I guess that this will be really simple for you guys, but i have no foundation in probability. Please, help me to find not only the answer but also what i need to learn in order to be able to solve this kind of problems by myself in the future.
2 boys decide to play dama. Every winning gives a point. They will stop when one of them gets 3 points ahead. How is it likely that the game won't stop before the 21st match If both have the same probability to win a match(50%)? Thanks for your patience.
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If the game doesn't stop before the 21st match, it may be useful to consider the complement, I.e. $1-P(\bar{A}),$ where $\bar{A}$ represents the event that the game stops before the 21st match. we see that the earliest possible support starts at $3$ games and ends at $20$. note that being $3$ points ahead implies that this can only occur on odd support, so we have the support $3,5,...,17,19.$ there are two ways to select a winner, so we multiply our answer by $2$. we then consider the fact that the winner $3 + i$, where $i$ is the index ( starting at 0) of our odd numbers (i.e. when $k = 3$, $i = 0$, and when $k = 5,$ $i = 1,$, etc.). Hence, there are $\binom{k}{3+i_k}$ ways of setting the wins for the winning player, where $i_k$ is the index of $k$ in our support sequence (starting the index at 0). We see that the probability of this win occuring is $(.5)^{3+i_k}(.5)^{k-3-i_k} = (.5)^k.$ Hence, the probability of this event occuring is $$1 - \sum_{k=3,k \text{ odd}}^{19} 2\binom{k}{3+i_k}(.5)^k.$$
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Hint: I would make a spreadsheet. Make rows with $0$ to $21$ games played and columns from $-3$ to $3$ for games won by player 1 minus games won by player 2. In each cell is the chance that after the row many games the differential is the column. The chance that the differential is $-1$ is half the chance it was $-2$ the round before plus half the chance it was $0$ before. If it gets to $\pm 3$ you stop. Then add the chance you get to $\pm 3$ each round. Copy down is your friend. I find the chance of getting to $\pm 3$ to be rather high.
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The game can only end aften an odd number of turns. Make the first move, which cannot end the game, and consider every two consecutive plays after that as one play of a new game whose states are the difference in score between the two players: -3, -1, +1, 3.
In this new doubled game, at every move from $\pm 1$ there is a $1/4$ chance of ending the game and $3/4$ chance to continue. The probability of at least $n$ moves before the game stops, which corresponds to $2n+1$ moves of the original game, is therefore $(3/4)^{n-1}$.
Let $P(n, k)$ be the probability after $n$ games that boy #1 has a $k$ game margin (i.e., has won $k$ more games than his opponent). After the $n + 1$st game, there is a probability of $$\frac{P(n, -2) + P(n, 2)}2$$ that the game will end, in addition to the probability that it had already ended before this game, while $$P(n+1, -2) = \frac{P(n, -1)}2\\P(n+1, -1) = \frac{P(n, -2)+P(n, 0)}2\\P(n+1, 0) = \frac{P(n, -1)+P(n, 1)}2\\P(n+1, 1) = \frac{P(n, 2)+P(n, 0)}2\\P(n+1, 2) = \frac{P(n, 1)}2$$ We can simplify this by noting that the probability for $-k$ should be equal to the probability for $k$. So: $$P(n+1, \text{ end}) = P(n,\text{ end}) + P(n, 2)\\P(n+1, 2) = {P(n,1)\over 2}\\P(n+1, 1) = \frac{P(n, 2)+P(n, 0)}2\\P(n+1, 0) = P(n,1)$$
Therefore $P(n, 0) = 2P(n,2)$ for all $n > 0$, and $$P(n + 2, 0) = \frac{P(n, 2)+P(n, 0)}2 = \frac 34P(n,0)$$ for $n \ge 2$. Now, $P(2,0) = \frac 12$, so letting $n = 2m$ $$P(2m, 0) = \frac 23\left(\frac 34\right)^m\\P(2m,2) = \frac 13\left(\frac 34\right)^m$$$$P(2m, \text{ end}) = \sum_{k=1}^{m-1}P(2k,2) = \frac 13\sum_{k=1}^{m-1}\left(\frac 34\right)^k = \frac 13\frac{\frac 34 -\left(\frac 34\right)^m}{1-\frac 34}= 1 - \left(\frac 34\right)^{m-1}$$
So after 20 games there will be a probability of $1 - \left(\frac 34\right)^9$ that one of the boys will have gained a margin of 3 and stop, and therefore a probability of $\left(\frac 34\right)^9 = \frac {3^9}{2^{18}} = \frac {19683}{262144} \approx 7.5\%$ that they will play at least 21 games.