3
$\begingroup$

The question:

Two persons $A$, $B$ simultaneously toss their individual coins, and win $1$ point if head is face up and $0$ points if tails is face up. The probability that the points of $A$ exceeds the points of $B$ after $3$ tosses is...

My attempt:

Because each person have $4$ available options of how many points they will have in the end $(0,1,2,$ and $3)$, the total amount of ways the score could be after $3$ tosses would be $4 \cdot 4 = 16$. The cases in which person $A$ wins would be $(1,0)$, $(2,0)$, $(2,1)$, $(3,0)$, $(3,1)$, and $(3,2)$. So the answer should be $\frac{6}{16}$ or $\frac{3}{8}$.

The actual answer is $\frac{11}{32}$

My question:

Where did I go wrong and how do I do it correctly?

Thank you!

$\endgroup$
4
$\begingroup$

Not all of the outcomes (0,0), (0,1), (0,2),... (3,2), (3,3) are equally likely. Although there are indeed sixteen possible outcomes, unless all outcomes are equiprobable you may not simply do the fraction of favorable outcomes compared to total outcomes regardless. For example, the outcome $(0,1)$ is three times as likely to occur as the outcome $(0,0)$.

Approach via a combination of the binomial theorem and the multiplication and addition principles of probability.

If $B$ has zero heads, $A$ wins with at least one head. The probability of $B$ having zero heads is $\frac{1}{8}$. The probability of $A$ having at least one head is $\frac{7}{8}$. The probability of $B$ having zero heads and $A$ having at least one head is then $\frac{7}{64}$

If $B$ has one head, $A$ wins with at least two heads. The probability of $B$ having one head is $\frac{3}{8}$. The probability of $A$ having at least two heads is $\frac{4}{8}$. The probability then of both occurring is $\frac{12}{64}$

If $B$ has two heads, $A$ wins only with getting three heads. The probability of $B$ having two heads is $\frac{3}{8}$. The probability of $A$ having three heads is $\frac{1}{8}$. The probability then of both occurring is $\frac{3}{64}$

If $B$ has three heads, $A$ cannot win.

As the sample space can be partitioned into the result of how many heads $B$ received as above, the probability that $A$ wins is the sum of the aforementioned probabilities:

$Pr(\text{A wins}) = \frac{7}{64}+\frac{12}{64}+\frac{3}{64}+0 = \frac{22}{64}=\frac{11}{32}$


(didn't see John's answer while typing this extra paragraph, the method is essentially the same)

An alternate way to view the solution: $1 = Pr(\text{A wins})+Pr(\text{A and B tie}) + Pr(\text{B wins})$

Due to the symmetry of their circumstances, we know that $Pr(\text{A wins})=Pr(\text{B wins})$. This leaves us with the question of finding $Pr(\text{A and B tie})$

This can occur with either zero heads each, one head each, two heads each, or three heads each.

These occur with probabilities $\frac{1}{64}, \frac{9}{64},\frac{9}{64},\frac{1}{64}$ respectively.

The probability that $A$ and $B$ tie is then $\frac{20}{64}$

This tells us that $Pr(A~\text{wins}) = \frac{1}{2}(1-\frac{20}{64})=\frac{22}{64}=\frac{11}{32}$

$\endgroup$
4
$\begingroup$

Here's another way to look at the problem.

What if they're tied after three coin tosses?

They could get no heads ($1$ possibility), one head each ($9$ possibilities), two heads each ($9$ possibilities), or three heads each ($1$ possibility).

(Why $9$ possibilities for one head each? $A$ could get a head on the first, second, or third toss, as could $B$. These are independent, so there are $3\cdot 3=9$ possibilities. The same reasoning can be applied to two heads each, as well as the total number of possible outcomes $8\cdot 8=64$.)

This adds to $20$ out of $64$ possibilities that they tie, which means $44/64 = 11/16$ probability that they don't.

As $A$ is just as likely to be ahead as $B$, the answer is half this, or $11/32$.

$\endgroup$
  • 1
    $\begingroup$ Just to add. using the identity $\sum_{k=0}^n \binom{n}{k}^2 = \binom{2n}{n}$, we can say generally that the possibilities for a tie after each player flips $n$ coins is $\binom{2n}{n}$. than the chance to win can be easily calculated in the same line and is: $0.5(1 - \dfrac{\binom{2n}{n}}{2^{2n}})$ $\endgroup$ – d_e Jul 7 '16 at 21:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.