it's "due" to come up and therefore has a higher probability, when the events are actually independent and the probability is unchanged. The name comes from the way that this type of thinking tends to show up in casinos and other gambling situations. For example, in roulette, if there's been a "streak" of red, many gamblers will bet on black because it's "overdue," even though the probability is equal for both red and black.

"But, Math Sister!" I hear you cry. "If the probability of getting 89 heads in a row is so small, how can the probability of getting another head still be 50%? How can we have a 50% chance of defying the odds like that?" It's true that the chances of getting 89 heads in a row are tiny: it's $(1/2)^{89}$, or about 1 in 619 septillion. Those odds are tiny. But the 89th coin flip itself isn't what defies the odds - it's the combination of the 89 coin flips that is so unbelievable. When you've already beaten ridiculous odds, beating them a little bit more isn't actually that hard.

To look at it another way: any particular combination of outcomes is equally unlikely. Think about it - if someone predicted "the next 89 coin flips will be HTHHH TTHGT HHTTT THHHH TTTTH THTTH HHTHT THTHT HHHTT HHTHT THHHT TTHTT HHHTT HHTHT HHHTT HTTTH THHTT THTH," the chances of getting that exact outcome would also be $(1/2)^{89}$ (and you’d be impressed with that person’s amazing psychic abilities). However, when you haven’t done the math, most people's gut reaction is that 89 heads would have a lower probability than that other outcome, even though the probabilities are actually the same. 

The issue is that, in our minds, we tend to interpret "all heads" or "all tails" as unique outcomes, but we lump together the outcomes that we see as "a bunch of heads and a bunch of tails." We know that getting "some heads and some tails" is more likely than getting "all heads," and we don't actually think about the uniqueness of that specific combination of heads and tails.

Similarly, the outcome "88 heads, then 1 tail" is also a unique result with a $(1/2)^{89}$ probability. So, if we've already gotten 88 heads in a row, then the streak after the next flip has only two possibilities: either "88 heads, then a tail," or "89 heads." Since there are two options and both are equally likely, the chances for each one is exactly the same as the probability would be for a single coin flip on its own: 50%.

"Coin tossing"by Scarygami is licensed under CC BY-SA 2.0