Is the probability of an event gets affected by previous events?

Question

Suppose we are tossing a coin and in first four trials only heads comes up. Then what will be the probability of getting tails in next trial? Is it $\frac{1}{2}$ or the probaibility should be more than it? Nature try to balance all outcomes. So to balance $4$ heads there are more chances of getting tails. So if this is the case how one would find the probability of it.

Answer 1

“Nature tries to balance all outcomes”

This statement is false. You are not more likely to get a tails after flipping four heads. These events are independent.

Answer 2

Previous outcomes have no bearing on the next outcome in any independent event (dice rolling, coin flipping, roulette, etc etc)

The mistaken belief that it does is called "Gambler's Fallacy"

Answer 3

We call a coin a fair coin if the probability of getting heads and tails in each throw is $\frac{1}{2}$. Thus, by definition, it doesn't matter how many times you had tails in the row, the next outcome is heads or tails, with the probability $\frac{1}{2}$ each.

Now you can ask three questions:

1. Are real coins fair?
2. Where is this intuition coming from, that, if you had a lot of tails in a row, you are more likely to have tails again?
3. What can in general be said about unfair coins?

Question 1: Pretty much so. Even where there is a discrepancy, it is not in whether $X_n$ (the outcome of $n$th throw) is dependent on $X_1,X_2,\ldots,X_{n-1}$ (it normally isn't), but whether $P(X_n\text{ is tails})$ is exactly $\frac{1}{2}$ or is it slightly(?) off.

Question 2: Not a mathematical question, but it seems that human beings, when estimating the probability, are born Bayesians, and automatically take into account the (small) probability that the coin is rigged (unfair). Assume, for instance, that we think that the coin may have both sides "tails" with probability $p$, and that it is fair with probability $q=1-p$. Now, we had "tails" $n$ times... what is now the probability that the coin is unfair? Let $T$ be the event that we got "tails" $n$ times, $F$ the event that the coin is fair and $U$ the event that it is unfair. We have:

$$P(U|T)=\frac{P(T|U)P(U)}{P(T|U)P(U)+P(T|F)P(F)}=\frac{p}{p+2^{-n}q}$$

and, similarly,

$$P(F|T)=\frac{P(T|F)P(F)}{P(T|U)P(U)+P(T|F)P(F)}=\frac{2^{-n}q}{p+2^{-n}q}$$

Based on that knowledge, we will now expect that the next throw will result in "tails" as $P(U|T)+\frac{1}{2}P(F|T)=\frac{p+2^{-n-1}q}{p+2^{-n}q}=1-\frac{1}{2}\frac{1}{1+2^n\frac{p}{q}}\to 1$ when $n\to\infty$, no matter how small $p$ is, to start with... and the convergence to $1$ is fast (exponential). Loosely speaking, we subconsciously take the evidence that we had $n$ tails in a row as evidence that the coin is unfair, and bias our guesses to accommodate that.

Question 3 is probably too broad and not very interesting in the scope of the OP's question, so will be skipped here.

Answer 4

There is a misunderstanding in the way the question is posed. When you flip a coin, the probability of getting heads is 1/2.

If you flip the coin once and get heads, you may ask yourself what is the probability of getting heads again. naturally, your intuition may suggest that since you already got heads, the next time you might expect tails. However, the problem is that this has now become a conditional probability problem, and what you are solving is not the question of what is the probability that after the next flipping I get heads, but the real question you are answering is: what is the probability that I get two heads in aa row given that the first time it was also heads.

Let's look at Bayes's theorem: $P(A \vert B) = \frac{P(A)*P(B \vert A)}{P(B)} \,$

What we are trying to solve is, what is the probability that I get two consecutive heads (A event) given that the first time I flipped the coin I got head (B event).

Now P(A) is, in fact 1/4 (it is harder to get two heads in a row than one), however P(B) is 1/2. What is P(B|A)? This is the probability of getting heads the first time assuming you will have two heads, and this is obviously 1, so P(A|B) = $\frac{1/4}{1/2} = 1/2$.

So, rather than asking whether the second flipping's outcomes is conditioned by the first, think of it as a conditional probability problem where "what is the probability I get n consecutive heads given that I know the first n-1 times I already always got heads?", and you will see that the solution is in fact always 1/2.