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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 231

A basketball player has a 50% free throw shooting average at his first two throws, which can be interpreted to mean that the probability of his hitting any single free throw (of the first two) is 0.5 or 5/10. Right thereafter, the probability to succeed to any new throw is equal to the true total average, of all throws from the first one up to the previous (of the one he is going to shoot now).

What are his chances

1. he succeeds to exactly 50 of all 100 throws he will shoot.

2. he succeeds to at least 50 of all 100 throws he will shoot.

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**anonimnystefy****Real Member**- From: Harlan's World
- Registered: 2011-05-23
- Posts: 16,015

Wait, so, if he succeeds at the first and second shot, he will succeed at all of the shots after those two as well?

Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment

The knowledge of some things as a function of age is a delta function.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

Just worked on this problem for a contest solution.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 231

anonimnystefy wrote:

Wait, so, if he succeeds at the first and second shot, he will succeed at all of the shots after those two as well?

No, he won't succeed at BOTH the first and second shot, only at one of them.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 231

bobbym wrote:

Can you explain?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

Hi;

I am unable to submit any working until the expiration time of the contest problem. That will be in 1 day and 19 hours from now. Since I submitted my solution he has been watching this forum, I guess to dq me.

I can say that it is done using "spot the pattern" and experimental mathematics. It yielded a small, tight formula for any number of heads.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 231

How did the contest go? Can you now explain your solution?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

Hi;

The contest is a continuing weekly for total points so it will never be over until I have the most points or die trying.

They want me to submit solutions but it is much easier for me to do the problems than to explain what I did. I think you will see how involved it is. I will try to explain my thoughts as I was doing it.

We call a hit H and a miss M.

The probabilty of H hits and M misses is always the same no matter how they are ordered. That is to say {H, M, H, T, H} has the same probability of occuring as {M,H, H, H, M}. No matter what arrangement of H,H,H,M,M,M,H,M... that produces 50 H's and 50 M's. This is because the numerators and denominators of the fractions produced are the same just in different order.

We only need to know how many arrangements are there.

The first two do not need to be figured. They are always {H,M} or {M,H}. So we are really only looking for 49 H's and 49 M's.

If this is not extremely confusing we can move on to the second question.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 231

bobbym wrote:

Hi;

The contest is a continuing weekly for total points so it will never be over until I have the most points or die trying.

They want me to submit solutions but it is much easier for me to do the problems than to explain what I did. I will try to explain my thoughts as I was doing it.

We call a hit H and a miss M.

The probabilty of H hits and M misses is always the same no matter how they are ordered. That is to say {H, M, H, T, H} has the same probability of occuring as {M,H, H, H, M}. No matter what arrangement of H,H,H,M,M,M,H,M... that produces 50 H's and 50 M's. This is because the numerators and denominators of the fractions produced are the same just in different order.

We only need to know how many arrangements are there.

The first two do not need to be figured. They are always {H,M} or {M,H}. So we are really only looking for 49 H's and 49 M's.

If this is not extremely confusing we can move on to the second question.

Well, it is quite a bit

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

Do not worry, it gets worse.

A shortcut consists of playing spot the pattern and going on from there proving it later by induction or whatever.

From some small problems of say 3 Hits and 4 Misses, I could guess at this formula:

This was followed by a large amount of empirical testing so that at least there is a chance it is correct.

Now for the last question. Since we have a formula we can just sum it:

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 231

anna_gg wrote:

bobbym wrote:Hi;

The contest is a continuing weekly for total points so it will never be over until I have the most points or die trying.

They want me to submit solutions but it is much easier for me to do the problems than to explain what I did. I will try to explain my thoughts as I was doing it.

We call a hit H and a miss M.

The probabilty of H hits and M misses is always the same no matter how they are ordered. That is to say {H, M, H, T, H} has the same probability of occuring as {M,H, H, H, M}. No matter what arrangement of H,H,H,M,M,M,H,M... that produces 50 H's and 50 M's. This is because the numerators and denominators of the fractions produced are the same just in different order.

We only need to know how many arrangements are there.

The first two do not need to be figured. They are always {H,M} or {M,H}. So we are really only looking for 49 H's and 49 M's.

If this is not extremely confusing we can move on to the second question.

Well, it is quite a bit

So what is this big number in the denominator of the second fraction? Where does it come from?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

That is from one single arrangement of 49 H's and 49 M's. Remember all those arrangements have the same probability.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**anna_gg****Member**- Registered: 2012-01-10
- Posts: 231

bobbym wrote:

Do not worry, it gets worse.

A shortcut consists of playing spot the pattern and going on from there proving it later by induction or whatever.

From some small problems of say 3 Hits and 4 Misses, I could guess at this formula:

This was followed by a large amount of empirical testing so that at least there is a chance it is correct.

Now for the last question. Since we have a formula we can just sum it:

Are you sure about the last part? Because my calculation (based on your calculations, actually!!!) results to

*Last edited by anna_gg (2013-07-09 03:45:55)*

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

Sorry, I did not follow the posts after the first one and lost sight of that condition. Based on post #4 where he must have a {H,M} or a {M,H} then yes, goto 99 instead of 100.

Also, you can use the formula which will be easier than the series.

where lower = 50 and higher = 99.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 538

Hello m;

That is a nice generalization of the problem from [site removed]. Was Math involved a lot? Why don't you start a thread for us beginners on how to use it?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

Hi EVW;

Please refer to the site as B and the CAS as M.

Why don't you start a thread for us beginners on how to use it?

That is what the computer math thread is for?! Also as you have been told, I am not an expert. Why don't you ask the Colonel or DL?

Was M involved a lot?

Of course, it is unthinkable to leave him out.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 538

Hi;

Was that supposed to be funny? Eric is much less talkative than you are and rarely even explains where those magical answers come from. He heads for the security elevators and goes down [removed by administrator]... When he returns he has the answers he needs. DL, knows nothing.

Please email me the code and I will bring over pizza.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

I thought it was very funny. But not as funny as your primitive attempts to weasel information out of me. Make it Boston Pizza and make sure there is none of that veggie garbage on top of it.

Code sent to your email.

He heads for the security elevators and goes down [removed by administrator]... When he returns he has the answers he needs.

So that is the reason for his superhuman capabilities... As if I did not know.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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**ElainaVW****Member**- Registered: 2013-04-29
- Posts: 538

Do I sense trepidation?

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 104,700

You would be well advised to start sensing Mozzarella cheese for those pizzas instead.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**

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