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Math Puzzles (from FiveThirtyEight - new puzzle every Friday)


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Ignoratio -- I just don't understand the math. Once the 538 answer is posted, please post here. That 2/3 number is really surprising to me

Will do. It looks like it's a weekly feature so maybe I'll bump this thread each Friday with a new puzzle.

(clip)

The insight to have here is that the denominators will always go 2, 3, 4, ... , (s-1). And the numerators will always include 1, 2, 3, ..., (m-1) (for each of the made shots, whatever order they come in) and 1, 2, 3, ... , (s-m-1) for each of the misses. That is, the probability of making exactly 3 of the 5 shots is the same, no matter what order they're made in. The same is true for larger cases - there are an absurdly large number of ways that you can make, say, 50 out of 99 shots, but the probability of every single one of those sequences is the same.

(clip)

Ok, this is where you are losing me.

I don't understand how the bolded can be true. The probability that he makes 49 shots in a row (after his 1/2 performance) followed by missing the next 49 shots has got to be really miniscule, right? Because that would mean he would be at 50 made shots out of 51 shots attempted. The likelihood that he'd make the next one is well over 95%, right? So the likelihood that he'd go on a run of 49 missed would be really really slim, right? I don't see how that would be "just as likely" as where he goes miss, hit, miss, miss, hit, hit, miss, hit, miss, etc." until he's at 50 out of 99.

In short, because the next basket made is dependent on the percentages made before it, I don't understand how "the probability of every single one of the sequences is the same."

The same way you can't say that a certain sequence of lottery numbers has a greater probability than any other. I mean the chances of the winning numbers being 1, 2, 3, 4, 5 and powerball 6 has got to be really minuscule, right?

Going 40 miss and 50 makes has the exact same probability as doing your hit, miss, miss, hit, miss......... in that exact sequence.

Edited by tonydead
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Ignoratio -- I just don't understand the math. Once the 538 answer is posted, please post here. That 2/3 number is really surprising to me

Will do. It looks like it's a weekly feature so maybe I'll bump this thread each Friday with a new puzzle.

(clip)

The insight to have here is that the denominators will always go 2, 3, 4, ... , (s-1). And the numerators will always include 1, 2, 3, ..., (m-1) (for each of the made shots, whatever order they come in) and 1, 2, 3, ... , (s-m-1) for each of the misses. That is, the probability of making exactly 3 of the 5 shots is the same, no matter what order they're made in. The same is true for larger cases - there are an absurdly large number of ways that you can make, say, 50 out of 99 shots, but the probability of every single one of those sequences is the same.

(clip)

Ok, this is where you are losing me.

I don't understand how the bolded can be true. The probability that he makes 49 shots in a row (after his 1/2 performance) followed by missing the next 49 shots has got to be really miniscule, right? Because that would mean he would be at 50 made shots out of 51 shots attempted. The likelihood that he'd make the next one is well over 95%, right? So the likelihood that he'd go on a run of 49 missed would be really really slim, right? I don't see how that would be "just as likely" as where he goes miss, hit, miss, miss, hit, hit, miss, hit, miss, etc." until he's at 50 out of 99.

In short, because the next basket made is dependent on the percentages made before it, I don't understand how "the probability of every single one of the sequences is the same."

I kept the example above small so it might not be apparent what's happening. Here's the way to think about it. Remember that the probability of making a shot at any point is simply (number of shots made so far)/(number of shots taken so far). It's not stated explicitly, but you can also see that this means the probability of missing a shot at any point is simply (number of shots missed so far)/(number of shots taken so far). For example, if after 10 shots he's made 7 of them (and thus missed 3), the probability of making the next one is 7/10 (and thus the probability of missing the next one is 3/10).

Ok, so let's say he makes 50 of the 99 shots. He makes the first and misses the second, we know that already. At some point, he'll make a second shot (this might come on his third try, or his fourth, or his 20th, whatever). When he does, we know that the probability of him making that shot was 1/(however many shots he's taken so far), because up to that point he'd only made 1 shot. At some point after that, he'll make another one. And again, even though we don't know when he'll make it, we know the probability of making the shot at that point is 2/(however many shots he's taken so far). And then at some point after that he'll make another one, with probability 3/(however many shots he's taken so far)... all the way up until he makes his 50th shot, with probability 49/(however many shots he's taken so far).

Similarly, at some point he'll miss. We don't know when, but whenever it happens we know that the probability of missing at that point was 1/(however many shots he's taken so far). At some later point, he'll miss again with probability 2/(however many shots he's taken so far)... all the way up until he misses his 49th shot with probability 48/(however many shots he's taken so far).

To calculate the total probability of the whole sequence occurring, we just multiply all of these individual probabilities together, and multiplication of fractions is associative in that we can multiply the numerators together in any order, and then multiply all the denominators together in any order, and then take the quotient. Like 4/9 * 6/7 is the same as 6/9 * 4/7. I can switch around the tops and bottoms and the answer is the same.

So, we've got a bunch of fractions where the numerators are 1, 2, 3, ..., (m-1) (for all the shots that were made) and 1, 2, 3, ..., (s-m-1) (for all the shots that were missed. If we multiply them all together, that first thing is just (m-1)! and the second thing is (s-m-1)!

And in these fractions, the denominators are just 1, 2, 3, ..., (s-1). Each one is just the number of shots that have come before. Multiplying all of these together is (s-1)!

So putting them together, we get (m-1)! * (s-m-1)! / (s-1)! If we have a sequence of s shots where exactly m of them are made, this is the probability of it occurring, regardless of the order in which they're made.

Not sure if that's clear and/or convincing enough so maybe a longer example will help. Pretend he took 10 shots and made 5 of them. Let's look at three ways this could happen (1 is make, 0 is miss):

1011110000 (he gets hot early then cools off)

1010101010 (he alternates making and missing shots)

1001000111 (he starts pretty cold then gets hot)

They all start with 10, we know that. Starting with the third shot then, let's look at the probabilities (and try to notice the pattern):

First sequence = 1/2 * 2/3 * 3/4 * 4/5 * 1/6 * 2/7 * 3/8 * 4/9.

Second sequence = 1/2 * 1/3 * 2/4 * 2/5 * 3/6 * 3/7 * 4/8 * 4/9.

Third sequence = 1/2 * 1/3 * 2/4 * 3/5 * 4/6 * 2/7 * 3/8 * 4/9.

The individual shot probabilities vary, but when you multiply them all together they're all the same answer. The bottom just goes from 2 up to 9 each time. And the tops go from 1 to 4 (for the 5 makes) and 1 to 4 (for the 5 misses) in some order.

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The odds after any given trial probably won't be 2 in 3. They could be anywhere from 98/99 to 2/99. You have no way of knowing exactly what the 96 shots between shot 2 and shot 99 looked like.

But if you add up all the ways things could have gone down and divide only by the number of trials that end with a hit, it's 2/3.

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Ignoratio -- I just don't understand the math. Once the 538 answer is posted, please post here. That 2/3 number is really surprising to me

Will do. It looks like it's a weekly feature so maybe I'll bump this thread each Friday with a new puzzle.

(clip)

The insight to have here is that the denominators will always go 2, 3, 4, ... , (s-1). And the numerators will always include 1, 2, 3, ..., (m-1) (for each of the made shots, whatever order they come in) and 1, 2, 3, ... , (s-m-1) for each of the misses. That is, the probability of making exactly 3 of the 5 shots is the same, no matter what order they're made in. The same is true for larger cases - there are an absurdly large number of ways that you can make, say, 50 out of 99 shots, but the probability of every single one of those sequences is the same.

(clip)

Ok, this is where you are losing me.

I don't understand how the bolded can be true. The probability that he makes 49 shots in a row (after his 1/2 performance) followed by missing the next 49 shots has got to be really miniscule, right? Because that would mean he would be at 50 made shots out of 51 shots attempted. The likelihood that he'd make the next one is well over 95%, right? So the likelihood that he'd go on a run of 49 missed would be really really slim, right? I don't see how that would be "just as likely" as where he goes miss, hit, miss, miss, hit, hit, miss, hit, miss, etc." until he's at 50 out of 99.

In short, because the next basket made is dependent on the percentages made before it, I don't understand how "the probability of every single one of the sequences is the same."

The same way you can't say that a certain sequence of lottery numbers has a greater probability than any other. I mean the chances of the winning numbers being 1, 2, 3, 4, 5 and powerball 6 has got to be really minuscule, right?

Going 40 miss and 50 makes has the exact same probability as doing your hit, miss, miss, hit, miss......... in that exact sequence.

I think I'm getting really off-track, but whatever. If we keep the same parameters that the next shot is dependent on the percentages of shots made before, I would beg to disagree.

Let's say twenty shots. Starting with the third shot (50/50 chance) that goes in. The next shot has a 66% chance of going on. Then 75% chance. Then 80% chance. Etc.

After ten shots, the percentage is 9/10. Would you agree that the next shot has a 90% chance of hitting, right?

So: If we had to predict the likelihood of the next 10 shots, wouldn't you agree that going 10 for 10 of the next shots is a LOT more likely than going 0 for 10?

I don't understand how you rectify that with the statement that all combinations have equal likelihood of happening.

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After 4 shots, he was either 2/3 to hit, or 1/3. If we tried this six times it would look something like this:

1011 (75% chance of hitting his next shot)

1011 (75%)

1010 (50%)

1001 (50%)

1000 (25%)

1000 (25%)

If you didn't watch the third shot and only saw him hit the fourth, then you can eliminate any trial that ended with a miss. The only ones left are:

1011 (75% of hitting the next shot)

1011 (75%)

1001 (50%)

75 + 75 + 50 = 200%

Divide that by 3 and it's 2/3.

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After 5 shots it looks like this

10111 (80% of hitting his next shot)

10111 (80%)

10111 (80%)

10110

10111 (80%)

10111 (80%)

10111 (80%)

10110

10101 (60%)

10101 (60%)

10100

10100

10011 (60%)

10011 (60%)

10010

10010

10000

10000

10000

10001 (40%)

10000

10000

10000

10001 (40%)

Add all those 80s and 60s and 40s up and you get 800% over 12 trials - again, exactly 2/3.

Edited by bostonfred
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The odds after any given trial probably won't be 2 in 3. They could be anywhere from 98/99 to 2/99. You have no way of knowing exactly what the 96 shots between shot 2 and shot 99 looked like.

But if you add up all the ways things could have gone down and divide only by the number of trials that end with a hit, it's 2/3.

ok, fred, that helps me conceptualize it a little bit better. Thanks!

But can you expand on it just a little more? How did you come to that conclusion? (or would you just point to Ignoratio's equation? And if so, is there an easier way to think about it?).

Edit: I think you expanded in an earlier post that I missed (I quoted below). So no need to answer. Thanks!

Edited by Sweet J
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After ten shots, the percentage is 9/10. Would you agree that the next shot has a 90% chance of hitting, right?

So: If we had to predict the likelihood of the next 10 shots, wouldn't you agree that going 10 for 10 of the next shots is a LOT more likely than going 0 for 10?

I don't understand how you rectify that with the statement that all combinations have equal likelihood of happening.

This is a different question. If he goes 10 for 10, then you're right, the probability that he makes the next 10 in a row is greater than the probability that he'll miss the next 10 in a row. But then in the first case, he's made 20/20 and in the second he's only made 10/20. The claim isn't that these are equally likely.

The claim is that IF he makes 10 out of 20 shots, then it doesn't matter if he makes the first 10 in a row and then misses the rest, or misses the first 10 in a row and then makes the rest, or alternates making and missing in some apparently random fashion. The probabilities of each possible sequence of 20 shots that includes exactly 10 makes and 10 misses is the same.

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He has a. 50/50 chance of hitting shot 3. There's only two possibilities

101

100

If the coach watches the third shot go in, though, we can eliminate 100. We know it went 101. So his probability of hooting the next shot is 2/3.

After 4 shots, he was either 2/3 to hit, or 1/3. If we tried this six times it would look something like this:

1011 (75% chance of hitting his next shot)

1011 (75%)

1010 (50%)

1001 (50%)

1000 (25%)

1000 (25%)

If you didn't watch the third shot and only saw him hit the fourth, then you can eliminate any trial that ended with a miss. The only ones left are:

1011 (75% of hitting the next shot)

1011 (75%)

1001 (50%)

75 + 75 + 50 = 200%

Divide that by 3 and it's 2/3.

After 5 shots it looks like this

10111 (80% of hitting his next shot)

10111 (80%)

10111 (80%)

10110

10111 (80%)

10111 (80%)

10111 (80%)

10110

10101 (60%)

10101 (60%)

10100

10100

10011 (60%)

10011 (60%)

10010

10010

10000

10000

10000

10001 (40%)

10000

10000

10000

10001 (40%)

Add all those 80s and 60s and 40s up and you get 800% over 12 trials - again, exactly 2/3.

Without continuing on, it seems logical that the trend would continue. The mathematical proof is a little more like what ig latin posted earlier, but this should help to make it clear.

The answer is 2/3.

ok, fred. I somehow missed this post earlier. I don't quite follow it the fist pass through, but I think if I sit with it a little while I'll get it. (i.e., I think I understand that the reasoning is there, even if I don't completely comprehend it).

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After ten shots, the percentage is 9/10. Would you agree that the next shot has a 90% chance of hitting, right?

So: If we had to predict the likelihood of the next 10 shots, wouldn't you agree that going 10 for 10 of the next shots is a LOT more likely than going 0 for 10?

I don't understand how you rectify that with the statement that all combinations have equal likelihood of happening.

This is a different question. If he goes 10 for 10, then you're right, the probability that he makes the next 10 in a row is greater than the probability that he'll miss the next 10 in a row. But then in the first case, he's made 20/20 and in the second he's only made 10/20. The claim isn't that these are equally likely.

The claim is that IF he makes 10 out of 20 shots, then it doesn't matter if he makes the first 10 in a row and then misses the rest, or misses the first 10 in a row and then makes the rest, or alternates making and missing in some apparently random fashion. The probabilities of each possible sequence of 20 shots that includes exactly 10 makes and 10 misses is the same.

I think I'm getting distracted by something that isn't relevant to working out the problem. Thanks for your patience buddy.

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The odds after any given trial probably won't be 2 in 3. They could be anywhere from 98/99 to 2/99. You have no way of knowing exactly what the 96 shots between shot 2 and shot 99 looked like.

But if you add up all the ways things could have gone down and divide only by the number of trials that end with a hit, it's 2/3.

ok, fred, that helps me conceptualize it a little bit better. Thanks!

But can you expand on it just a little more? How did you come to that conclusion? (or would you just point to Ignoratio's equation? And if so, is there an easier way to think about it?).

The equation is the "right" way to solve the problem but it's too hard. The number of trials uses an exclamation point and nobody even remembers what those are after high school's over. If he takes one shot there are two outcomes. (2x1=2). 2 shots has six outcomes (3x2x1=6). 3 shots have twenty four outcomes (4x3x2x1 =24). And so on.

There are 99x98x97x96x....2x1 outcomes that could have led up to that 100th shot, and all are equally likely (although there are a whole lot of duplicates). That may literally be more outcomes than there are atoms in the entire universe. So I'm not going to list them all out. But it's a lot

But you can list them out for 3, 4 or 5 shots and the answer is always exactly the same. That helps to visualize it. Then you can work on the proof like ig Latin did.

The math is the boring part. The fun part is understanding why.

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What are all the possible outcomes of shot 3?

(Hit miss) hit

Or

(Hit miss) miss.

They're equally likely because he had a 50/50 chance.

What are all the outcomes of the fourth shot? If he hit shot 3, he had a 2 in 3 chance of hitting shot four. If he missed it was one in three. So we can list out all those chances like this:

(Hit miss hit) hit

(Hit miss hit) hit

(Hit miss hit) miss

Or

(Hit miss miss) miss

(Hit miss miss) miss

(Hit miss miss) hit

For each one of the above, can you tell me the odds that he hits his next shot? For example, if he went hit miss hit miss, he's 50% to hit the next shot.

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What are all the possible outcomes of shot 3?

(Hit miss) hit

Or

(Hit miss) miss.

They're equally likely because he had a 50/50 chance.

What are all the outcomes of the fourth shot? If he hit shot 3, he had a 2 in 3 chance of hitting shot four. If he missed it was one in three. So we can list out all those chances like this:

(Hit miss hit) hit 75%

(Hit miss hit) hit 75%

(Hit miss hit) miss 50%

Or

(Hit miss miss) miss 25%

(Hit miss miss) miss 25%

(Hit miss miss) hit 50%

For each one of the above, can you tell me the odds that he hits his next shot? For example, if he went hit miss hit miss, he's 50% to hit the next shot.

Is that right? Ok, now what?

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Now let's assume the coach walks in and sees the player hit the last shot he took.

That means we only need to consider these three:

(Hit miss hit) hit 75%

(Hit miss hit) hit 75%

(Hit miss miss) hit 50%

Because in these three scenarios, he didn't hit the last shot.

(Hit miss hit) miss 50%

(Hit miss miss) miss 25%

(Hit miss miss) miss 25%

What's (75%+75%+50%) divided by three?

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After 5 shots it looks like this

10111 (80% of hitting his next shot)

10111 (80%)

10111 (80%)

10110

10111 (80%)

10111 (80%)

10111 (80%)

10110

10101 (60%)

10101 (60%)

10100

10100

10011 (60%)

10011 (60%)

10010

10010

10000

10000

10000

10001 (40%)

10000

10000

10000

10001 (40%)

Add all those 80s and 60s and 40s up and you get 800% over 12 trials - again, exactly 2/3.

1 is a hit and 0 is a miss. See if this makes sense now

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If the coach sees him hit his third shot then he knows that the odds are 2/3.

If he doesn't watch the third shot but sees the player make the fourth shot, he knows the odds were either 75%, 75% or 50% - which averages out to exactly 2/3.

If he doesn't watch the third or fourth shots, but sees the player make the fifth, then it's some combination of 40%, 60% and 80% that again average out to exactly 2/3.

It doesn't matter how many shots there were from the time he leaves the gym until the time he sees the guy make that last shot - the answer will always average out to 2/3.

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If the coach sees him hit his third shot then he knows that the odds are 2/3.

If he doesn't watch the third shot but sees the player make the fourth shot, he knows the odds were either 75%, 75% or 50% - which averages out to exactly 2/3.

If he doesn't watch the third or fourth shots, but sees the player make the fifth, then it's some combination of 40%, 60% and 80% that again average out to exactly 2/3.

It doesn't matter how many shots there were from the time he leaves the gym until the time he sees the guy make that last shot - the answer will always average out to 2/3.

Weighted averages, but yup

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After 5 shots it looks like this

10111 (80% of hitting his next shot)

10111 (80%)

10111 (80%)

10110

10111 (80%)

10111 (80%)

10111 (80%)

10110

10101 (60%)

10101 (60%)

10100

10100

10011 (60%)

10011 (60%)

10010

10010

10000

10000

10000

10001 (40%)

10000

10000

10000

10001 (40%)

Add all those 80s and 60s and 40s up and you get 800% over 12 trials - again, exactly 2/3.

1 is a hit and 0 is a miss. See if this makes sense now

Ok, all this together does help. You are right. The fun isn't in the equation, but in getting the theory behind it. I'm not fully getting it yet, but only because my brain is wicked distracted with other stuff. I probably need to be at work so I have something that I'm trying to avoid, thereby driving me to understand a meaningless internet problem.

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I think Sweet's point is that the odds of any particular sequence is only the same if the event is Independant. But since these events are not Independant (in this case the odds are dependent upon what happened in the previous events), then you can not say the odds of any particular sequence are all equivalent. I agree if that is his point.

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I think Sweet's point is that the odds of any particular sequence is only the same if the event is Independant. But since these events are not Independant (in this case the odds are dependent upon what happened in the previous events), then you can not say the odds of any particular sequence are all equivalent. I agree if that is his point.

The probability of any particular sequence that includes exactly m made shots is the same. There are (96 choose 48) ~= 6.4 octillion different ways that the shooter can make exactly 50 free throws, but the probability of each of those sequences occurring is exactly the same. There are (96 choose 1) = 96 different ways the shooter can make exactly 97 free throws, and the probability of each of those sequences is also the same (but not the same as the probability of a sequence containing exactly 50 made free throws).

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I think Sweet's point is that the odds of any particular sequence is only the same if the event is Independant. But since these events are not Independant (in this case the odds are dependent upon what happened in the previous events), then you can not say the odds of any particular sequence are all equivalent. I agree if that is his point.

The probability of any particular sequence that includes exactly m made shots is the same. There are (96 choose 48) ~= 6.4 octillion different ways that the shooter can make exactly 50 free throws, but the probability of each of those sequences occurring is exactly the same. There are (96 choose 1) = 96 different ways the shooter can make exactly 97 free throws, and the probability of each of those sequences is also the same (but not the same as the probability of a sequence containing exactly 50 made free throws).

I am not convinced, but you are probably right. If I have time tonight I may run a few smaller examples like 6 and see if every combination of 4 is equivalent.

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I think you guys are missing the point. The coach knows he made shot 99. The coach knows his tendencies. So, from his point of view, he knows the player is either on heater and making them all, or somewhere around 50/50. He's ruling out that he's on the cold streak because he saw him hit shot 99. So, the coach is probably putting it @ 75% that he hits shot 100.

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Trying to set up sigma notation....

Let n = number of attempts, ie 99

Let k = number of makes (2 to 98)

Number of arrangements = Comb( n-3, k-2) = (n-3)!/(k-2)!(n-k-1)!

Probabilty of each permutation = (k-1)!(n-k-1)!/(n-1)!

Probability of each combination = (k-1)/(n-1)(n-2)

EV = k(k-1)/(n-1)(n-2) need to divide this by the sum of the probabilities as we omitting half of them...

Sum of probs = (1/98)(1/97) Sigma (k-1) 1 to 98 = (1/98)(1/97) * ((98 * 99/2) - 98) = (1/98)(1/97) ( 98*97)/2 = 1/2

1/(98)(97) Sigma (k^2-k) from 2 to 98.....k 0 1 case is zero so evakluate from 1 to 98

1/(98)(97) (98)(99)(197)/6 - (98)(99)/2

(99/97) (197/6 -1/2)

(99/97)(194/6) = 33

Messing up something obvious, should be 66

Edited by Short Corner
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EV = k(k-1)/(n-1)(n-2)

...

Messing up something obvious, should be 66

You're missing the denominator:

So the expected value of m, the number of shots made, after s shots are taken, is:

Sum from m=2 to s-1 of: (s-3 choose m-2) * (m-1)! * (s-m-1)! / (s-1)! * m

divided by

Sum from m=2 to s-1 of: (s-3 choose m-2) * (m-1)! * (s-m-1)! / (s-1)!

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I think you guys are missing the point. The coach knows he made shot 99. The coach knows his tendencies. So, from his point of view, he knows the player is either on heater and making them all, or somewhere around 50/50. He's ruling out that he's on the cold streak because he saw him hit shot 99. So, the coach is probably putting it @ 75% that he hits shot 100.

Right, but he doesn't know anything about shots 3-98. He could have missed every one of those shots and there would still be a probability > 0 of hitting 99. It would be highly improbable, but not impossible.

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Right, but he doesn't know anything about shots 3-98. He could have missed every one of those shots and there would still be a probability > 0 of hitting 99. It would be highly improbable, but not impossible.

He knows something about shots 3-98. If he didn't see the 99th shot, he'd have no idea what happened. But given that the player made shot 99 (and the probability of making shot 99 depends on how many shots he made from 3-98), is it more likely that he had made most of his shots from 3-98, or missed most of them?

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This problem is not generating as much controversy as the 'Let's Make a Deal' problem. But in some ways it is similar.

I think the difference is in the Monty Hall problem, the naive, "common sense" answer is 1/2 and the actual answer is 2/3.

In this problem the naive answer is 2/3, and the actual answer is 2/3 (though it's not exactly trivial to prove it). So there's no real controversy, many people seem to quickly land on the right answer even if they don't really understand why it's right.

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Right, but he doesn't know anything about shots 3-98. He could have missed every one of those shots and there would still be a probability > 0 of hitting 99. It would be highly improbable, but not impossible.

He knows something about shots 3-98. If he didn't see the 99th shot, he'd have no idea what happened. But given that the player made shot 99 (and the probability of making shot 99 depends on how many shots he made from 3-98), is it more likely that he had made most of his shots from 3-98, or missed most of them?

He knows only what is more likely to have occurred, not what actually occurred.

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Right, but he doesn't know anything about shots 3-98. He could have missed every one of those shots and there would still be a probability > 0 of hitting 99. It would be highly improbable, but not impossible.

He knows something about shots 3-98. If he didn't see the 99th shot, he'd have no idea what happened. But given that the player made shot 99 (and the probability of making shot 99 depends on how many shots he made from 3-98), is it more likely that he had made most of his shots from 3-98, or missed most of them?
He knows only what is more likely to have occurred, not what actually occurred.

Which is what probability is all about. If you know the actual result, you don't need it.

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No disagreement on that - I was only pointing out the shaky foundation of hulk's comment that he saw him hit 99 meant that the coach could rule out that he was on a cold streak. It's almost unfortunate that the actual binary tree calculations work out to the intuitive result that is based on faulty logic.

Edited by elbowrm
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This problem is not generating as much controversy as the 'Let's Make a Deal' problem. But in some ways it is similar.

I think the difference is in the Monty Hall problem, the naive, "common sense" answer is 1/2 and the actual answer is 2/3.

In this problem the naive answer is 2/3, and the actual answer is 2/3 (though it's not exactly trivial to prove it). So there's no real controversy, many people seem to quickly land on the right answer even if they don't really understand why it's right.

I actually found the 2/3 answer to be somewhat surprising. My instinct immediately after reading the question was that the percentage would be in the 90%+ range. Maybe other people found the 2/3 answer intuitive. :shrug:

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I bet you didn't think it's definitely 90%.

I bet what you really thought it might be 90%, then thought it might also be like 2%, but it probably wasn't.

And that is correct. It probably isn't 90%, although it might be. It probably isn't 2%, although that's technically possible. It's the average of all those probablies and might bes.

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I bet you didn't think it's definitely 90%.

I bet what you really thought it might be 90%, then thought it might also be like 2%, but it probably wasn't.

And that is correct. It probably isn't 90%, although it might be. It probably isn't 2%, although that's technically possible. It's the average of all those probablies and might bes.

No, I really thought it was 90%+. My thought process was basically this --

1) due to the way this problem works, either the percentage over the 95 unseen shots should steadily rise over time to a very high number or steadily decline to a very low number.

2) because he made shot 99, the overwhelming likelihood is that we're in the "very high number" scenario. So we have a very high likelihood of a high number or a very low likelihood of a low percentage.

3) The average of these probabilities would therefore be a really high number.

The I got lazy and decided just to read the posts instead of trying to figure it out for myself.

In hindsight, I guess my mistake was just in overestimating how high the high number would be and how low the low number would be. The percentages wouldn't rise or fall in such a steady way, even if the guy starts with a high percentage, he'll miss some along the way.

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After ten shots, the percentage is 9/10. Would you agree that the next shot has a 90% chance of hitting, right?

So: If we had to predict the likelihood of the next 10 shots, wouldn't you agree that going 10 for 10 of the next shots is a LOT more likely than going 0 for 10?

I don't understand how you rectify that with the statement that all combinations have equal likelihood of happening.

This is a different question. If he goes 10 for 10, then you're right, the probability that he makes the next 10 in a row is greater than the probability that he'll miss the next 10 in a row. But then in the first case, he's made 20/20 and in the second he's only made 10/20. The claim isn't that these are equally likely.

The claim is that IF he makes 10 out of 20 shots, then it doesn't matter if he makes the first 10 in a row and then misses the rest, or misses the first 10 in a row and then makes the rest, or alternates making and missing in some apparently random fashion. The probabilities of each possible sequence of 20 shots that includes exactly 10 makes and 10 misses is the same.

I think I'm getting distracted by something that isn't relevant to working out the problem. Thanks for your patience buddy.

If I read you right, I think you're mixing situations of what information is available to the person we're asking the odds of while pondering that the odds should eventually converge towards 0% or 100% if it goes long enough. Try taking a step back and look at the difference of the situation from the coach's perspective and from the player's perspective.

Let's say the player makes 9 of the first 10 as you say. The player knows this, the coach does not.

Using the information the player knows, if we asked him after shot 10, the player would not expect the odds of him making the 20th shot to be 50%. As you are thinking, the odds are heavily skewed towards him making most of the shots because he's starting at a 90% chance of making shot 11, etc.

But we are not asking what the player thinks the odds are, we're asking the coach.

The coach does not know 9 out of 10 were made. He still only knows that the first 2 shots were made. Using the information the coach knows, the player was just as likely to go 1 for the first 10 as he was to go 9 for the first 10. While the player has information that rules out other possibilities and leads him to expect a high success rate on shot 20, the coach does not have that information. Based on the info the coach can work from, all the scenarios in the 1 out of 10 possibility are still as likely as the 9 out of 10. This is where IE keeps talking about the odds are just as likely. At that point, the coach only knows that the first two shots were a make and a miss.

I think you keep getting lost in the fact that though the odds will eventually converge towards either 0% or 100% over enough trials, our coach is given no information which way it is diverging with just knowing the first 2 shots. The coach might agree it is likely the player made most or missed most rather than making 50%. But based on the info he can work from, those are equally likely so leave him at 50% odds of making the next basket... until such time as he knows the outcome of additional baskets such as learning the result of shot 99.

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He has a. 50/50 chance of hitting shot 3. There's only two possibilities

101

100

If the coach watches the third shot go in, though, we can eliminate 100. We know it went 101. So his probability of hooting the next shot is 2/3.

After 4 shots, he was either 2/3 to hit, or 1/3. If we tried this six times it would look something like this:

1011 (75% chance of hitting his next shot)

1011 (75%)

1010 (50%)

1001 (50%)

1000 (25%)

1000 (25%)

If you didn't watch the third shot and only saw him hit the fourth, then you can eliminate any trial that ended with a miss. The only ones left are:

1011 (75% of hitting the next shot)

1011 (75%)

1001 (50%)

75 + 75 + 50 = 200%

Divide that by 3 and it's 2/3.

After 5 shots it looks like this

10111 (80% of hitting his next shot)

10111 (80%)

10111 (80%)

10110

10111 (80%)

10111 (80%)

10111 (80%)

10110

10101 (60%)

10101 (60%)

10100

10100

10011 (60%)

10011 (60%)

10010

10010

10000

10000

10000

10001 (40%)

10000

10000

10000

10001 (40%)

Add all those 80s and 60s and 40s up and you get 800% over 12 trials - again, exactly 2/3.

Without continuing on, it seems logical that the trend would continue. The mathematical proof is a little more like what ig latin posted earlier, but this should help to make it clear.

The answer is 2/3.

Thanks for this. Even after reading it, I still couldn't figure out why my original thoughts were wrong.

I took a look at your 5 shots scenario and instead of the ones where he makes the 5th, look at the ones where he misses the 5th. Adding up those 60s and 40s and 20s gets you 400% over 12 trials, exactly 1/3.

It's interesting, that no matter how many shots the coach misses, once he comes back and watches the player take a shot, the odds of the player making the next is 2/3 if he makes it and 1/3 if he doesn't according to the coach.

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For me, the clue was in the way the problem was written. With the limited information the coach has, the answer is pretty clearly 2/3, but he leaves the gym, so his limited-information conclusion has to be wrong, right? Turns out that no, it's right whether he left or not.

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Is this true?

The information the coach has leans more towards a 9/10 streak than a 1/10 streak, right?

Not for the moment of shots 10 or 20 as Sweet J and IE are talking, no.

His only information is a make and a miss on the first 2 shots. Going 8-for-8 and going 0-for-8 from there are equally likely to someone with that information. He gets no additional info until shot 99.

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For me, the clue was in the way the problem was written. With the limited information the coach has, the answer is pretty clearly 2/3, but he leaves the gym, so his limited-information conclusion has to be wrong, right? Turns out that no, it's right whether he left or not.

If he stayed at the gym, and watched every single shot up to shot 99, the chances that it would be exactly 2/3 at that point is miniscule.

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For me, the clue was in the way the problem was written. With the limited information the coach has, the answer is pretty clearly 2/3, but he leaves the gym, so his limited-information conclusion has to be wrong, right? Turns out that no, it's right whether he left or not.

If he stayed at the gym, and watched every single shot up to shot 99, the chances that it would be exactly 2/3 at that point is miniscule.

No. See this from Greg:

"I think you keep getting lost in the fact that though the odds will eventually converge towards either 0% or 100% over enough trials, our coach is given no information which way it is diverging with just knowing the first 2 shots. The coach might agree it is likely the player made most or missed most rather than making 50%. But based on the info he can work from, those are equally likely so leave him at 50% odds of making the next basket... until such time as he knows the outcome of additional baskets such as learning the result of shot 99."

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For me, the clue was in the way the problem was written. With the limited information the coach has, the answer is pretty clearly 2/3, but he leaves the gym, so his limited-information conclusion has to be wrong, right? Turns out that no, it's right whether he left or not.

If he stayed at the gym, and watched every single shot up to shot 99, the chances that it would be exactly 2/3 at that point is miniscule.

No. See this from Greg:

"I think you keep getting lost in the fact that though the odds will eventually converge towards either 0% or 100% over enough trials, our coach is given no information which way it is diverging with just knowing the first 2 shots. The coach might agree it is likely the player made most or missed most rather than making 50%. But based on the info he can work from, those are equally likely so leave him at 50% odds of making the next basket... until such time as he knows the outcome of additional baskets such as learning the result of shot 99."

I disagree on the convergence

The odds of making n shots in a row (after shot 2) is 1/(n-1). Which means the odds of converging to 100% decrease over time

Similarly the odds of converging to 0 decrease over time

Edited by Rove!
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