A quarter turns up with heads with the probability of 0.6 and a fair dime is tossed. The quarter is flipped until a head occurs, Independently the dime is flipped until a head occurs. Find probability that number of flips is the same for both coins.
Let $P_Q$ be the probability of quarter turning heads and $P_D$ probability of dime turning heads.
$P_Q = \frac{6}{10}$.
Probability of both turning heads is
$P_Q \cdot P_D = \frac{3}{5} \cdot \frac{1}{2} = \frac{3}{10}$
Is this the answer?
Do one round of each. Four outcomes are possible, namely: $(H,H), \,(T,H),\,(H,T),\,(T,T)$. Here let's say the first slot denotes the quarter and the second the dime. the first is a win, the second and third are losses and the fourth restarts the game. Thus your answer, $P$, satisfies, $$P=\frac 3{10}\times 1+\frac 4{10}\times \frac 12 \times 0 +\frac 6{10}\times \frac 12\times 0 + \frac 4{10}\times \frac 12\times P= \frac 3{10}+\frac 2{10}P$$
We deduce that $$10P=3+2P\implies P=\frac 38$$
The probability that it takes exactly $n$ flips to flip a head when heads have probability $p$ is $$ (1-p)^{n-1}p$$ since there needs to be $n-1$ tails then a head. So the probability distribution for $N_Q$ that the number of flips for the quarter is $$ P(N_Q = n) = \left(\frac{4}{10}\right)^{n-1}\frac{6}{10}$$ and for the fair dime $$ P(N_D=n) = \left(\frac{1}{2}\right)^n.$$
Since they're independent, the probability they both take $n$ tries is just the product: $$P(N_D = N_Q = n) = \left(\frac{1}{2}\right)^n\left(\frac{4}{10}\right)^{n-1}\frac{6}{10}.$$
The probability they take the same number of flips is just the sum of the probability they both take one, both take two, etc, so $$ P(N_D=N_Q) = \sum_{n=1}^\infty P(N_D = N_Q = n)= \sum_{n=1}^\infty \left(\frac{1}{2}\right)^n\left(\frac{4}{10}\right)^{n-1}\frac{6}{10} = \frac{3}{8}$$
No, you want the probability that, when you flip the two coins, the first head for both coins occurs at the same flip count.
That is: Flip the biased coin, counting the flips until the first head. Flip the fair coin, counting the flips until the first head. What is the probability that these counts will be the same?
Think: Unto what kind of distribution do each of these counts belong?
Alternatively: What is the probability that both coins are heads when given that this is the very first flip when at least one coin is not tails.
$$P(E) = \dfrac{P_DP_Q}{P_D+P_Q-P_DPQ}$$
The word "odds" is so often misused that it is good to find a question where its proper use simplifies the analysis.
In each "round", flip both coins simultaneously , and continue until both show heads (win) or only one shows a head (loss)
Odds in favor of a "win" = $\left(\dfrac 35\cdot\dfrac12\right):\left(\dfrac25\cdot\dfrac12 + \dfrac35\cdot\dfrac12\right) = 3:5$
Thus P("win") $= \dfrac3{3+5} = \dfrac38$