Tricky problem from Probability

[Physics lover]

This may not help. If B has two extra shots: $n + 2$ against $n$. Let $x$ be the probability that they are equal after $n$ shots, $y$ the probability that either is one duck ahead and $z$ the probability that either is two or more ducks ahead. First we have:
$$x + 2y + 2z = 1$$
The probablity that B shoots more ducks than A is:
$$ p = y + z + \frac 3 4 x + \frac 1 4 y$$
We can eliminate $z$ to get:
$$p = \frac 1 2 + \frac 1 4(x + y)$$
This is better than the way I did it previously. But, I don't know how easy it will be to calculate $x$ and $y$ in general.

yes i see it.Your method seems to be applicable for 1 duck only.But i think 2 ducks will be a difficult one and quite complicated.

 

[WWGD]

For the 1-3 case. We look at the probabilities of A getting $0$ and $1$ ducks:

$p(A =0) = p(A = 1) = \frac 1 2$

If A gets $0$ ducks, then B needs more than zero with 3 shots. And $p(B > 0) = \frac 7 8$.

If A gets $1$ duck, then B needs more than one. $p(B > 1) = p(B =2) + p(B = 3) = \frac 3 8 + \frac 1 8 = \frac 1 2$.

The probability that B outshoots A is:

$p = (\frac 1 2)(\frac 7 8) + (\frac 1 2)(\frac 1 2) = \frac{11}{16}$

In the limit, as $n$ increases this probability will reduce towards $\frac 1 2$. E.g. if they have 1000 shots to 1002 it's not often that the extra two shots will make much difference. They only make a difference if it's equal after 1000 shots each or if A is only one duck ahead.

For a million shots, the probability will be very close to $\frac 1 2$ as it's very unlikely that the extra two shots will matter.

Ok, my method was, for case 1-3:
If A shoots 0 B shoots anything other than 000, B wins, same if A shoots 1. And this extends to any case. So I correct to 1-(1/2)^n , where n is the number of additional shots by B, since the only string that will not grant B more shots will be the 0..0 string. I had the right idea but did not implement it correctly.

 

[PeroK]

yes i see it.Your method seems to be applicable for 1 duck only.But i think 2 ducks will be a difficult one and quite complicated.

I couldn't resist the challenge:
$$ x = (\frac 1 2)^{2n} \frac{(2n)!}{(n!)^2}, \ \ y = (\frac 1 2)^{2n} \binom{2n}{n-1}$$
For example, with $n = 1$, we get $x = \frac 1 2, y = \frac 1 4$ and:
$$p = \frac 1 2 + \frac 1 4(x + y) = \frac{11}{16}$$
And, for $n = 50$, we have $p \approx 0.54$.

 

[Physics lover]

I couldn't resist the challenge:
$$ x = (\frac 1 2)^{2n} \frac{(2n)!}{(n!)^2}, \ \ y = (\frac 1 2)^{2n} \binom{2n}{n-1}$$
For example, with $n = 1$, we get $x = \frac 1 2, y = \frac 1 4$ and:
$$p = \frac 1 2 + \frac 1 4(x + y) = \frac{11}{16}$$
And, for $n = 50$, we have $p \approx 0.54$.

But this will involve lengthy calculation for large values of n.

 

[PeroK]

But this will involve lengthy calculation for large values of n.

That's an exact answer. You can't do better than that. You might need to use an approximation for large $n$, but for large enough $n$ the answer is approximately $\frac 1 2$ in any case.