Flip three coins
Three friends (let's call them A,B,C) flip coins. By getting 3 heads or 3 tails, they flip again until one of the coins is different from the other.
Denite Y:=number of rounds.
Find :
$1. P(Y=3)$
$2. P(Y\geq4)$
$3.$ $P$(The different coin is C)
My solution:
$P$(All coins are heads or tails)=$1\cdot\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}$
$P$(one of the coins is different from the others)=$1-P$(all coins are heads or tails) $\implies 1-\frac{1}{4}=\frac{3}{4}$
$1. P(Y=3)=(1\cdot \frac{1}{2} \cdot \frac{1}{2})^2\cdot(3\cdot \frac{1}{4})$
$2. P(Y\geq 4)=1-P(Y\leq 4)$
$P(Y\leq 4)=\frac{3}{4}+\frac{1}{4}\cdot \frac{3}{4}+(\frac{1}{4})^2\cdot \frac{3}{4}+(\frac{1}{4})^3\cdot \frac{3}{4}=\frac{255}{256} \implies P(Y\geq 4)=1-\frac{255}{256}=\frac{1}{256}$
$3.$ I think the answer is $\{T,T,H\}+\{H,T,T\}=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$ but I am really not sure since I don't know how many rounds exist.
Is my solution correct?
I'd be grateful for feedback\help!
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