"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."
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).
"Use this answer to calculate the probability of someone winning (getting more heads than the other person)."
Now we have:
- $\Pr\left(\text{tie}\right)=\Pr\left(H+H'=4\right)=2^{-8}\binom{8}{4}$.
- $1=\Pr\left(\text{you win}\right)+\Pr\left(\text{tie}\right)+\Pr\left(\text{friend wins}\right)$
- $\Pr\left(\text{you win}\right)=\Pr\left(\text{friend wins}\right)$
leading to: $$\Pr\left(\text{you win}\right)=\frac{1}{2}\left(1-2^{-8}\binom{8}{4}\right)$$
"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."
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 $\frac12$.
Best Answer
No. Your answer is not correct. First, note that you included THHH as a "success," but that case violates your condition that "First flip is heads."
Given that the first flip is heads, there are only $2^3 = 8$ equally likely scenarios:
HHHH, HHHT, HHTH, HHTT, HTHH, HTHT, HTTH, and HTTT.
Now do your counting...