How can i solve this question with Geometric distribution or random variables? I tried have tried using a 'Geometric distribution', but it hasn't worked.
John and Ron play basketball 10 times.
The probability that John wins in single round = $0.4$.
The probability that Ron wins in single round = $0.3$.
The probability that there is equality between them in a single round = $0.3$.
The "Winner" is defined to be the first to win a single round.
The rotations are different and independent.  
What is the probability that John is the Winner?
 A: Given that there are exactly $10$ rounds, the probability of having no winner by the end of the $10$ rounds is $0.3^{10}$. Then we know that the probability of having a winner is $1-0.3^{10}$. 
Such probability should be apportioned between John and Ron in the ratio $0.4/0.3$. So the probability that John wins is 
$$\frac{4}{7} (1-0.3^{10})$$
A: Since they only play $10$ times, the probability that John wins is equal to the sum of probabilities that John wins in the $i$-th round:
$$P=\sum_{i=1}^{10}\left ( 0.3^{i-1}\times 0.4\right )=0.4\times\frac{1-0.3^{10}}{0.7}$$
where the probability that John wins in the $i$-th round is $0.3^{i-1}\times 0.4$ because the first $i-1$ rounds were all draws.
A: Let J be the probability that John wins a round. (0.3)
Let T be the probability that there is no winner for the round. (0.3)
Let WJn be the probability that John wins in round n or later.
WJ1 = J + T*WJ2
WJ2 = J + T*WJ3
:
WJ10 = J (since there are no later rounds)
Substituting and expanding
WJ1 = J + JT + JT^2 + ... + JT^9 
WJ1 = J * (sum i=0 to 9)T^i
WJ1 = 0.3 * 1.43 = 0.43
