Maximizing probability of a sequence of coin tosses for a biased coin

Question

Consider the following sequence created by flipping a biased coin five times $$ D={H,H,T,H,H} $$

where H denotes heads and T denotes tails.

The probability of observing D (in the specific ordering given), assuming it was generated by flipping a coin X with an unequal probability of heads (H) and tails (H), where the distribution is:

$$ P(X=H) = 0.75 $$ $$ P(X=T) = 0.25$$

is simply:

$$ P(D) = \dfrac 3 4 \times \dfrac 34 \times \dfrac 14 \times \dfrac 34 \times \dfrac 34 = \dfrac {81}{1024} $$

However, my aim is to maximize this probability value by varying the individual $P(X=H)$ or in other words for what value of $P(X=H)$ do I maximize $P(D)$ ?

Answer 1

Let $p=\mathsf P(H)$ and $(1-p)=\mathsf P(T)$.

Then we have: $\mathsf P(D)=p^4(1-p)$

You seek to find the value of $p$ that maximises the probability for $D$.

That is done by finding the value(s) which makes the derivative vanish and checking that this is in fact a maximum rather than a minimum or inflection. Sketching the graph helps.$$\dfrac{\mathrm d\,\mathsf P(D)}{\mathrm d\,p\hspace{4ex}}=0$$

.

Answer 2

The probability of $D$ would be $p^h(1-p)^t$ where $p$ is the probability of heads, $h$ is the number of heads in $D$, and $t$ is the number of tails in $D$. In your example $h = 4$ and $t=1$. Clearly, if $D$ consists of only heads, $p = 1$, while if $D$ consists of only tails, $p = 0$. Otherwise, we can use the derivative.

To maximize $\mathbb{P}(D)$, $$\frac{d}{dp} \mathbb{P}(D) = 0$$

The derivative is equal to $$(1-p)^t\frac{d}{dp}p^h + p^h \frac{d}{dp}(1-p)^t = h(1-p)^tp^{h-1} - tp^h(1-p)^{t-1} = 0$$

Solving for $p$ yields $p = 0, p=1$, and $p = \frac{h}{h+t}$. The $p = 0$ and $p=1$ cases will be extraneous unless $D$ is only heads or only tails (in which case $\frac{h}{h+t}$ would give the same result).