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Subject: New Friday Math Problem (Jan. 15th) rss

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The Neurotic Basketball Player

A basketball player is in the gym practicing free throws. He makes his first shot, then misses his second. This player tends to get inside his own head a little bit, so this isn’t good news.

Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that he’s made thus far. (His neuroses are very exacting.)
His coach, who knows his psychological tendency and saw the first two shots, leaves the gym and doesn’t see the next 96 shots. The coach returns, and sees the player make shot No. 99. What is the probability, from the coach’s point of view, that he makes shot No. 100?


http://fivethirtyeight.com/features/will-the-neurotic-basket...
And, in case you missed it, the website recommended that people check out our BGG thread about the Car Salesman problem from last week!
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Re: New Friday Math Problem
Another interesting problem! Thanks for posting.
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Re: New Friday Math Problem
Here's my intuitive first thoughts.
Spoiler (click to reveal)
I imagine there will be three main cases.

1) He makes a lot of early shots, which means he'll end up making the majority of subsequent shots. By the 99th shot he's virtually guaranteed to make the shot.

2) He misses a lot of early shots, which means he'll end up missing the majority of subsequent shots. By the 99th shot he's virtually guaranteed to miss the shot.

3) He'll break even on early shots... which I don't think is sustainable over time and will probably degenerate into either making or missing shots, so by the time he gets to the 99th shot, he'll be in a situation similar to 1 or 2.

Since he made the 99th shot, it seems likely that 1 is the case and the chance that he makes the 100th shot are almost perfect.
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Re: New Friday Math Problem
MABBY wrote:
Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that he’s made thus far

Quote:
His coach...saw the first two shots, ... and sees the player make shot No. 99. What is the probability, from the coach’s point of view
so the coach sees three shots but I'm guessing the real answer has to factor in each shot after the first two, but why say from the coach's point of view?

Quote:
And, in case you missed it, the website recommended that people check out our BGG thread about the Car Salesman problem from last week!
awesome!
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Re: New Friday Math Problem
mafman6 wrote:
Quote:
His coach...saw the first two shots, ... and sees the player make shot No. 99. What is the probability, from the coach’s point of view
so the coach sees three shots but I'm guessing the real answer has to factor in each shot after the first two, but why say from the coach's point of view?
Because if you used the player's point of view, he'd know the exact percentage chance as he knows how many he made and how many he missed. The point of the problem is that you have to deduce the answer purely on the basis of the three shots that the coach saw.
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Re: New Friday Math Problem
From the coach's point of view is problem speak for a conditional probability.

With a fairly obvious notation we are being asked for P(n100=1 | n1=1 and n2=0 and n99=1) with n3 to n100 as defined in the problem.

Now just a matter of determining that If I decide to waste time on this one I'll probably start either with a small number of shots then increasing to see the pattern, or a simple simulation (I have a program I can do that with) to get a numerical result, which may show something. Or I may not.
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Re: New Friday Math Problem
My thoughts below:

Spoiler (click to reveal)
First approach:

After the 2nd shot, the coach knows the probability of the next is 50%. This will skew heavily towards one side or the other at the next shot, but since the coach doesn't see it, he must assume the 50% rate for the next 96 shots.

So the coach comes back to see shot 99 made. Assuming the 50% rate on the 96 he did not see, the coach guesses about 48 (.5 * 96) + 2 (shots seen) which gives a probability of 50/99 for shot 100.

Second approach:

M(1) = 1
M(2) = 0

M(3) = .5
M(4|3) = .75
M(4|!3) = .25
M(!4|3) = .25
M(!4|!3) = .75

M(3 AND 4) = M(3) * M(4|3) = .5 * .75 = .375
M(3 AND !4) = M(3) * M(!4|3) = .5 * .25 = .125
M(!3 AND 4) = M(!3) * M(4|!3) = .5 * .25 = .125
M(!3 AND !4) = M(!3) * M(!4|!3) = .5 * .75 = .375

M(4) = M(4 AND 3) + M(4 AND !3) = .375 + .125 = .5

...So my first answer is right? The coach has the probability of each shot he didn't see as 50%.
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Re: New Friday Math Problem
OK, this is "obviously" another Pascal's Triangle based problem. I really will have to get going with Python …
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Re: New Friday Math Problem
Anonymouse512 wrote:
My thoughts below:

Spoiler (click to reveal)
So the coach comes back to see shot 99 made. Assuming the 50% rate on the 96 he did not see, the coach guesses about 48 (.5 * 96) + 2 (shots seen) which gives a probability of 50/99 for shot 100.

Spoiler (click to reveal)
This would be true if the coach didn't witness the 99th shot, but the fact that the player makes the 99th shot gives us information about the past history. (It heavily suggests that he's been more successful than not in the last 96 shots...)
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Re: New Friday Math Problem
Edit: Ignore this. I found a stupid error. Recomputing.
Spoiler (click to reveal)
I did it as a Monte Carlo simulation, repeated 10,000 times. Each time I run that the probability for the 100th shot averages roughly 50.3%. (I see numbers between 50.2% and 50.4%. Not an exact solution, but that gives an idea of what to look for.

Obviously the dependence on previous shots doesn't kick in until after the second shot, which makes the problem slightly less satisfying.
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Re: New Friday Math Problem
Edit: Ignore this. Found a stupid error. Recomputing.
Spoiler (click to reveal)
Interestingly, if you turn it around and say the coach saw him miss the 99th shot, the probability only falls to about 49.65% for the last shot. Further, each time I run 10,000 trials the minimum probability reached by the shooter is about 39%. You'd think--as Bryan points out--you might get a case where he misses a bunch early and the odds go almost to zero.
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Re: New Friday Math Problem
Quote:
Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that he’s made thus far. (His neuroses are very exacting.)

Then he couldn't have missed the second shot, he had 100% chance of making it?

Edit: I still plan to think about it, but this part bothered me with its sloppiness.
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Re: New Friday Math Problem
Like all math problems, this one is impossible to figure out on paper.

So, I went out back and put it to the test and actually performed the experiment.

I made my first shot, thank goodness, and I missed the second shot on purpose in keeping with the test, which was way more difficult than it sounds because I am the type that never gives up.

After that, getting into my own head was easier than expected so I think I nailed the neuroses part.

My percentage after 98 shots was 75%.

I missed shot 99
Spoiler (click to reveal)
which invalidated the entire experiment. Either the setup is wrong or this is a hypothetical problem. Either way, it is unsolvable.
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Re: New Friday Math Problem
qzhdad wrote:
Quote:
Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that he’s made thus far. (His neuroses are very exacting.)

Then he couldn't have missed the second shot, he had 100% chance of making it?

Edit: I still plan to think about it, but this part bothered me with its sloppiness.

I guess he only gets into his own head after the second shot... I'm just using P(I1=1, I2=0) = 1 in all of my calculations as the base case.

Here's my start:
Spoiler (click to reveal)
Define I_k to be the indicator that he made the kth shot.

We want
P(I_100=1 | I_1=1, I_2=0, I_99=1),
which is
P(I_1=1, I_2=0, I_99=1, I_100=1)/P(I_1=1, I_2=0, I_99=1).

But the denominator is just 1/2 (since the distribution of his "situation" after the 98th shot is symmetric given the 1,0 start to the sequence, the expected position after the 98th shot is 49/98=1/2).

So we want to calculate
2*P(I_1=1, I_2=0, I_99=1, I_100=1),
which is just
2*P(I_99=1, I_100=1 | I_1=1, I_2=0)*P(I_1=1, I_2=0)
which simplifies to
2*P(I_99=1, I_100=1 | I_1=1, I_2=0)
(Since the first two shots are "fixed" at 1,0)

The tricky part about this is the asymmetry: making two shots in a row when he's in a shot deficit does not exactly balance out two-in-a-row when he's at a surplus. (Since the first success improves his odds in all cases). However, it should be enumerable.
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Re: New Friday Math Problem
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Now I think I've got it. I'm running a Monte Carlo simulation with 10,000 trials. Each time the chances averaged over 10,000 trials of him making the 100th shot end up being just about exactly 2/3.

About 1% of the time he doesn't make another shot after the first one.

If the coach had seen him miss the 99th shot, his odds of making the 100th would be about 1/3.
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Re: New Friday Math Problem
Here are my thoughts:

Spoiler (click to reveal)

Shot 1: Made

Shot 2: Miss

Shot 3: 50% chance

Shot 4: If he made shot 3, there is a 2/rds chance of making it, and if he missed there is a 1/3rd chance of making it. Splitting evenly as shot 3 is a 50% shot, we have .5(1/3) + .5(2/3), which is 50%.

Shots 5-98: Since shot 4 was 50%, and shot 3 was 50%, we're just going to assume this pattern holds up 'cause I'm lazy.

Shot 99: Made

Shot 100: Coach assumes he made 50% of his shots from 3 to 98, which is 96/2 shots = 48. 48 plus the two he made is 50. He's attempted 99 shots, so 50/99, or a 50.51% chance he makes the 100th shot.
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Re: New Friday Math Problem
I created a simulation to test it out.

Spoiler (click to reveal)
First, I ran a normal simulation 1000 times where in each run the player takes 100 shots just to see the normal situation. Running that simulation it appears that the odds of him making the next shot approach 50% on average:

Simulation 985: Odds are 0.7171717171717171, Average odds are: 0.5013997846485158
Simulation 986: Odds are 0.32323232323232326, Average odds are: 0.5012190874259842
Simulation 987: Odds are 0.8585858585858586, Average odds are: 0.5015811611556295
Simulation 988: Odds are 0.45454545454545453, Average odds are: 0.5015335541651333
Simulation 989: Odds are 0.5555555555555556, Average odds are: 0.5015881770179043
Simulation 990: Odds are 0.5656565656565656, Average odds are: 0.5016528925619838
Simulation 991: Odds are 0.30303030303030304, Average odds are: 0.5014524661346057
Simulation 992: Odds are 0.9090909090909091, Average odds are: 0.50186339198436
Simulation 993: Odds are 0.21212121212121213, Average odds are: 0.5015716073117888
Simulation 994: Odds are 0.1717171717171717, Average odds are: 0.5012397618031423
Simulation 995: Odds are 0.9292929292929293, Average odds are: 0.5016699659915743
Simulation 996: Odds are 0.1919191919191919, Average odds are: 0.5013589712384896
Simulation 997: Odds are 0.42424242424242425, Average odds are: 0.5012816226457152
Simulation 998: Odds are 0.020202020202020204, Average odds are: 0.5007995789558919
Simulation 999: Odds are 0.020202020202020204, Average odds are: 0.5003185003185007
Simulation 1000: Odds are 0.8282828282828283, Average odds are: 0.500646464646465


I then went and tweaked the program so that it only took into account cases where the 99th shot was successfully made. Interestingly, the average odds for the 100th shot rose to roughly 66% (2/3):

Simulation 963: Odds are 0.8282828282828283, Average odds are: 0.6637354758440349
Simulation 968: Odds are 0.6161616161616161, Average odds are: 0.6636363636363632
Simulation 970: Odds are 0.6565656565656566, Average odds are: 0.6636216636216632
Simulation 972: Odds are 0.7070707070707071, Average odds are: 0.6637118068653334
Simulation 973: Odds are 0.8585858585858586, Average odds are: 0.6641152728109245
Simulation 975: Odds are 0.8383838383838383, Average odds are: 0.6644753318307033
Simulation 976: Odds are 0.7676767676767676, Average odds are: 0.6646881182963653
Simulation 977: Odds are 0.797979797979798, Average odds are: 0.6649623810117633
Simulation 978: Odds are 0.98989898989899, Average odds are: 0.6656296019745709
Simulation 981: Odds are 0.98989898989899, Average odds are: 0.6662940884252356
Simulation 983: Odds are 0.5959595959595959, Average odds are: 0.66615025510731
Simulation 985: Odds are 0.9090909090909091, Average odds are: 0.6666460523603378
Simulation 987: Odds are 0.1919191919191919, Average odds are: 0.6656791952107631
Simulation 988: Odds are 0.7676767676767676, Average odds are: 0.6658865073499217
Simulation 991: Odds are 0.5858585858585859, Average odds are: 0.6657241789087629
Simulation 992: Odds are 0.9292929292929293, Average odds are: 0.6662577188892976
Simulation 993: Odds are 0.797979797979798, Average odds are: 0.6665238240995816
Simulation 994: Odds are 0.3434343434343434, Average odds are: 0.6658724340175951
Simulation 995: Odds are 0.5656565656565656, Average odds are: 0.6656707924313557
Simulation 996: Odds are 0.35353535353535354, Average odds are: 0.6650440144416047
Simulation 999: Odds are 0.797979797979798, Average odds are: 0.6653104188174327


edit: Now, I'm not sure why this is the answer. I leave that for you guys to speculate on.
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Re: New Friday Math Problem
My thoughts...

Spoiler (click to reveal)
Consider the situation from the coach's perspective prior to the player making the 99th shot.

He knows the player has made 1 shot and missed 1 shot.
Calculate the probability of a given sequence of makes and misses such that the player makes n shots out of the 96 that the coach didn't see:

This is a product of 96 probabilities:
(# of makes + 1) / (Total # of shots + 2) at the time of each make
and
(# of misses + 1) / (Total # of shots + 2) at the time each miss

If you re-order the numerator and denominator of this product you can show that this gives:

(1 * 2 * 3 * ... * (# of makes ) ) * (1 * 2 * 3 * ... * (# of misses ) ) / (2 * 3 * 4 * ... * 97)

# of makes = n
# of misses = 96 - n

so the probability of this sequence of makes and misses is

n! (96 - n)! / 97!

1 / 97 * 1 / C(96,n)

where C(96,n) is the combination function

This is independent of the order of the misses and makes, i.e. any combination that makes n out of the 96 shots occurs with this probability. There are C(96,n) possible sequences that gets you to n makes each occurring with this probability, so the probability of the player making n out of the 96 shots is: 1 / 97

i.e. From the coach's perspective, before the 99th shot, there is an equal change of him having made 1,2,3,...,98 total shots.

The coach needs to adjust these probabilities after he sees the 99th shot. i.e. he needs to calculate P(n | 99th make)

Using Bayes Theorem:
P(n | 99th make) = P(99th make | n) * P(n) / P(99th make)

where
P(n) = 1/97
P(99th make) = 1/2
P(99th make | n) = (n + 1) / 98

so

P(n | 99th make) = (n + 1) / (97 * 49) = (n + 1) / 4753

We can then calculate the expected make probability as:

sum (n + 2) / 99 * P(n | 99th make) {n = 0 to 96}

1 / 470547 * sum (n+1)*(n+2) {n = 0 to 96}

2 / 3
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Re: New Friday Math Problem
park0523 wrote:
My thoughts...

Spoiler (click to reveal)
...
We can then calculate the expected make probability as:

sum (n + 2) / 99 * P(n | 99th make) {n = 0 to 96}

1 / 470547 * sum (n+1)*(n+2) {n = 0 to 96}

2 / 3

Well, this lines up with the simulation that I ran above...
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Re: New Friday Math Problem
Spoiler (click to reveal)
And about 1% of the time he ends up making every shot after the 2nd one.
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Re: New Friday Math Problem
Spoiler (click to reveal)

Two things.

First, with 4 or 5 shots, working the probabilities in my head I got exactly the same answer, 2/3. I'd need some paper to try more cases, but I don't think that would be enlightening.

Second, I tried ten million results (no half measures here. I've got some software I can do this with and ten million results - or rather five million as I throw half away - took just over a minute). The answer is clearly very close to 2/3, in fact my 95% confidence interval is about .6665 to .6673.

So I'm going to be very surprised if the answer isn't 2/3. It's just proving it.
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Re: New Friday Math Problem
I think the key to the problem is

Spoiler (click to reveal)

because of the particular structure of the problem, the probability of making n out of the 96 shots that the coach didn't see is exactly 1/97 regardless of what n is.

i.e. the probability of a given sequence with n makes scales exactly the same way as the number of sequences with n makes.

For example, there is only one sequence where he makes all 96 shots, and that occurs with probability 1/97. Any given sequence where he makes 95 out of 96 is far less likely, 1 / (96 * 97), but there are 96 ways that he can do it so the probability of making 95 out of 96 is exactly the same.
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Re: New Friday Math Problem
park0523 wrote:
I think the key to the problem is

Spoiler (click to reveal)

because of the particular structure of the problem, the probability of making n out of the 96 shots that the coach didn't see is exactly 1/97 regardless of what n is.

i.e. the probability of a given sequence with n makes scales exactly the same way as the number of sequences with n makes.

For example, there is only one sequence where he makes all 96 shots, and that occurs with probability 1/97. Any given sequence where he makes 95 out of 96 is far less likely, 1 / (96 * 97), but there are 96 ways that he can do it so the probability of making 95 out of 96 is exactly the same.

That's a really good observation that I hadn't spotted. (I've proved it though, in the general case). Not sure if it's the key to a solution yet. (I had an idea, I'm now trying to decide if to pursue it, or try building on this. Or doing something else.)
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Re: New Friday Math Problem

andyholt wrote:
OK, this is "obviously" another Pascal's Triangle based problem. I really will have to get going with Python …

OK, so it isn't. Matt has clearly got the correct answer with a mathematically purer approach than I have in mind.

Spoiler (click to reveal)

Once you realise that the solution cannot depend on the individual sequence of intermediate throws, essentially all the information the coach has is two hits and one miss - thus as far as the coach is concerned the probability has to be 2/3 (or, more generally, if the coach as seen h hits and m misses the probability is h/(h+m) - the number of unobserved trials is irrelevant)
The player may have (almost certainly does have) a different estimate.
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Re: New Friday Math Problem
Matt's insight leads directly to a solution. Write down the conditional probability you want, apply Bayes Theorem, sum over number of hits, use Matt's result, and it drops out. To the answer everyone thought.

However unfortunately my proof is various scribbles literally on the back of an envelope. And I lack the motivation to try to reproduce here (including proving Matt's result).

So, less time wasted this week.
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