Math Insight

First of all, what valid options for $N$? The smallest value that $N$ could be is
. Is there a largest possible value of $N$? yes no
The probability that the first $13,462,305$ flips result in a head before you see your first tail is greater than zero equal to zero greater than 0.1 less than zero equal to 0.1
. Therefore, $N$ could be any non-negative integer. It would just be extremely unlikely to be a large number.

So far, we've introduced two events, which we labeled in terms of the random variable $N$.

• $N=0$: the event that the first coin flip was $T$, so that we obtained zero heads.
• $N \ge 1$: the event that the first coin flip was $H$, so that we know we obtained at least one head.

Since we keep flipping coins only if we obtain an $H$, we continue with more events only conditioned on the event $N \ge 1$. In that case, we consider two more events.

• 
  : the event that the second coin flip was $T$, so we obtained exactly
  head.
• 
  : the event that the second coin flip was $H$, so we know we obtained at least
  heads.

(Online, enter $\ge$ as >= or as the symbol ≥.)

The probabilities of these two events, conditioned on the event $N \ge 1$, are simple since they involve just one more coin flip.

• $P(N=1\,|\,N \ge 1) =$
  
• $P(N \ge 2 \,|\,N \ge 1) =$
  

From the definition of conditional probability, recall that $P(A,B) = P(A\,|\,B)P(B)$. Substituting the event $N \ge 1$ for $B$ and either event $N=1$ or $N \ge 2$ for $A$, we can calculate that

• $P(N=1, N \ge 1) = P(N=1\,|\,N \ge 1) P(N \ge 1) = $
  $\times$
  $=$
  
• $P(N \ge 2, N \ge 1) = P(N \ge 2\,|\,N \ge 1) P(N \ge 1) = $
  $\times$
  $=$
  

But now, the notation is a bit silly. If we've obtained exactly one head ($N=1$), then obviously we've obtained at least one head ($N \ge 1$). Similarly, if we've obtained at least two heads ($N \ge 2$), then obviously we've obtained at least one head ($N \ge 1$). The events $N=1$ and $N \ge 2$ cannot occur without the event $N \ge 1$. Hence, $P(N=1)=P(N=1, N \ge 1)$ and $P(N \ge 2) = P(N \ge 2, N \ge 1)$. We summarize by giving the probabilities of obtaining exactly one head and of obtaining at least two heads.

• $f_N(1) = P(N=1) = $
  
• $P(N \ge 2) = $
  

We can repeat this procedure to include one more coin flip and calculate the probability that $N=2$ (if third coin flip is $T$) and that $N \ge 3$ (if third coin flip is $H$).

Calculate the probabilities of these events conditioned on the fact that we already flipped heads twice in a row.

• $P(N=2 \,|\, N \ge 2)=$
  
• $P(N \ge 3 \,|\, N \ge 2)=$
  

Multiply by $P(N \ge 2)$ to determine the probabilities of obtaining exactly two heads or obtaining at least three heads.

• $f_N(2) = P(N=2)=$
  
• $P(N \ge 3)=$
  

Notice to get from $P(N \ge 1)$ to $P(N \ge 2)$ we multiply by $\frac{1}{2}$, the probability of getting one more $H$. Similarly, we multiply by $\frac{1}{2}$ to get from $P(N \ge 2)$ to $P(N \ge 3)$. In general, the event $N \ge n$ corresponds to flipping $H$ $n$ times in a row, so the probability $P(N \ge n)$ is the product of $n$ $\frac{1}{2}$'s. $P(N \ge n)$ is the exponential function
$P(N \ge n) = $
.
(Online enter $a^b$ as a^b.)

To know that we obtained exactly $n$ heads, we need to flip a coin one more time and this time obtain a $T$. We need to multiply by one more $\frac{1}{2}$ (the probability of $T$) to determine the probability distribution function $f_N(n)$.
$f_N(n) = P(N=n) =$

(Online, you'll need to use parentheses after the ^ to put a n+1 in the exponent.)

Use the below applet to sketch the probability distribution $f_N(n)$ of the number $N$ of consecutive heads. The distribution continues for all non-negative integers $n$, but we just plot $0 \le n \le 5$.