# Probability question - Bernoulli trials vs Binomial distribution

I am handed a fair coin. I understand that the probability of H/T is 0.5.

Trial 1.
I flip and observe H. I expect that on my 2nd toss, the probability of H is still 0.5

Trial 2.
I flip and observe H. Again I expect that upon my 3rd toss, p(H) = 0.5.

Why is it that I should expect p(H) = 0.5 on my 3rd trial? If I use the binomial distribution, I know that observing 3 successes in 3 trials is actually p=0.125. I did not set out to flip the coin 3 times as a set, yet I've done so.

What am I missing or forgetting?

• Be careful about using the binomial distribution. Only in this case is it calculating a specific outcome, and it's only because there is only one possible combination to have all heads. Conversely, if you wanted to know the probably of getting a HHT sequence that is also p=.125. However, the probably of any outcome with two heads and a tails from 3 flips is p=(3C2)*.5^2*.5^1=.125*3=.375. Jun 22 at 15:44
• The motto says: Fate has no memory.
– Ale
Sep 1 at 18:27

When you flip the third coin, you’ve already observed the first two flips. Mathematically, $$p(\text{flip}_3=H \mid \text{flip}_2=H, \text{flip}_1=H) = 0.5$$. This comes from the independence among coin flips.

The value 0.125 is the joint probability $$p(\text{flip}_3=H, \text{flip}_2=H, \text{flip}_1=H)$$. This is the probability of flipping three heads in a row.

By the way, the two are related. The joint probability can be decomposed into each coin flip:

\begin{align} p(\text{flip}_3=H, \text{flip}_2=H, \text{flip}_1=H) =& p(\text{flip}_3=H \mid \text{flip}_2=H, \text{flip}_1=H) p(\text{flip}_2=H \mid \text{flip}_1=H) p( \text{flip}_1=H) \\ =& p(\text{flip}_3=H) p(\text{flip}_2=H) p( \text{flip}_1=H) \\ =& 0.5 \times 0.5 \times 0.5 \\ =& 0.125 \end{align} The first step is the chain rule of probability, and the second comes from independence amongst coin flips.

If it is a fair coin why would suddenly the probability of the heads change? Just think of physics of the phenomenon, it can’t be possibly affected by what happened in previous coin tosses. A coin can't remember what was a previous toss, and can't know when did you start observing its tosses.

Besides, the binomial distribution assumes that the tosses are independent. If you make coin tosses dependent then you can’t use binomial distribution. No eating the cake and having it too

Why is it that I should expect p(H) = 0.5 on my 3rd trial?

If you subscribe to the assumption the flips are iid, then this just follows. The real question is "are the flips iid" and since it is a thought experiment only you can answer that.

If I use the binomial distribution, I know that observing 3 successes in 3 trials is actually p=0.125. I did not set out to flip the coin 3 times as a set, yet I've done so.

Yes, this is nuanced but important.

The binomial distribution is essentially the distribution of the sum of $$n$$ bernoulli random variables. However, the assumption here is that the result of any prior bernoulli trial does not affect $$n$$. Essentially, you set out to flip $$n$$ times and nothing stops or alters that.

This is a classic gambler's fallacy. There is no law of averages for i.i.d. processes. See Regression to the mean vs gambler's fallacy