Tricky problem from Probability

[Physics lover]

Homework Statement

two Hunters A and B set out to hunt ducks. each of them hits as often as he misses when shooting at ducks. the probability that B bags more ducks than A can be expressed as p/q in its lowest form.
Find the value of p+q.

Relevant Equations

P(A intersection B)=P(A)×P(B)
when A and B are independent events.

I decided to take cases.For example-:A gets one 1 duck and B gets 2,3,4,...,51.So i can write this as
50C1(1/2)⁵⁰51C2(1/2)⁵¹+50C1(1/2)⁵⁰51C3(1/2)⁵¹+...

But i was unable to solve it further.
please help.

 

[PeroK]

Are you sure that's the whole question?

 

[Physics lover]

Are you sure that's the whole question?

no i made further more cases like A gets 2 and B gets 3,4,5...,51.
Similarly A gets 3 and B gets 4,5,6,...51.And i did the same think upto A gets 50 amd B gets 51.
Also i saw one more case that I left earlier that A gets 0 and B gets 1,2,...,51.
But the sum became quite complicated.

 

[PeroK]

no i made further more cases like A gets 2 and B gets 3,4,5...,51.
Similarly A gets 3 and B gets 4,5,6,...51.And i did the same think upto A gets 50 amd B gets 51.
Also i saw one more case that I left earlier that A gets 0 and B gets 1,2,...,51.
But the sum became quite complicated.

Does the number ##51## feature in the problem somewhere?

 

[Physics lover]

Does the number ##51## feature in the problem somewhere?

oh sorry for that i wrote the question incompletely.
Now i have edited it.

 

[Physics lover]

The complete questions is
two Hunters A and B set out to hunt ducks. each of them hits as often as he misses when shooting at ducks.A shoots at 50 ducks and B shoots at 51 ducks.The probability that B bags more ducks than A is p/q in its lowest form.Find the value of p+q.

 

[PeroK]

The complete questions is
two Hunters A and B set out to hunt ducks. each of them hits as often as he misses when shooting at ducks.A shoots at 50 ducks and B shoots at 51 ducks.The probability that B bags more ducks than A is p/q in its lowest form.Find the value of p+q.

I can guess the answer! Can you?

 

[Physics lover]

I can guess the answer! Can you?

no sir i was not able to solve it.
I am struggling with it for hours.
Please help.

 

[PeroK]

no sir i was not able to solve it.
I am struggling with it for hours.
Please help.

I guessed the answer and now I can prove it. First, try for 0-1, 1-2 and 2-3 instead of 50-51.

Next, think about an equal shooting match of 50 shots each. Then, think about the last shot (the 51st for B).

 

[Physics lover]

I guessed the answer and now I can prove it. First, try for 0-1, 1-2 and 2-3 instead of 50-51.

Next, think about an equal shooting match of 50 shots each. Then, think about the last shot (the 51st for B).

At last i solved it.
I decided to take cases like if the difference between the ducks bagged by them is 1,solved for it and it found it to be 101C51.Similary for second case i got it as 101C52 and for the last case that is for difference 50 i got is as 101C101.I solved this sum and it simplified to 2¹⁰⁰.For total cases I doubled these cases and added an extra case when they bag equal amount of ducks.And to my surprise the sum simplified to 2¹⁰¹.And thus i got the answer as 1/2.Thanks for the help.

 

[PeroK]

At last i solved it.
I decided to take cases like if the difference between the ducks bagged by them is 1,solved for it and it found it to be 101C51.Similary for second case i got it as 101C52 and for the last case that is for difference 50 i got is as 101C101.I solved this sum and it simplified to 2¹⁰⁰.For total cases I doubled these cases and added an extra case when they bag equal amount of ducks.And to my surprise the sum simplified to 2¹⁰¹.And thus i got the answer as 1/2.Thanks for the help.

Here's the quick way for any number of shots, where B has an extra shot:

Before B takes the last shot, the probability that A has more than B and that B has more than A are the same. Let's call this ##x##. And there is a probability, ##y##, that they are equal. Now we calculate the probability that B outshoots A:

If A has more than B, then B cannot overtake A with the last shot, so fails to outshoot A.

If B has already more than A then he doesn't need his last shot. That's ##x## for outshooting A.

If they are level, then B outshoots A with 50-50 probability on his last shot, so that's ##y/2## for outshooting A.

The total probability of B outshooting A is, therefore, ##x + y/2 = 1/2##. (As ##2x + y = 1##.)

 

[Physics lover]

Here's the quick way for any number of shots, where B has an extra shot:

Before B takes the last shot, the probability that A has more than B and that B has more than A are the same. Let's call this ##x##. And there is a probability, ##y##, that they are equal. Now we calculate the probability that B outshoots A:

If A has more than B, then B cannot overtake A with the last shot, so fails to outshoot A.

If B has already more than A then he doesn't need his last shot. That's ##x## for outshooting A.

If they are level, then B outshoots A with 50-50 probability on his last shot, so that's ##y/2## for outshooting A.

The total probability of B outshooting A is, therefore, ##x + y/2 = 1/2##. (As ##2x + y = 1##.)

Thanks a lot.That was a brilliant solution.
Can you tell me how will this method in the case B has more than 1 extra shot.Because i think that this method would be longer for more than 1 extra shot.

 

[PeroK]

Thanks a lot.That was a brilliant solution.
Can you tell me how will this method in the case B has more than 1 extra shot.Because i think that this method would be longer for more than 1 extra shot.

That would be trickier. It would depend on how many shots there are.

 

[WWGD]

My take:
Consider the first 50 attempts by B. They will be exactly the same as those by A. And only 1/2 of the choices on the 51st attempt will be successful in producing a string with more successes for B.

 

[Physics lover]

My take:
Consider the first 50 attempts by B. They will be exactly the same as those by A. And only 1/2 of the choices on the 51st attempt will be successful in producing a string with more successes for B.

but B could have more shots successful even if they have 50 shots each.

 

[WWGD]

but B could have more shots successful even if they have 50 shots each.

List all strings for B of length 51 containing 0s for fail and 1s for success. Ignore the 51st. This will give you all strings by A. Now all entries with a 1 on the 51st row will have more wins by B, those with a 0 will not. Half the strings will have a 0, half will have a 1.

Example : A has 2 chances, B has 3:
List of strings for B.

00|0
00|1
01|0
10|0
01|1
10|1
11|0
11|1

Every string for B equals a string for A with either a 0 or 1 attached to the right. Those with a 1 attached-- half the total-- will represent more successes for B, those with a 0 attached to the right will not represent more successes for B.

 

[PeroK]

List all strings for B of length 51 containing 0s for fail and 1s for success. Ignore the 51st. This will give you all strings by A. Now all entries with a 1 on the 51st row will have more wins by B, those with a 0 will not. Half the strings will have a 0, half will have a 1.

Example : A has 2 chances, B has 3:
List of strings for B.

00|0
00|1
01|0
10|0
01|1
10|1
11|0
11|1

Every string for B equals a string for A with either a 0 or 1 attached to the right. Those with a 1 attached-- half the total-- will represent more successes for B, those with a 0 attached to the right will not represent more successes for B.

What if ##B## has two extra shots?

 

[WWGD]

What if ##B## has two extra shots?

But from what I read, game is designed so that B has exactly one more shot than A. That is my assumption here.

 

[PeroK]

But from what I read, game is designed so that B has exactly one more shot than A. That is my assumption here.

Why doesn't your method extend to two extra shots?

 

[Physics lover]

But from what I read, game is designed so that B has exactly one more shot than A. That is my assumption here.

I think Perok is asking you what if the question was changed and B would have 2 shots more.

 

[WWGD]

I think Perok is asking you what if the question was changed and B would have 2 shots more.

Well, I can't give you the answer to every possible question, just the one you asked. How can I give a universal answer of sorts?

 

[WWGD]

Why doesn't your method extend to two extra shots?

Because the number of possibilities for 2 shots is different and does not lend itself to a simple analysis as the difference by one does. Do you see a flaw in my argument?

 

[PeroK]

Because the number of possibilities for 2 shots is different.

I'd don't see it. Your argument, as you have set it out, contains no element that limits the extra shots to 1. What part of your argument fails if there are two extra shots? In which case you get an answer of ##3/4##.

 

[WWGD]

I'd don't see it. Your argument, as you have set it out, contains no element that limits the extra shots to 1. What part of your argument fails if there are two extra shots? In which case you get an answer of ##3/4##.

It actually extends, as you said. There are 4 choices for the last 2. All except the 00 will grant more wins for B than for A. Again, list all strings up till 52 ( for definiteness; it generalizes to a difference of 2). Then cut of at the 50th place to get all strings for A. Then the ensuing 2 rows, the 51st and 52nd will consist of either of 00, 01,10 or 11. The last 3 will give B a larger amount of successes, the first will not. So 3/4 will give B more successes.

 

[PeroK]

So 3/4 will give B more successes.

Which is, of course, wrong! That only applies in the case of 0-2 shots. For more shots the answer is less than ##3/4##. For ##n## and ##n + 2## shots for A and B:

##p = \frac 3 4 - \frac a 4 - \frac b 2##

Where ##a## is the probabilty that after ##n## shots A is one duck ahead and ##b## is the probability that A is two ducks ahead.

Your argument assumes they are always level after ##n## shots each, which is only true for ##n =0##.

 

[WWGD]

Which is, of course, wrong! That only applies in the case of 0-2 shots. For more shots the answer is less than ##3/4##. For ##n## and ##n + 2## shots for A and B:

##p = \frac 3 4 - \frac a 4 - \frac b 2##

Where ##a## is the probabilty that after ##n## shots A is one duck ahead and ##b## is the probability that A is two ducks ahead.

Your argument assumes they are always level after ##n## shots each, which is only true for ##n =0##.

Yes, I was answered for cases when the difference was of 2 shots and probability of success is 1/2. But it generalizes to larger differences and different probability of success. I am answering the problem that was asked, but my solution generalizes.

 

[WWGD]

Which is, of course, wrong! That only applies in the case of 0-2 shots. For more shots the answer is less than ##3/4##. For ##n## and ##n + 2## shots for A and B:

##p = \frac 3 4 - \frac a 4 - \frac b 2##

Where ##a## is the probabilty that after ##n## shots A is one duck ahead and ##b## is the probability that A is two ducks ahead.

Your argument assumes they are always level after ##n## shots each, which is only true for ##n =0##.

That's not what my argument assumes. It just assumes that the restriction of all shots by B to the previous 1, 2,,n shots describes all shots by player A.

 

[WWGD]

Which is, of course, wrong! That only applies in the case of 0-2 shots. For more shots the answer is less than ##3/4##. For ##n## and ##n + 2## shots for A and B:

##p = \frac 3 4 - \frac a 4 - \frac b 2##

Where ##a## is the probabilty that after ##n## shots A is one duck ahead and ##b## is the probability that A is two ducks ahead.

Your argument assumes they are always level after ##n## shots each, which is only true for ##n =0##.

So please show me a case where the difference of shots is 2 , probability of success is 1/2 and B having more successes with probability different from 3/4. Maybe we are talking at cross from each other.

 

[PeroK]

So please show me a case where the difference of shots is 2 , probability of success is 1/2 and B having more successes with probability different from 3/4. Maybe we are talking at cross from each other.

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.

 

[PeroK]

Thanks a lot.That was a brilliant solution.
Can you tell me how will this method in the case B has more than 1 extra shot.Because i think that this method would be longer for more than 1 extra shot.

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.

 

[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.