Math Puzzles (from FiveThirtyEight - new puzzle every Friday) (2 Viewers)

[the moops]

2/3

 

[Ignoramus]

50% You either make it or you don't

Yep. You guys are way over thinking this.

 

[Sweet J]

I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.

And the only reason I haven't shown my work is that I have no idea how to approach this.

 

[Psychopav]

2/3 from the coache's perspective, since he's seen the player miss 1 and make 2 of his previous shots.

 

[the moops]

2/3 from the coache's perspective, since he's seen the player miss 1 and make 2 of his previous shots.

Exactly

The coach leaves the gym, and you just know the dude sat on his ### and did nothing,. No ####### way he is gonna shoot a hundred free throws if nobody is watching. Coach comes back in, he hurries to the line, makes the free throw, which is actually only his third shot taken.

So, he went, make, miss, make.

2/3

 

[Dickies]

I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.

And the only reason I haven't shown my work is that I have no idea how to approach this.

Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.

I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.

 

[IC FBGCav]

2/3 from the coache's perspective, since he's seen the player miss 1 and make 2 of his previous shots.

Exactly

The coach leaves the gym, and you just know the dude sat on his ### and did nothing,. No ####### way he is gonna shoot a hundred free throws if nobody is watching. Coach comes back in, he hurries to the line, makes the free throw, which is actually only his third shot taken.

So, he went, make, miss, make.

2/3

He knows the 3rd shot was 50/50.

 

[IC FBGCav]

.625

 

[Dinsy Ejotuz]

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 isnt good news. Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that hes 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 doesnt see the next 96 shots. The coach returns, and sees the player make shot No. 99. What is the probability, from the coachs point of view, that he makes shot No. 100?

From here: http://fivethirtyeight.com/features/will-the-neurotic-basketball-player-make-his-next-free-throw/

Without doing a lot of thinking...

50/99? (.505)

ETA: as written if he makes the first shot his future % would be 100%.

 

[belljr]

What is the shooters lifetime free throw shooting %?

 

[Hoosier16]

2/3

 

[the moops]

2/3 from the coache's perspective, since he's seen the player miss 1 and make 2 of his previous shots.

ExactlyThe coach leaves the gym, and you just know the dude sat on his ### and did nothing,. No ####### way he is gonna shoot a hundred free throws if nobody is watching. Coach comes back in, he hurries to the line, makes the free throw, which is actually only his third shot taken.

So, he went, make, miss, make.

2/3

He knows the 3rd shot was 50/50.

Yea, and it went in. So his 4th shot had a 2/3 chance.

 

[IC FBGCav]

What is the shooters lifetime free throw shooting %?

On or off the court?

 

[skol asylum]

48

16

11

 

[rick6668]

Having a hard time writing it out, but I am thinking if we try it for the 3rd or 4th or 5th shot we get the following probabilities.

Makes 3rd, probablity for 4th is 2/3

Makes 4th, probablity for 5th is 5/8

Makes 5th, probablity for 6th is 6/10

Makes 6th, probablity for 7th is 7/12

This follows as (n+1)/2n

Edited, he makes the 99th not 98th, so it should be

(99+1)/198 = 100/198 = 50/99

 

[tommyboy]

2/3 from the coache's perspective, since he's seen the player miss 1 and make 2 of his previous shots.

this is correct.

it would be an interesting question if you were told upfront that Player is 50% free throw shooter. In that scenario what would the odds be he makes the final shot after coach walks back in.

 

[JaxBill]

48

16

11

OMA-HA!

 

[tonydead]

Anyone who's trying to take into account that it's the 99th shot versus any other number is over thinking it. Although you might have to do some math to convince yourself otherwise.

2/3

 

[IC FBGCav]

If you look at the answer explanation to last week question no way 2/3's will be correct.

 

[tommyboy]

get Shick! in here, he'll tell us definitely

 

[Sweet J]

I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.

And the only reason I haven't shown my work is that I have no idea how to approach this.

Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.

I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.

All of you who are saying 50/99 need to show your work.

I think that they may be right. But I would approach it like this:

There are three ways (and only three ways) that I see the free throw shooting playing out. Let's say, instead of 100 free throws, its 1,000 or even 10,000. Either:

1. Early misses come quickly, and the number eventually approaches 0%. Three or four misses in a row, and the shooter is suddenly 1-for-5, and the 20% number is too hard to recover from, so even more misses come . . . and more. So that an occasional hit can't salvage it, and we will never have 4 or 5 hits in a row when there is a 20% chance to hit. So: The number will approach 0% (but never reach it of course).

2. Same reasoning if the early shots made come. The number approaches 100% (but never reach it, of course).

As far as I see it, there is equal chance of either 1 or 2 coming true (although I couldn't tell you what that would be). But then, maybe we have this as a possibility:

3. The results aren't dominated by streaks of hits OR misses. (hit, miss, hit, hit, miss, miss, hit, miss, hit, miss, etc.). Before long, the shooter has taken 10 or 15 or even 20 shots, and the percentage is very close to 50%. Even something like 11-for-20 or 9-for-20 isn't enough to keep the percentages too far away from 50%. So no matter if the shooter goes on a mini-streak or mini-fail, that won't be enough to keep him from righting himself. So . . . generally a 50/50 over time (and the longer shooter goes about 50/50, the greater a chance it STAYS at 50/50.

So. . . . If 1 and 2 are equal probability (whatever that is), they cancel eachother out. That leaves #3, which is 50/50.

So, I'd say that by the time the guy gets to 98 shots, he's most likely at 49/98. If he makes #99, that puts him at 50/99. So that's the likelihood that he hits number 100.

Anyway, that's my reasoning.

 

[Pigskin Fanatic]

I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.

And the only reason I haven't shown my work is that I have no idea how to approach this.

Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.

I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.

All of you who are saying 50/99 need to show your work.

thought i already did...

can't be 50-50 from the op

Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that hes made thus far.

there have been 99 shots attempted, first and last were made, 50/99 is a good assumption. the third shot probability was 50-50 and that was the only shot that had a 50-50 chance by this premise. From there forward every other shot will be either one above 50% or one below. i'd go with 50/99 and slightly more likely he makes it than miss.

 

[Sweet J]

Having a hard time writing it out, but I am thinking if we try it for the 3rd or 4th or 5th shot we get the following probabilities.

Makes 3rd, probablity for 4th is 2/3

Makes 4th, probablity for 5th is 5/8

Makes 5th, probablity for 6th is 6/10

Makes 6th, probablity for 7th is 7/12

This follows as (n+1)/2n

Edited, he makes the 99th not 98th, so it should be

(99+1)/198 = 100/198 = 50/99

I don't understand your underlying logic. I'm assuming when you say "makes 4th, makes 5th, makes 6th," you mean no misses? Otherwise you would have indicated such?

If he makes the 4th, that means he's made 3 out of 4 shots. Isn't the probability for the 5th 3/4 or 75%?

If he makes the 5th, that means he's made 4 of 5 shots (again, assuming only the one miss). Probability for 6th is 4/5 or 80%.

Makes the 6th (again, assuming no misses), probability for 7th is 5/6.

I have no idea where you got your other numbers.

 

[rick6668]

Having a hard time writing it out, but I am thinking if we try it for the 3rd or 4th or 5th shot we get the following probabilities.

Makes 3rd, probablity for 4th is 2/3

Makes 4th, probablity for 5th is 5/8

Makes 5th, probablity for 6th is 6/10

Makes 6th, probablity for 7th is 7/12

This follows as (n+1)/2n

Edited, he makes the 99th not 98th, so it should be

(99+1)/198 = 100/198 = 50/99

Let's assume the coach just watches the third shot and he makes it. He's now 2/3 and the odds he makes the 4th is 2/3.

Now let's assume the coach walks out and returns for the 4th shot.

Now the player makes the third shot 1/2 the time to now be 2/3 and 1/2 the time misses to be 1/3.

He then makes the 4th. 1/2 the time he was 2/3 and 1/2 the time was 1/3 so now 1/2 the time he's 3/4 and 1/2 the time he's 2/4. 3/8 + 2/8 = 5/8

Now let's assume the coach walks out and returns for the 5th shot.

From above the player makes the third shot 1/2 the time to now be 2/3 and 1/2 the time misses to be 1/3.

Now for the 4th.

1/2 the time he was 2/3 (1/2 of this he goes 3/4 and 1/2 he goes 2/4)

The other 1/2 the time he was 1/3 (1/2 of this he goes 2/4 and the other half 1/4)

So we have

1/4 time he's 3/4

1/4 time he's 2/4

1/4 time he's 2/4

1/4 time he's 1/4

He makes the 5th, now-

1/4 time he's 4/5

1/4 time he's 3/5

1/4 time he's 3/5

1/4 time he's 2/5

=12/20 = 6/10

Extrapolating this out, I get 7/12 for the 7th if he makes the 6th.

So (n+1)/2n for n=99, 100/198 = 50/99

 

[Sweet J]

I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.

And the only reason I haven't shown my work is that I have no idea how to approach this.

Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.

I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.

All of you who are saying 50/99 need to show your work.

thought i already did...

can't be 50-50 from the op

Specifically, the probability he hits any subsequent shot is equal to the overall percentage of shots that hes made thus far.

there have been 99 shots attempted, first and last were made, 50/99 is a good assumption. the third shot probability was 50-50 and that was the only shot that had a 50-50 chance by this premise. From there forward every other shot will be either one above 50% or one below. i'd go with 50/99 and slightly more likely he makes it than miss.

I guess there is a little something missing for me. Sure, the third shot probability was 50-50, but when he makes or misses, he probability soars to 66/33. One more in a row, and we are zooming forward with 75/25. Don't you have to account for a few early misses/makes at the beginning to drive the number to 0% or 100%? (i.e., although 3-for-4 is technically "one above 50%" it is still a hard-to-recover from 66.6%. I don't know how often we'd get to 50%, but I think the 0% and 100% numbers are not insignificant.

 

[Drunken Cowboy]

I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.

And the only reason I haven't shown my work is that I have no idea how to approach this.

Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.

I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.

All of you who are saying 50/99 need to show your work.

I think that they may be right. But I would approach it like this:

There are three ways (and only three ways) that I see the free throw shooting playing out. Let's say, instead of 100 free throws, its 1,000 or even 10,000. Either:

1. Early misses come quickly, and the number eventually approaches 0%. Three or four misses in a row, and the shooter is suddenly 1-for-5, and the 20% number is too hard to recover from, so even more misses come . . . and more. So that an occasional hit can't salvage it, and we will never have 4 or 5 hits in a row when there is a 20% chance to hit. So: The number will approach 0% (but never reach it of course).

2. Same reasoning if the early shots made come. The number approaches 100% (but never reach it, of course).

As far as I see it, there is equal chance of either 1 or 2 coming true (although I couldn't tell you what that would be). But then, maybe we have this as a possibility:

3. The results aren't dominated by streaks of hits OR misses. (hit, miss, hit, hit, miss, miss, hit, miss, hit, miss, etc.). Before long, the shooter has taken 10 or 15 or even 20 shots, and the percentage is very close to 50%. Even something like 11-for-20 or 9-for-20 isn't enough to keep the percentages too far away from 50%. So no matter if the shooter goes on a mini-streak or mini-fail, that won't be enough to keep him from righting himself. So . . . generally a 50/50 over time (and the longer shooter goes about 50/50, the greater a chance it STAYS at 50/50.

So. . . . If 1 and 2 are equal probability (whatever that is), they cancel eachother out. That leaves #3, which is 50/50.

So, I'd say that by the time the guy gets to 98 shots, he's most likely at 49/98. If he makes #99, that puts him at 50/99. So that's the likelihood that he hits number 100.

Anyway, that's my reasoning.

I think #1 is where the info about the 99th shot comes into play. If he is at a near zero chance of hitting the shot, There is almost no way he could have hit shot 99. Therefore this isn't a likely choice.

 

[rick6668]

Having a hard time writing it out, but I am thinking if we try it for the 3rd or 4th or 5th shot we get the following probabilities.

Makes 3rd, probablity for 4th is 2/3

Makes 4th, probablity for 5th is 5/8

Makes 5th, probablity for 6th is 6/10

Makes 6th, probablity for 7th is 7/12

This follows as (n+1)/2n

Edited, he makes the 99th not 98th, so it should be

(99+1)/198 = 100/198 = 50/99

I don't understand your underlying logic. I'm assuming when you say "makes 4th, makes 5th, makes 6th," you mean no misses? Otherwise you would have indicated such?

If he makes the 4th, that means he's made 3 out of 4 shots. Isn't the probability for the 5th 3/4 or 75%?

If he makes the 5th, that means he's made 4 of 5 shots (again, assuming only the one miss). Probability for 6th is 4/5 or 80%.

Makes the 6th (again, assuming no misses), probability for 7th is 5/6.

I have no idea where you got your other numbers.

No, see above. I had a hard time writing it out. See if the above makes sense. It was much easier to visualize using a probability tree, but too hard to draw here.

 

[Pigskin Fanatic]

I guess there is a little something missing for me. Sure, the third shot probability was 50-50, but when he makes or misses, he probability soars to 66/33. One more in a row, and we are zooming forward with 75/25. Don't you have to account for a few early misses/makes at the beginning to drive the number to 0% or 100%? (i.e., although 3-for-4 is technically "one above 50%" it is still a hard-to-recover from 66.6%. I don't know how often we'd get to 50%, but I think the 0% and 100% numbers are not insignificant.

right, so from shot 3 forward the % should fluctuate between 1/3 (33%) and 2/3 (66%) with the number progressing to 50% after each shot but never going below 33% and above 66%. A graph would have an arc getting smaller and smaller to a flat line. On top of that each shot really has 50-50 chance of going in, the fact that each shot depends on the previous ones is really just a numbers game and only AFTER the 100th shot, so essentially the 101st shot, can be a mathematical 50-50 proposition.

 

[Dickies]

I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.

And the only reason I haven't shown my work is that I have no idea how to approach this.

Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.

I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.

All of you who are saying 50/99 need to show your work.

I think that they may be right. But I would approach it like this:

There are three ways (and only three ways) that I see the free throw shooting playing out. Let's say, instead of 100 free throws, its 1,000 or even 10,000. Either:

1. Early misses come quickly, and the number eventually approaches 0%. Three or four misses in a row, and the shooter is suddenly 1-for-5, and the 20% number is too hard to recover from, so even more misses come . . . and more. So that an occasional hit can't salvage it, and we will never have 4 or 5 hits in a row when there is a 20% chance to hit. So: The number will approach 0% (but never reach it of course).

2. Same reasoning if the early shots made come. The number approaches 100% (but never reach it, of course).

As far as I see it, there is equal chance of either 1 or 2 coming true (although I couldn't tell you what that would be). But then, maybe we have this as a possibility:

3. The results aren't dominated by streaks of hits OR misses. (hit, miss, hit, hit, miss, miss, hit, miss, hit, miss, etc.). Before long, the shooter has taken 10 or 15 or even 20 shots, and the percentage is very close to 50%. Even something like 11-for-20 or 9-for-20 isn't enough to keep the percentages too far away from 50%. So no matter if the shooter goes on a mini-streak or mini-fail, that won't be enough to keep him from righting himself. So . . . generally a 50/50 over time (and the longer shooter goes about 50/50, the greater a chance it STAYS at 50/50.

So. . . . If 1 and 2 are equal probability (whatever that is), they cancel eachother out. That leaves #3, which is 50/50.

So, I'd say that by the time the guy gets to 98 shots, he's most likely at 49/98. If he makes #99, that puts him at 50/99. So that's the likelihood that he hits number 100.

Anyway, that's my reasoning.

I think you are wrong here. Given that we know he hit shot 99, the number of scenarios where he hits the 99th shot will be particularly skewed to the scenarios where he hits shots in the beginning. You cannot simply cancel out the two extreme paths.

 

[Sweet J]

I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.

And the only reason I haven't shown my work is that I have no idea how to approach this.

Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.

I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.

All of you who are saying 50/99 need to show your work.

I think that they may be right. But I would approach it like this:

There are three ways (and only three ways) that I see the free throw shooting playing out. Let's say, instead of 100 free throws, its 1,000 or even 10,000. Either:

1. Early misses come quickly, and the number eventually approaches 0%. Three or four misses in a row, and the shooter is suddenly 1-for-5, and the 20% number is too hard to recover from, so even more misses come . . . and more. So that an occasional hit can't salvage it, and we will never have 4 or 5 hits in a row when there is a 20% chance to hit. So: The number will approach 0% (but never reach it of course).

2. Same reasoning if the early shots made come. The number approaches 100% (but never reach it, of course).

As far as I see it, there is equal chance of either 1 or 2 coming true (although I couldn't tell you what that would be). But then, maybe we have this as a possibility:

3. The results aren't dominated by streaks of hits OR misses. (hit, miss, hit, hit, miss, miss, hit, miss, hit, miss, etc.). Before long, the shooter has taken 10 or 15 or even 20 shots, and the percentage is very close to 50%. Even something like 11-for-20 or 9-for-20 isn't enough to keep the percentages too far away from 50%. So no matter if the shooter goes on a mini-streak or mini-fail, that won't be enough to keep him from righting himself. So . . . generally a 50/50 over time (and the longer shooter goes about 50/50, the greater a chance it STAYS at 50/50.

So. . . . If 1 and 2 are equal probability (whatever that is), they cancel eachother out. That leaves #3, which is 50/50.

So, I'd say that by the time the guy gets to 98 shots, he's most likely at 49/98. If he makes #99, that puts him at 50/99. So that's the likelihood that he hits number 100.

Anyway, that's my reasoning.

I think #1 is where the info about the 99th shot comes into play. If he is at a near zero chance of hitting the shot, There is almost no way he could have hit shot 99. Therefore this isn't a likely choice.

Wow, then that changes things, doesn't it.

In my original thought, I had three possibilities:

#1

#2

#3

So if he hits, then maybe we can't discount #1, right? But we can discount #2? So would the probabilities for #1 and #3 be added?

So, for example, if #1 and #2 each happened 10% of the time, and #3 happened 80% of the time, then . . . .

Forget it, my head hurts.

This hurts my head.

 

[Sweet J]

Having a hard time writing it out, but I am thinking if we try it for the 3rd or 4th or 5th shot we get the following probabilities.

Makes 3rd, probablity for 4th is 2/3

Makes 4th, probablity for 5th is 5/8

Makes 5th, probablity for 6th is 6/10

Makes 6th, probablity for 7th is 7/12

This follows as (n+1)/2n

Edited, he makes the 99th not 98th, so it should be

(99+1)/198 = 100/198 = 50/99

I don't understand your underlying logic. I'm assuming when you say "makes 4th, makes 5th, makes 6th," you mean no misses? Otherwise you would have indicated such?

If he makes the 4th, that means he's made 3 out of 4 shots. Isn't the probability for the 5th 3/4 or 75%?

If he makes the 5th, that means he's made 4 of 5 shots (again, assuming only the one miss). Probability for 6th is 4/5 or 80%.

Makes the 6th (again, assuming no misses), probability for 7th is 5/6.

I have no idea where you got your other numbers.

No, see above. I had a hard time writing it out. See if the above makes sense. It was much easier to visualize using a probability tree, but too hard to draw here.

Right. Got it. I don't follow it necessarily, but I understand the reasoning behind your numbers.

 

[Sweet J]

I'm curious to hear the OP's reasoning. I've yet to hear from anybody's reasoning that I understand. Sure, anyone can throw out a number, like 50%. Maybe it's right, but show your work, man.

And the only reason I haven't shown my work is that I have no idea how to approach this.

Yea, I hear you. I am thinking it over, but also don't really know how to approach it in a formula, but if I had some time I am confident I could map it out by hand. The tricky part is that the 3rd free throw creates a drastically different set of results depending on if he makes or misses the shot.

I am tending to think if we mapped it out in tree form, all 99 free throws that details the probabilities of each shot being made/missed then at the end we would have all of the probabilities of what shot 100 could potentially be. We would know that shot 99 was made, so we could cross off all the probable outcomes that result from him missing the shot and average the remaining probabilities. I am too far removed from college math classes and creating a formula for this is beyond my current capabilities.

All of you who are saying 50/99 need to show your work.

I think that they may be right. But I would approach it like this:

There are three ways (and only three ways) that I see the free throw shooting playing out. Let's say, instead of 100 free throws, its 1,000 or even 10,000. Either:

1. Early misses come quickly, and the number eventually approaches 0%. Three or four misses in a row, and the shooter is suddenly 1-for-5, and the 20% number is too hard to recover from, so even more misses come . . . and more. So that an occasional hit can't salvage it, and we will never have 4 or 5 hits in a row when there is a 20% chance to hit. So: The number will approach 0% (but never reach it of course).

2. Same reasoning if the early shots made come. The number approaches 100% (but never reach it, of course).

As far as I see it, there is equal chance of either 1 or 2 coming true (although I couldn't tell you what that would be). But then, maybe we have this as a possibility:

3. The results aren't dominated by streaks of hits OR misses. (hit, miss, hit, hit, miss, miss, hit, miss, hit, miss, etc.). Before long, the shooter has taken 10 or 15 or even 20 shots, and the percentage is very close to 50%. Even something like 11-for-20 or 9-for-20 isn't enough to keep the percentages too far away from 50%. So no matter if the shooter goes on a mini-streak or mini-fail, that won't be enough to keep him from righting himself. So . . . generally a 50/50 over time (and the longer shooter goes about 50/50, the greater a chance it STAYS at 50/50.

So. . . . If 1 and 2 are equal probability (whatever that is), they cancel eachother out. That leaves #3, which is 50/50.

So, I'd say that by the time the guy gets to 98 shots, he's most likely at 49/98. If he makes #99, that puts him at 50/99. So that's the likelihood that he hits number 100.

Anyway, that's my reasoning.

I think you are wrong here. Given that we know he hit shot 99, the number of scenarios where he hits the 99th shot will be particularly skewed to the scenarios where he hits shots in the beginning. You cannot simply cancel out the two extreme paths.

EXACTLY!! Doesn't that mean we have to account for the one extreme path? (i.e., hitting 10 in a row to start and driving the number to 100%?). So, if you account for that small percentage of times, doesn't that make the number slightly higher than 50%? Or maybe not, I don't know. It's all confusing to me.

 

[Sweet J]

This is why I'm not allowed to post in logic threads.

 

[BroadwayG]

What is the highest level math course the coach has completed?

 

[Sweet J]

What is the highest level math course the coach has completed?

Seventeen.

 

[Ignoratio Elenchi]

So heres my solution. For the sake of brevity, and because it's kinda hard to format all the math stuff nicely, I'm not showing all the work but I encourage you to fill in the gaps if you're not convinced that the math works out. For the record, the solution won't be posted on fivethirtyeight until Friday so I'm not 100% sure this is correct (but I'm 99% sure ).

We want to calculate the probability of the 100th shot. Another way of framing this question is, what is the expected number of shots made in the first 99?

Let s be the number of shots taken (in this case its 99, but it could be anything).

Let m be the number of these shots that are made. This can be anywhere from 2 to s-1 (since we know he made at least two and missed at least one).

OK, so:

For any possible value of m, there are (s-3 choose m-2) possible orderings of the shots (we know what three of the shots are, and that two of them were made, that's where the -3 and -2 come from).

For each of these possible orderings, the probability of that particular ordering occurring is the product of the probability of each individual shot in the sequence, which is, quite simply,

(m-1)! (s-m-1)! / (s-1)!

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)!

A little algebra and this reduces easily to:

Sum from m=2 to s-1 of: m * (m-1)

divided by

Sum from m=2 to s-1 of: (m-1)

The numerator there turns out to be s(s-1)(s-2)/3, and the denominator is (s-1)(s-2)/2.

Dividing these shows that the expected number of makes after s shots is 2s/3. Therefore the probability of making the (s+1)th shot is 2/3. Note that it doesnt matter how many shots the coach missed, whether he walks back in to see the 9th, or 99th, or 99999th shot being made, the probability of the next one being made is 2/3.

 

[the moops]

MOP chime in yet?

 

[Sweet J]

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.

 

[the moops]

This has a Monty Hall vibe, IMO

If the coach saw him make the first, miss the second, then make the third (instead of the 99th),the coach would assume he had a 2/3 chance of making the 4th,right?

 

[Uruk-Hai]

This has a Monty Hall vibe, IMO

If the coach saw him make the first, miss the second, then make the third (instead of the 99th),the coach would assume he had a 2/3 chance of making the 4th,right?

That's what I'm getting from what I think the wording in the OP is saying.

 

[Officer Pete Malloy]

Too many variables IMO. The player is human.

 

[tonydead]

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.

Link

 

[Ignoratio Elenchi]

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.

I think the interesting thing about this puzzle is that a simple, first take is, "The coach only saw three shots, and he made two of them, so from the coach's perspective it's 2/3." And then I thought, "Well that's way too simple, it obviously can't be that easy." And at one point I hit on the 50/99 answer that some others posted above, but that wasn't sitting right with me. And I kept plugging away until I worked it all out and was honestly surprised to find that the answer actually was just 2/3 (or more generally, the answer really does rely on just the shots the coach saw and not at all on what happened when he left the room).

To help visualize the problem, let's consider a much simpler example. Imagine the problem is exactly the same, except instead of walking in to see the 99th shot, he walks in to see the 5th shot. We know he made the 1st and 5th, and missed the 2nd, so there are only four possible sequences that could've possibly happened (let 1 represent a make and 0 represent a miss):

10001 (2 total makes)

10011 (3 total makes)

10101 (3 total makes)

10111 (4 total makes)

* This corresponds to the first part of my solution above - for each possible number of made shots, there are (s-3 choose m-2) ways of it occurring. Here, s = 5, so the outcomes are (2 choose 0) = 1, (2 choose 1) = 2 and (2 choose 2) = 1.

What is the probability of each of these four sequences occurring? Consider the first sequence 10001. The probability that he would've missed the third shot is 1/2. The probability that he then would've missed the 4th shot is 2/3. And the probability that he then would've made the 5th shot is 1/4. Combine them all to get 1/2 * 2/3 * 1/4 = 1/12.

Now consider the second sequence 10011. The probabilities for the 3rd, 4th and 5th shots respectively are 1/2, 1/3, and 2/4. 1/2 * 1/3 * 2/4 = 1/12.

Now consider the third sequence 10101. The probabilities are 1/2, 1/3, and 2/4. 1/2 * 1/3 * 2/4 = 1/12.

Finally the fourth sequence 10111. The probabilities are 1/2, 2/3, and 3/4. 1/2 * 2/3 * 3/4 = 1/4.

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.

So back to our 5-shot example...

There's only one way of making exactly 2 out of 5 shots, with probability 1/12.

There are two ways of making exactly 3 out of 5 shots, with probability 1/12 each (so total probability 2/12).

There's only one way of making exactly 4 out of 5 shots, with probability 1/4.

So the expected number of made shots is (2 * 1/12) + (3 * 2/12) + (4 * 1/4) / (1/12 + 2/12 + 1/4)

Which equals 3.333. So the expected probability of making the 6th shot is 3.333 / 5 = 2/3.

 

[rick6668]

So heres my solution. For the sake of brevity, and because it's kinda hard to format all the math stuff nicely, I'm not showing all the work but I encourage you to fill in the gaps if you're not convinced that the math works out. For the record, the solution won't be posted on fivethirtyeight until Friday so I'm not 100% sure this is correct (but I'm 99% sure ).

We want to calculate the probability of the 100th shot. Another way of framing this question is, what is the expected number of shots made in the first 99?

Let s be the number of shots taken (in this case its 99, but it could be anything).

Let m be the number of these shots that are made. This can be anywhere from 2 to s-1 (since we know he made at least two and missed at least one).

OK, so:

For any possible value of m, there are (s-3 choose m-2) possible orderings of the shots (we know what three of the shots are, and that two of them were made, that's where the -3 and -2 come from).

For each of these possible orderings, the probability of that particular ordering occurring is the product of the probability of each individual shot in the sequence, which is, quite simply,

(m-1)! (s-m-1)! / (s-1)!

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)!

A little algebra and this reduces easily to:

Sum from m=2 to s-1 of: m * (m-1)

divided by

Sum from m=2 to s-1 of: (m-1)

The numerator there turns out to be s(s-1)(s-2)/3, and the denominator is (s-1)(s-2)/2.

Dividing these shows that the expected number of makes after s shots is 2s/3. Therefore the probability of making the (s+1)th shot is 2/3. Note that it doesnt matter how many shots the coach missed, whether he walks back in to see the 9th, or 99th, or 99999th shot being made, the probability of the next one being made is 2/3.

This is likely more correct than my answer. I'm liking missing something in my simplistic approach. Don't have time to digest now.

 

[Ignoratio Elenchi]

This has a Monty Hall vibe, IMO

If the coach saw him make the first, miss the second, then make the third (instead of the 99th),the coach would assume he had a 2/3 chance of making the 4th,right?

Yes. After the first two shots, the probability of making the next shot is always (number made so far)/(number taken so far). In your case, the coach sees all of the shots so he wouldn't have to assume anything, he'd know that the probability was 2/3.

 

[Angry Beavers]

100% - for him to get that deep into this. I would think he only missed shot #2.

 

[jon_mx]

If this guy ever gets to the point where he can make his first two shots, he will have an awesome career.

 

[Leroy Hoard]

If this guy ever gets to the point where he can make his first two shots, he will have an awesome career.

Right now I'm thinking he is either Shaq or Andre Drummond.

 

[Sweet J]

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."

 

[bostonfred]

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.