Me and friend flip a coin 4 times, whoever gets most heads wins. Each coin flip is independent of each other and the coins are fair. (So the probability of a flip is 1/2)
Show that the probability of a tie (we get the same number of heads) is the same as getting exactly 4 heads and 4 tails on 8 coin flips. Use this answer to calculate the probability of someone winning (getting more heads than the other person).
Also, If I toss the coin 5 times, while my friend only tosses hers 4 times, calculate the probability that I will get strictly more heads than my friend.
So the possible outcomes for the first question is ((HHHH),(HHHH)),((HTHH)(HTHH), ((HHTH),(HHTH)), ((HHHT),(HHHT)),…, ((TTTT),(TTTT)). Am I thinking of this right? So I am trying to prove that the probaility of the above outcomes is equal to the probability of getting (HHHHHHHH)? Still not sure how to figure out the second and third question.
Best Answer
Let $H,T$ denote the number of heads and tails repectively flipped by you in $4$ flips .
Let $H',T'$ denote the number of heads and tails respectively flipped by your friend in $4$ flips.
Then the probability of a tie is $\Pr\left(H=H'\right)$.
But $T'$has the same distribution as $H'$ and there is independence, so we observe:
$\Pr\left(H=H'\right)=\Pr\left(H=T'\right)=\Pr\left(H=4-H'\right)=\Pr\left(H+H'=4\right)$
The last probability can be recognized as the probability of $4$ heads by $8$ flips (the flips of you and your friend taken together).
Now we have:
leading to: $$\Pr\left(\text{you win}\right)=\frac{1}{2}\left(1-2^{-8}\binom{8}{4}\right)$$
If you toss $5$ times then think of it as a match1 as described above that is followed by an extra toss of you, and call the whole thing match2.
Now apply that:
$$\Pr\left(\text{you win match2}\right)=$$$$\Pr\left(\text{you win match1}\right)+\Pr\left(\text{match1 ends in a tie}\wedge\text{extra toss is a head}\right)=\Pr\left(\text{you win match1}\right)+\Pr\left(\text{match1 ends in a tie}\right)\Pr\left(\text{extra toss is a head}\right)=$$$$\frac{1}{2}\left(1-2^{-8}\binom{8}{4}\right)+2^{-8}\binom{8}{4}\times\frac12=\frac12$$
There is another (more elegant) route to this result.
Let $H,T$ denote the number of heads and tails repectively flipped by you in $5$ flips .
Let $H',T'$ denote the number of heads and tails respectively flipped by your friend in $4$ flips.
The probability of winning for you is $\Pr(H>H')$ and just as above we find:
$\Pr\left(H>H'\right)=\Pr\left(H>T'\right)=\Pr\left(H>4-H'\right)=\Pr\left(H+H'>4\right)$
The RHS is the probability that by $9$ flips there are more heads than tails. Symmetry then tells us that this equals $\frac12$.