Probability of 5 coin flips having strictly more heads than 4 coin flips with the probability of landing heads being 1/3

Question

Person A has 5 fair coins and Person B has 4 fair coins. Person A wins only if he flips more heads than B does. What is the probability of A winning? I know how to solve this classic problem but my question is what if the coin is unfair with the probability of getting a head being 1/3. then is the probability of A winning still 1/2?

if we list out the possible outcomes like the following plot, it seems the probability of A winning still 1/2, no matter the coin is fair or not. Here's the plot.

but i found a contradiction in the following part:

consider the first 4 coin tosses first:

p = P(person A has more heads)=P(person B has more heads) they are equal because they are symmetric

P(A and B tossed equal number)=$1-2p$

so the probability that A won is P(person A has more heads)+P(A and B tossed equal number)*P(A tossed head)=$p+(1-2p)*\frac{1}{3}=p+\frac{1}{3}-\frac{2}{3}p=\frac{1}{3}p+\frac{1}{3}$

but since based on the plot this $\frac{1}{3}p+\frac{1}{3}$ should equal $\frac{1}{2}$, therefore we get p is $\frac{1}{2}$, but this gives 1-2p=0, however, 1-2p should be greater than 0 since there's a chance that A B tossed equal in the first 4 tosses, which leads to a contradiction. How should I understand this? is the conclusion that A won is still $\frac{1}{2}$ wrong? what should be the correct answer?

Answer 1

The plot, to me, seems meaningless… there are an equal number of orange and grey dots, yes, but each dot has a different probability. So what does it prove?

On the other hand, your algebraic analysis is exactly right. Suppose each coin is heads with probability $p$. If the probability of $A$ getting more heads than $B$ with $4$ coins each is $q$ (which can be expressed as a function of $p$), then the probability of $A$ getting more heads than $B$ when they have $5$ and $4$ coins, respectively, is $$f(p)=q+(1-2q)p=p+q(1-2p).$$ So $f(0)=0$, $f(1/2)=1/2$, and $f(1)=1$ (since the second term vanishes in all those cases). But to find $f(1/3)=1/3+q/3$, you actually need to calculate the value of $q$ when $p=1/3$… I don’t think a symmetry argument helps. But since $q$ is certainly greater than $0$ and less than $1/2$, it’s certain that $1/3 < f(1/3) < 1/2$.

Answer 2

Personally, since the numbers are small enough, it is routine to simply apply the binomial distribution formula, $~\displaystyle \binom{n}{k} p^k q^{n-k}.$

Let $~p = 1/3, ~q = 1-p.$

For $~k \in \{1,2,3,4,5\}, ~$ let:

• $f(k)~$ denote the probability that person A has exactly $~k~$ heads.

• $g(k)~$ denote the probability that person B has less than $~k~$ heads.

Then, the desired computation is

$$\sum_{k=1}^5 [ ~f(k) \times g(k) ~],$$

where

$$f(k) = \binom{5}{k} p^k q^{5-k}, ~~~~g(k) = \sum_{i = 0}^{k-1} \binom{4}{i} p^i q^{4-i}.$$

Answer 3

Note that the argument in the answer you linked to relies crucially on the symmetry of the binomial distribution $\ B\left(n,p\right)\ $ for $\ \color{red}{p=\frac{1}{2}}\ :$ \begin{align} P\big(H_A=a\big)&=P\big(H_A=5-a\big)\\ P\big(H_B=b\big)&=P\big(H_B=4-b\big)\ , \end{align} where $\ H_A\ $ and $\ H_B\ $ are the numbers of heads tossed by A and B, respectively. This symmetry no longer holds when $\ p=\frac{1}{3}\ .$ The formula for the probability that A wins if he or she has $\ c_A\ $ coins that land heads with probability $\ p_A\ ,$ and B has $\ c_B\le c_A\ $ coins which land heads with probability $\ p_B\ $ is \begin{align} P(\text{A wins})&=\sum_{b=0}^{c_B}{c_B\choose b}p_B^b(1-p_B)^{c_B-b}\sum_{a=b+1}^{c_A}{c_A\choose a}p_A^a(1-p_A)^{c_A-a}\ . \end{align} Evaluating this formula for $\ c_A=5,c_B=4\ $ and $\ p_A=p_B=\frac{1}{3}\ $ gives $$ P(\text{A wins})=\frac{8881}{19683}<\frac{1}{2}\ . $$

Answer 4

The best way is to use the idea in the answer you linked to with suitable modifications.

We survey the situation after each has tossed $4$ coins, and take advantage of the symmetry that exists at this point, just as it did in your linked answer.

For clarity, I shall use the notation $d$ for the probability of a draw after both have tossed $4$ coins

We can compute this probability easily as

$$d= \sum_{0}^4\left[\binom4 k p^k q^{4-k}\right]^2 = \frac{1921}{6561}$$

P(A wins)
= P(A wins by the $4th$ toss) + P(wins after $4th$ toss)

$ =\frac{1-d}2 + \frac13d = \frac12- \frac{d}6$,

and plugging in the value of $d$,
$$P(A\; wins) = \frac12 - \frac16\frac{1921}{6561} = \frac{8881}{19683}$$

Answer 5

Let $B\sim\mathcal{Bin}(4,p)$, $C\sim\mathcal{Bin}(4,p)$, $D\sim\mathcal{Bern}(p)$, and $B,C,D$ are mutually independent, with $A=C+D$. That is, $D$ is the extra coin's head count.

Player A wins when their extra coin show tails but the other coins show strictly more heads than has player B, or when their extra coin shows heads and their other coins contain at least as many heads as has player B.

$\qquad\begin{align}\mathsf P(A>B) &= \mathsf P(D=0)\,\mathsf P(C>B)+\mathsf P(D=1)\,\mathsf P(C\geq B)\\[1ex]&=\tfrac 12\mathsf P(D=0)\,(1-\mathsf P(C=B))+\tfrac 12\mathsf P(D=1)\,(1+\mathsf P(C=B))\\[1ex]&=\tfrac 12(1+(\mathsf P(D=1)-\mathsf P(D=0))\,\mathsf P(C=B))\\&=\tfrac 12+(p-\tfrac 12)\sum_{x=0}^4\binom 4x^2p^{2x}(1-p)^{2(4-x)}\\&~~\vdots\end{align}$