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So here is a first guess to create a lower limit on the probability.

Firstly, we're taking 9 shots at 4 targets with equal probability. One of the targets has a score of 0. We need to assign a score to each of the other three to maximize our probability of getting 44.

It seems to me that we should look at divisors of 44 since there is no advantage to getting close to 44. So 2, 4 or 11 are the obvious cases. And since we are expecting to be hitting the scoring target 6 or 7 times (more likely 7, I guess - if we distribute the 8 equally, the ninth is more likely to hit than not), so we're targeting an average of around 6 or 7.

So here are a couple of possible strategies:

1. Have one of the target areas be worth 22, and the others 0. We would need to hit that target area twice to win.
2. Have one of the target areas be worth 4, and the others 8. I'll divide everything through by 4. So we are targeting a score of 11 now, and have scores of 2,2, and 1. We would need:

• 5x2 and 1x1, or
• 4x2 and 3x1, or
• 3x2 and 5x1, or
• 2x2 and 7x1.

The first of these has probability of $(0.25)^2(0.75)^7\binom{9}{2}\approx0.300$.

The second of these I can't see a neat way to calculate so I'll do it the painful way, which is to add up the four terms:

• $(0.5)^5(0.25)^1(0.25)^3\left(\frac{9!}{5!1!3!}\right)\approx0.0615$
• $(0.5)^4(0.25)^3(0.25)^2\left(\frac{9!}{4!3!2!}\right)\approx0.0769$
• $(0.5)^3(0.25)^5(0.25)^1\left(\frac{9!}{3!5!1!}\right)\approx0.0153$
• $(0.5)^2(0.25)^7(0.25)^0\left(\frac{9!}{2!7!0!}\right)\approx0.0005$

This is about 0.15. (Having two 1s and one 2 is worse).

I'll be interested to see if there are better strategies than this. But current leading strategy is

to have 1 with a score of 22 and no scores for anything else (equivalent strategy is to have 44/7 on all three targets).

However, this solution is no longer allowed with the new rules, so I would go for the second option:

One area worth 4 and two worth 8. This has approx 15% chance.

So here is a first guess to create a lower limit on the probability.

Firstly, we're taking 9 shots at 4 targets with equal probability. One of the targets has a score of 0. We need to assign a score to each of the other three to maximize our probability of getting 44.

It seems to me that we should look at divisors of 44 since there is no advantage to getting close to 44. So 2, 4 or 11 are the obvious cases. And since we are expecting to be hitting the scoring target 6 or 7 times (more likely 7, I guess - if we distribute the 8 equally, the ninth is more likely to hit than not), so we're targeting an average of around 6 or 7.

So here are a couple of possible strategies:

1. Have one of the target areas be worth 22, and the others 0. We would need to hit that target area twice to win.
2. Have one of the target areas be worth 4, and the others 8. I'll divide everything through by 4. So we are targeting a score of 11 now, and have scores of 2,2, and 1. We would need:

• 5x2 and 1x1, or
• 4x2 and 3x1, or
• 3x2 and 5x1, or
• 2x2 and 7x1.

The first of these has probability of $(0.25)^2(0.75)^7\binom{9}{2}\approx0.300$.

The second of these I can't see a neat way to calculate so I'll do it the painful way, which is to add up the four terms:

• $(0.5)^5(0.25)^1(0.25)^3\left(\frac{9!}{5!1!3!}\right)\approx0.0615$
• $(0.5)^4(0.25)^3(0.25)^2\left(\frac{9!}{4!3!2!}\right)\approx0.0769$
• $(0.5)^3(0.25)^5(0.25)^1\left(\frac{9!}{3!5!1!}\right)\approx0.0153$
• $(0.5)^2(0.25)^7(0.25)^0\left(\frac{9!}{2!7!0!}\right)\approx0.0005$

This is about 0.15. (Having two 1s and one 2 is worse).

I'll be interested to see if there are better strategies than this. But current leading strategy is

to have 1 with a score of 22 and no scores for anything else (equivalent strategy is to have 44/7 on all three targets).

However, this solution is no longer allowed with the new rules, so I would go for the second option:

One area worth 4 and two worth 8. This has approx 15% chance.

@Nopalaa used a computer to consider all the possibilities and confirmed that this is the optimal solution. He also confirmed the probability as being $\frac{40464}{262144}=\frac{2529}{16384}\approx15.44%$, as shown in the wolfram alpha link above.

So here is a first guess to create a lower limit on the probability.

Firstly, we're taking 9 shots at 4 targets with equal probability. One of the targets has a score of 0. We need to assign a score to each of the other three to maximize our probability of getting 44.

It seems to me that we should look at divisors of 44 since there is no advantage to getting close to 44. So 2, 4 or 11 are the obvious cases. And since we are expecting to be hitting the scoring target 6 or 7 times (more likely 7, I guess - if we distribute the 8 equally, the ninth is more likely to hit than not), so we're targeting an average of around 6 or 7.

So here are a couple of possible strategies:

1. Have one of the target areas be worth 22, and the others 0. We would need to hit that target area twice to win.
2. Have one of the target areas be worth 4, and the others 8. I'll divide everything through by 4. So we are targeting a score of 11 now, and have scores of 2,2, and 1. We would need:

• 5x2 and 1x1, or
• 4x2 and 3x1, or
• 3x2 and 5x1, or
• 2x2 and 7x1.

The first of these has probability of $(0.25)^2(0.75)^7\binom{9}{2}\approx0.300$.

The second of these I can't see a neat way to calculate so I'll do it the painful way, which is to add up the four terms:

• $(0.5)^5(0.25)^1(0.25)^3\left(\frac{9!}{5!1!3!}\right)\approx0.0615$
• $(0.5)^4(0.25)^3(0.25)^2\left(\frac{9!}{4!3!2!}\right)\approx0.0769$
• $(0.5)^3(0.25)^5(0.25)^1\left(\frac{9!}{3!5!1!}\right)\approx0.0153$
• $(0.5)^2(0.25)^7(0.25)^0\left(\frac{9!}{2!7!0!}\right)\approx0.0005$

This is about 0.15. (Having two 1s and one 2 is worse).

I'll be interested to see if there are better strategies than this. But current leading strategy is

to have 1 with a score of 22 and no scores for anything else (equivalent strategy is to have 44/7 on all three targets).

So here is a first guess to create a lower limit on the probability.

Firstly, we're taking 9 shots at 4 targets with equal probability. One of the targets has a score of 0. We need to assign a score to each of the other three to maximize our probability of getting 44.

It seems to me that we should look at divisors of 44 since there is no advantage to getting close to 44. So 2, 4 or 11 are the obvious cases. And since we are expecting to be hitting the scoring target 6 or 7 times (more likely 7, I guess - if we distribute the 8 equally, the ninth is more likely to hit than not), so we're targeting an average of around 6 or 7.

So here are a couple of possible strategies:

1. Have one of the target areas be worth 22, and the others 0. We would need to hit that target area twice to win.
2. Have one of the target areas be worth 4, and the others 8. I'll divide everything through by 4. So we are targeting a score of 11 now, and have scores of 2,2, and 1. We would need:

• 5x2 and 1x1, or
• 4x2 and 3x1, or
• 3x2 and 5x1, or
• 2x2 and 7x1.

The first of these has probability of $(0.25)^2(0.75)^7\binom{9}{2}\approx0.300$.

The second of these I can't see a neat way to calculate so I'll do it the painful way, which is to add up the four terms:

• $(0.5)^5(0.25)^1(0.25)^3\left(\frac{9!}{5!1!3!}\right)\approx0.0615$
• $(0.5)^4(0.25)^3(0.25)^2\left(\frac{9!}{4!3!2!}\right)\approx0.0769$
• $(0.5)^3(0.25)^5(0.25)^1\left(\frac{9!}{3!5!1!}\right)\approx0.0153$
• $(0.5)^2(0.25)^7(0.25)^0\left(\frac{9!}{2!7!0!}\right)\approx0.0005$

This is about 0.15. (Having two 1s and one 2 is worse).

I'll be interested to see if there are better strategies than this. But current leading strategy is

to have 1 with a score of 22 and no scores for anything else (equivalent strategy is to have 44/7 on all three targets).

However, this solution is no longer allowed with the new rules, so I would go for the second option:

One area worth 4 and two worth 8. This has approx 15% chance.

So here is a first guess to create a lower limit on the probability.

Firstly, we're taking 9 shots at 4 targets with equal probability. One of the targets has a score of 0. We need to assign a score to each of the other three to maximize our probability of getting 44.

It seems to me that we should look at divisors of 44 since there is no advantage to getting close to 44. So 2, 4 or 11 are the obvious cases. And since we are expecting to be hitting the scoring target 6 or 7 times (more likely 7, I guess - if we distribute the 8 equally, the ninth is more likely to hit than not), so we're targeting an average of around 6 or 7.

So here are a couple of possible strategies:

1. Have one of the target areas be worth 22, and the others 0. We would need to hit that target area twice to win.
2. Have one of the target areas be worth 4, and the others 8. I'll divide everything through by 4. So we are targeting a score of 11 now, and have scores of 2,2, and 1. We would need:

• 5x2 and 1x1, or
• 4x2 and 3x1, or
• 3x2 and 5x1, or
• 2x2 and 7x1.

The first of these has probability of $(0.25)^2(0.75)^7\binom{9}{2}\approx0.300$.

The second of these I can't see a neat way to calculate so I'll do it the painful way, which is to add up the four terms:

• $(0.5)^5(0.25)^1(0.25)^3\left(\frac{9!}{5!1!3!}\right)\approx0.0615$
• $(0.5)^4(0.25)^3(0.25)^2\left(\frac{9!}{4!3!2!}\right)\approx0.0769$
• $(0.5)^3(0.25)^5(0.25)^1\left(\frac{9!}{3!5!1!}\right)\approx0.0153$
• $(0.5)^2(0.25)^7(0.25)^0\left(\frac{9!}{2!7!0!}\right)\approx0.0005$

This is about 0.15. (Having two 1s and one 2 is worse).

I'll be interested to see if there are better strategies than this. But leading current strategy is have 1 with a score of 22 and no scores for anything else.

So here is a first guess to create a lower limit on the probability.

Firstly, we're taking 9 shots at 4 targets with equal probability. One of the targets has a score of 0. We need to assign a score to each of the other three to maximize our probability of getting 44.

It seems to me that we should look at divisors of 44 since there is no advantage to getting close to 44. So 2, 4 or 11 are the obvious cases. And since we are expecting to be hitting the scoring target 6 or 7 times (more likely 7, I guess - if we distribute the 8 equally, the ninth is more likely to hit than not), so we're targeting an average of around 6 or 7.

So here are a couple of possible strategies:

1. Have one of the target areas be worth 22, and the others 0. We would need to hit that target area twice to win.
2. Have one of the target areas be worth 4, and the others 8. I'll divide everything through by 4. So we are targeting a score of 11 now, and have scores of 2,2, and 1. We would need:

• 5x2 and 1x1, or
• 4x2 and 3x1, or
• 3x2 and 5x1, or
• 2x2 and 7x1.

The first of these has probability of $(0.25)^2(0.75)^7\binom{9}{2}\approx0.300$.

The second of these I can't see a neat way to calculate so I'll do it the painful way, which is to add up the four terms:

• $(0.5)^5(0.25)^1(0.25)^3\left(\frac{9!}{5!1!3!}\right)\approx0.0615$
• $(0.5)^4(0.25)^3(0.25)^2\left(\frac{9!}{4!3!2!}\right)\approx0.0769$
• $(0.5)^3(0.25)^5(0.25)^1\left(\frac{9!}{3!5!1!}\right)\approx0.0153$
• $(0.5)^2(0.25)^7(0.25)^0\left(\frac{9!}{2!7!0!}\right)\approx0.0005$

This is about 0.15. (Having two 1s and one 2 is worse).

I'll be interested to see if there are better strategies than this. But current leading strategy is

to have 1 with a score of 22 and no scores for anything else (equivalent strategy is to have 44/7 on all three targets).