[Prof. Templeton]

Mrs. Abernathy and Mrs. Beaumont live across the street from each other and like to look out their front windows to see what each other are up to. Just to make sure nothing untoward is happening, one would presume. Mrs. Abernathy will look out her window six random times in one day at equal durations for a total of three hours per day, while Mrs. Beaumont will look out six times a day in equal durations for a total of two hours per day.

What is the chance that they will look out their windows and see the other looking back?

If the total time spent looking out their windows stays the same, but the number of times doubles for both, does the chance of them seeing each other change? To what?

[bonanova]

Mrs. Abernathy and Mrs. Beaumont live across the street from each other and like to look out their front windows to see what each other are up to. Just to make sure nothing untoward is happening, one would presume. Mrs. Abernathy will look out her window six random times in one day at equal durations for a total of three hours per day, while Mrs. Beaumont will look out six times a day in equal durations for a total of two hours per day.

What is the chance that they will look out their windows and see the other looking back?

If the total time spent looking out their windows stays the same, but the number of times doubles for both, does the chance of them seeing each other change? To what?

Assuming they do this over a 24-hour day, A is at her window 1/8 of the time; B 1/12 of the time.

The chance that at any given moment they are both there seems to be 1/96.

But that's not the question.

Suppose they do their looking in a single chunk of time each.

Further, suppose A's time is from midnight to 3 am.

Then if B's time starts at 10 pm or a later time up to 3 am [five hour window] they see each other.

So in that case, the probability is 5/24.

OP says A has 6 random 30-minute appearances; B has 6 random 20-minute appearances.

Fix one of A's appearances to start at midnight.

By the above reasoning, each of B's appearances has a 50-minute window of start times to produce overlap.

That is, from 20 minutes before, until 30 minutes after, midnight.

This is true for each of A's appearances.

This clearly gives a higher likelihood of overlap.

It's tempting to say the probability is thus 6 x 6 x [5/6] / 24.

But that product is greater than unity.

First conclusion: the probability increases with number of trips to the window.

Second conclusion: since 5/24 obtains for one visit each, that is the minimum value.

More later.

[Guest]

Assuming they do this over a 24-hour day, A is at her window 1/8 of the time; B 1/12 of the time.

The chance that at any given moment they are both there seems to be 1/96.

But that's not the question.

Suppose they do their looking in a single chunk of time each.

Further, suppose A's time is from midnight to 3 am.

Then if B's time starts at 10 pm or a later time up to 3 am [five hour window] they see each other.

So in that case, the probability is 5/24.

OP says A has 6 random 30-minute appearances; B has 6 random 20-minute appearances.

Fix one of A's appearances to start at midnight.

By the above reasoning, each of B's appearances has a 50-minute window of start times to produce overlap.

That is, from 20 minutes before, until 30 minutes after, midnight.

This is true for each of A's appearances.

This clearly gives a higher likelihood of overlap.

It's tempting to say the probability is thus 6 x 6 x [5/6] / 24.

But that product is greater than unity.

First conclusion: the probability increases with number of trips to the window.

Second conclusion: since 5/24 obtains for one visit each, that is the minimum value.

More later.

wow!

im impressed

[Guest]

Mrs. Abernathy and Mrs. Beaumont live across the street from each other and like to look out their front windows to see what each other are up to. Just to make sure nothing untoward is happening, one would presume. Mrs. Abernathy will look out her window six random times in one day at equal durations for a total of three hours per day, while Mrs. Beaumont will look out six times a day in equal durations for a total of two hours per day.

What is the chance that they will look out their windows and see the other looking back?

If the total time spent looking out their windows stays the same, but the number of times doubles for both, does the chance of them seeing each other change? To what?

I don't know if my math is right here, but I figure Abernathy has a 1 in 8 chance of looking out her window, Beaumont has a 1 in 12. Since the times are random and could therefore overlap in any myriad of ways, it would only be necessary for them to be looking at the same time for a fraction of a second. Therefore, I would go with the odds of one seeing the other are equal to within a negligible variation to the odds of the least likely person to be looking out, or 1 in 12.

[Guest]

Mrs. Abernathy and Mrs. Beaumont live across the street from each other and like to look out their front windows to see what each other are up to. Just to make sure nothing untoward is happening, one would presume. Mrs. Abernathy will look out her window six random times in one day at equal durations for a total of three hours per day, while Mrs. Beaumont will look out six times a day in equal durations for a total of two hours per day.

What is the chance that they will look out their windows and see the other looking back?

If the total time spent looking out their windows stays the same, but the number of times doubles for both, does the chance of them seeing each other change? To what?

[Prof. Templeton]

Assuming they do this over a 24-hour day, A is at her window 1/8 of the time; B 1/12 of the time.

The chance that at any given moment they are both there seems to be 1/96.

But that's not the question.

Suppose they do their looking in a single chunk of time each.

Further, suppose A's time is from midnight to 3 am.

Then if B's time starts at 10 pm or a later time up to 3 am [five hour window] they see each other.

So in that case, the probability is 5/24.

OP says A has 6 random 30-minute appearances; B has 6 random 20-minute appearances.

Fix one of A's appearances to start at midnight.

By the above reasoning, each of B's appearances has a 50-minute window of start times to produce overlap.

That is, from 20 minutes before, until 30 minutes after, midnight.

This is true for each of A's appearances.

This clearly gives a higher likelihood of overlap.

It's tempting to say the probability is thus 6 x 6 x [5/6] / 24.

But that product is greater than unity.

First conclusion: the probability increases with number of trips to the window.

Second conclusion: since 5/24 obtains for one visit each, that is the minimum value.

More later.

I have a slightly lower minimum.

A can look out her window starting at one of 43 different start times (midnight until 9PM), while B can look out her windows one of 67 different start times (midnight until 10PM).

If B's block is the first 2 hours in the day or the last 2, A has 4 choices for a start time that will overlap B.

If B leaves a 20 minute gap either at the begining of the day or the end of the day, A has 5 choices.

If B leaves a 40 minute gap or an hour gap, A has 6 choices.

If B leaves a 80 minute gap, A has 7 choices.

Any other block B chooses, A has 8 choices that will produce an overlap.

I'll leave off there, but it's a lower percentage then 5/24 (20.8%)

[Guest]

I have a slightly lower minimum.

A can look out her window starting at one of 43 different start times (midnight until 9PM), while B can look out her windows one of 67 different start times (midnight until 10PM).

If B's block is the first 2 hours in the day or the last 2, A has 4 choices for a start time that will overlap B.

If B leaves a 20 minute gap either at the begining of the day or the end of the day, A has 5 choices.

If B leaves a 40 minute gap or an hour gap, A has 6 choices.

If B leaves a 80 minute gap, A has 7 choices.

Any other block B chooses, A has 8 choices that will produce an overlap.

I'll leave off there, but it's a lower percentage then 5/24 (20.8%)

19/96=19.8 is that right?

[Guest]

Well, Mrs. Abernathy and Mrs. Beaumont can’t be looking at each other when it’s dark out, right?

So let’s pretend there’re 12 hours of darkness, and 12 hours of light in one day. The sun starts shining at noon, and plus they spend 12 hours in darkness, sleeping or doing whatever else that isn’t looking at their neighbors.

Since Mrs. Abernathy looks out of her window 6 times in 3 hours, we assume that she has 9 hours left to do her own things. And since these are done in regular intervals, 9/6 or 1.5 hours of hers are spent doing her own things, and then .5 hours of hers are spent looking over at Mrs. Beaumont.

Likewise, Mrs. Beaumont looks out of her window 6 times in 2 hours, so she has 10 hours left in a day. 10/6 is 1.666 repeating. She spends 1.666 repeating hours doing her own things, and then .333 of an hour looking over at Mrs. Abernathy.

(DOT stands for Doing Own Things, LAN stands for Looking At Neighbor)

For all intervals, Mrs. Abernathy spends 3/2 hours DOT, ½ hours LAN

For all intervals, Mrs. Beaumont spends 5/3 hours DOT, 1/3 hours LAN

Mrs. Abernathy Mrs. Beaumont

12:00-1:30 DOT 12:00-1:40 DOT

1:30-2:00 LAN 1:40-2:00 LAN

2:00- 3:30 DOT 2:00- 3:40 DOT

3:30-4:00 LAN 3:40-4:00 LAN

4:00-5:30 DOT 4:00-5:40 DOT

5:30-6:00 LAN 5:40-6:00 LAN

6:00-7:30 DOT 6:00-7:40 DOT

7:30-8:00 LAN 7:40-8:00 LAN

8:00-9:30 DOT 8:00-9:40 DOT

9:30-10:00 LAN 9:40-10:00 LAN

10:00- 11:30 DOT 10:00-11:40 DOT

11:30-12:00 LAN 11:40-12:00 LAN

And that's the end of the visible 12 hours, we assume they either sleep or do something else besides watching their neighbors during that time. My chart isn't supposed to repeat itself until after the 12 hours of darkness have passed.

[Guest]

but I get a 73% chance for 6 times and a 93% chance for 12 times.

Mrs. Abernathy has a 1/12 chance of seeing Mrs. Beaumont when she first looks out the window.

Mrs. Beaumont has a 1/8 chance of seeing Mrs. Abernathy when she first looks out the window.

So, the probability that each will NOT see the other is 11/12 and 7/8.

They each repeat 6 times, so:

The probability that Mrs. A will not see Mrs. B when she first looks out the window is 11/12^6

The probability that Mrs. B will not see Mrs. A when she first looks out the window is 7/8^6

Thus, the probability that both will not see the other is (11/12^6)*(7/8^6)=27%. The chance that they will see each other is 1-.27=73%

Same process for 12 times yields 93%

[Prof. Templeton]

Well, Mrs. Abernathy and Mrs. Beaumont can’t be looking at each other when it’s dark out, right?

So let’s pretend there’re 12 hours of darkness, and 12 hours of light in one day. The sun starts shining at noon, and plus they spend 12 hours in darkness, sleeping or doing whatever else that isn’t looking at their neighbors.

Since Mrs. Abernathy looks out of her window 6 times in 3 hours, we assume that she has 9 hours left to do her own things. And since these are done in regular intervals, 9/6 or 1.5 hours of hers are spent doing her own things, and then .5 hours of hers are spent looking over at Mrs. Beaumont.

Likewise, Mrs. Beaumont looks out of her window 6 times in 2 hours, so she has 10 hours left in a day. 10/6 is 1.666 repeating. She spends 1.666 repeating hours doing her own things, and then .333 of an hour looking over at Mrs. Abernathy.

(DOT stands for Doing Own Things, LAN stands for Looking At Neighbor)

For all intervals, Mrs. Abernathy spends 3/2 hours DOT, ½ hours LAN

For all intervals, Mrs. Beaumont spends 5/3 hours DOT, 1/3 hours LAN

Mrs. Abernathy Mrs. Beaumont

12:00-1:30 DOT 12:00-1:40 DOT

1:30-2:00 LAN 1:40-2:00 LAN

2:00- 3:30 DOT 2:00- 3:40 DOT

3:30-4:00 LAN 3:40-4:00 LAN

4:00-5:30 DOT 4:00-5:40 DOT

5:30-6:00 LAN 5:40-6:00 LAN

6:00-7:30 DOT 6:00-7:40 DOT

7:30-8:00 LAN 7:40-8:00 LAN

8:00-9:30 DOT 8:00-9:40 DOT

9:30-10:00 LAN 9:40-10:00 LAN

10:00- 11:30 DOT 10:00-11:40 DOT

11:30-12:00 LAN 11:40-12:00 LAN

And that's the end of the visible 12 hours, we assume they either sleep or do something else besides watching their neighbors during that time. My chart isn't supposed to repeat itself until after the 12 hours of darkness have passed.

For our purposes let's make some assumptions.

1. Either person will look out of their window at completely random times throughout the day (they have no set schedules and possibly suffer some some ailment that causes them to sleep/ be awake at random times of the day).
   
2. They live on a well lit street and always have the lights on in the houses if they are awake.
   

[Guest]

but I get a 73% chance for 6 times and a 93% chance for 12 times.

Mrs. Abernathy has a 1/12 chance of seeing Mrs. Beaumont when she first looks out the window.

Mrs. Beaumont has a 1/8 chance of seeing Mrs. Abernathy when she first looks out the window.

So, the probability that each will NOT see the other is 11/12 and 7/8.

They each repeat 6 times, so:

The probability that Mrs. A will not see Mrs. B when she first looks out the window is 11/12^6

The probability that Mrs. B will not see Mrs. A when she first looks out the window is 7/8^6

Thus, the probability that both will not see the other is (11/12^6)*(7/8^6)=27%. The chance that they will see each other is 1-.27=73%

Same process for 12 times yields 93%

You are over-thinking it. you have accounted for random chance and repetition by using the fractions you chose. To do the math your way you would need to choose fractions of 47/48 and 71/72. That is the likelihood of them not seeing each other if they each look out the window only once per day, Mrs. A for 30 minutes and Mrs. B for 20.

[CaptainEd]

Mrs. Abernathy and Mrs. Beaumont live across the street from each other and like to look out their front windows to see what each other are up to. Just to make sure nothing untoward is happening, one would presume. Mrs. Abernathy will look out her window six random times in one day at equal durations for a total of three hours per day, while Mrs. Beaumont will look out six times a day in equal durations for a total of two hours per day.

What is the chance that they will look out their windows and see the other looking back?

If the total time spent looking out their windows stays the same, but the number of times doubles for both, does the chance of them seeing each other change? To what?

Yes, the chance goes up, as far as I can tell. It doesn't feel right to me, but...

Imagine A's 3 hour viewing block as starting at midnight. Now, divide the block into 6 blocks, and imagine 6 gaps, totalling 21 hours, one added in front of each view.

Now, let's answer an easier question about B's views: Suppose B's view is a point in the 24 hours. What is the probability that one point lies inside of an A block or within 10 minutes on either side of a gap?

Total size of A's blocks = 180 minutes.

Each block is 30 mins long, and we add 10 minutes on either side = 50 mins.

There are 6 such blocks, so there are 300 mins in which B's point could lie (these are the same 300 mins formed by totalling A's and B's times)

What is that prob? 300/1440 = .208

What is the prob of a miss? 1-.208 = .792

So I get pretty much the same as Voltage gets.

What is the prob of 6 misses? .792 ^ 6 = .2468

Thus prob of at least 1 hit = 1 - .2468 = .7532

(Notice this is an underestimate, because this analysis would allow B to overlap her own view attempts, which is not the case. A more accurate reading would happen if we "sampled with replacement", in some sense)

What if you divide into 12 hits? The total size of the blocks is the same: 300 minutes (12 blocks of size 15 mins, adding 5 mins on either side = 25 mins, times 12)

So prob of a miss is still .792, so prob of 12 misses is .792^12 = .061

So prob of at least 1 hit in 12 tries = .939

I'm uncomfortable with this result. It's as if the more views, the higher the probability of a hit.

Edited March 18, 2009 by CaptainEd

[CaptainEd]

Yes, the chance goes up, as far as I can tell. It doesn't feel right to me, but...

Imagine A's 3 hour viewing block as starting at midnight. Now, divide the block into 6 blocks, and imagine 6 gaps, totalling 21 hours, one added in front of each view.

Now, let's answer an easier question about B's views: Suppose B's view is a point in the 24 hours. What is the probability that one point lies inside of an A block or within 10 minutes on either side of a gap?

Total size of A's blocks = 180 minutes.

Each block is 30 mins long, and we add 10 minutes on either side = 50 mins.

There are 6 such blocks, so there are 300 mins in which B's point could lie (these are the same 300 mins formed by totalling A's and B's times)

What is that prob? 300/1440 = .208

What is the prob of a miss? 1-.208 = .792

So I get pretty much the same as Voltage gets.

What is the prob of 6 misses? .792 ^ 6 = .2468

Thus prob of at least 1 hit = 1 - .2468 = .7532

(Notice this is an underestimate, because this analysis would allow B to overlap her own view attempts, which is not the case. A more accurate reading would happen if we "sampled with replacement", in some sense)

What if you divide into 12 hits? The total size of the blocks is the same: 300 minutes (12 blocks of size 15 mins, adding 5 mins on either side = 25 mins, times 12)

So prob of a miss is still .792, so prob of 12 misses is .792^12 = .061

So prob of at least 1 hit in 12 tries = .939

I'm uncomfortable with this result. It's as if the more views, the higher the probability of a hit.

Suppose A spends her time as before (in any number of intervals totalling 180 minutes), and suppose B only makes a few glances of essentially no duration.

Then prob of B hitting one of A's intervals in any one glance is 180/1440 = .125

Prob of miss = .875

Prob of 6 misses = .449

Prob of at least 1 hit in 6 tries = .551

Prob of 12 misses = .20

Prob of at least 1 hit in 12 tries = .80

So the key to the rise in probability is the number of views by B.

[Prof. Templeton]

Suppose A spends her time as before (in any number of intervals totalling 180 minutes), and suppose B only makes a few glances of essentially no duration.

Then prob of B hitting one of A's intervals in any one glance is 180/1440 = .125

Prob of miss = .875

Prob of 6 misses = .449

Prob of at least 1 hit in 6 tries = .551

Prob of 12 misses = .20

Prob of at least 1 hit in 12 tries = .80

So the key to the rise in probability is the number of views by B.

in 2 minute intervals. 99.98% chance of catching each other spying.

Also I posted the chances of overlap for solid blocks or time. Someone just needs to finish the math. Unless I'm mistaken it's a lower minimum then what's been suggested so far.

[bonanova]

The probability can be arbitrarily close to unity if only one of the neighbors looks in shorter time intervals.

Suppose A looks once for 3 hours.

Then if B looks at equally spaced times, the spacing only has to be less than 3 hours.

That is, she looks 8 times a day 15 minutes at a time.

That guarantees overlap.

If you want to get picky, B looks at 9 equally spaced times for 13 1/3 minutes.

But B looks at randomly spaced times. No problem.

Increase the number of visits to the window and decrease the interval accordingly.

B could look for 1 minute at 120 random times.

Or for 1 second at 7200 random times.

Or for 1 millisecond at 7 200 000 random times.

And so forth, with the probability of a 3 hour gap going to zero.

That is, for any given probability, there is a number of times that gives a smaller one.

As it stands, the probability of a 3 hour gap in 7200000 randomly chosen times is very small.

But the OP requires both neighbors to look more often for shorter time intervals.

This analysis breaks down in that case.

I suspect the probability approaches not unity but something with an e in it.

[bonanova]

Assuming the following

For one trip [each] to the window of intervals A and B minutes respectively,

the probability the intervals do not overlap is p₁ = [1440 - (A + B)]/1440.

If the watching times are divided by N,

this becomes p_N = [1440 - (A + B)/N]/1440.

But now there are N² ways for overlap

[each of A's visits to the window vs. each of B's visits to the window.]

So the total avoidance probability becomes p_Tot = [p_N]^NxN,

which goes to zero quickly for even moderate values of N.

So it seems the likelihood of A and B seeing each other [1-p_Tot] approaches certainty.

Specifically, responding to the OP, for N=6 we get .71979, and for N=12 we get .91970.

For N=50, we get .99997

We get these

[b]180 120 1 .79167 .79167 .20833[/b] <- each looks once for full time
180 120 2 .89583 .64403 .35597
180 120 3 .93056 .52322 .47678
180 120 4 .94792 .42494 .57506
180 120 5 .95833 .34508 .65492
[b]180 120 6 .96528 .28021 .71979[/b] <- each looks 6 times for 20 and 30 minutes
180 120 7 .97024 .22753 .77247
180 120 8 .97396 .18475 .81525
180 120 9 .97685 .15001 .84999
180 120 10 .97917 .12180 .87820
180 120 11 .98106 .09890 .90110
[b]180 120 12 .98264 .08030 .91970 [/b]<- each looks 12 times or 10 and 15 minutes
180 120 13 .98397 .06520 .93480
180 120 14 .98512 .05294 .94706
180 120 15 .98611 .04298 .95702
180 120 16 .98698 .03490 .96510
180 120 17 .98775 .02834 .97166
180 120 18 .98843 .02301 .97699
180 120 19 .98904 .01868 .98132
180 120 20 .98958 .01517 .98483

TimeA TimeB  N  PMissN  PMissTOT    SEE

Here are some larger values of N.

120 180 10 .97917 .12180 .87820 <- 10 looks of 12 and 18 minutes
120 180 20 .98958 .01517 .98483 <- 20 looks of 6 and 9 minutes
120 180 30 .99306 .00189 .99811 <- 30
120 180 40 .99479 .00024 .99976
[b] 120 180 50 .99583 .00003 .99997[/b] <- 50 looks of 2.4 and 3.6 minutes - almost certainty
120 180 60 .99653 .00000 1.00000
120 180 70 .99702 .00000 1.00000
120 180 80 .99740 .00000 1.00000
120 180 90 .99769 .00000 1.00000
120 180 100 .99792 .00000 1.00000 <- 100 looks of 1.2 and 1.8 minutes

TimeA TimeB  N  PMissN  PMissTot    SEE

[Prof. Templeton]

I had a lower minimum then yours for consecutive viewing times

Solid blocks of time]As I said earlier,

A can look out her window starting at one of 43 different start times (midnight until 9PM), while B can look out her windows one of 67 different start times (midnight until 10PM).

If B's block is the first 2 hours in the day or the last 2, A has 4 choices for a start time that will overlap B.

If B leaves a 20 minute gap either at the begining of the day or the end of the day, A has 5 choices.

If B leaves a 40 minute gap or an hour gap, A has 6 choices.

If B leaves a 80 minute gap, A has 7 choices.

Any other block B chooses, A has 8 choices that will produce an overlap.

Using that I have 4/43 * 2/67 + 5/43 * 2/67 + 6/43 * 4/67 + 7/43 * 2/67 + 8/43 * 58/67 (made a mistake earlier, here)

so..

268/2881 * 86/2881 + 335/2881 *86/2881 + 402/2881 * 172/2881 + 469/2881 * 86/2881 + 536/2881 * 2494/2881

then...

23,048/8,300,161 + 28,810/8,300,161 + 69,144/8,300,161 + 40,334/8,300,161 + 1,336,784/8,300,161

equals...

1,498,120/8,300,161

or...

18% is the minimum chance of A and B seeing each other given that A views for 3 hours and B views for 2.

Don't forget that we are working with a fixed day which limit the potential start times for A in relation to B around the begining and end of the day. I should have been more clear on the definition of "a day", but I'm figuring it to be from 12AM to 12AM. If A is at her window starting at 12AM(midnight) and is there until 3AM, B can only coincide with A starting also at 12AM(midnight). All the start times prior to 12AM (like 11:40,11:20,11:00,10:40,10:20) that may overlap are excluded because they are in a different day.

[bonanova]

1. I didn't read the OP as requiring A or B to leave their windows at midnight if the 1/N duration had not expired. I interpreted the OP as saying they went to their windows, daily, at 6 random times, staying there for the 1/N duration. That would permit A's last viewing time to line up with [part of] B's first, or v.v.
   
   
   
2. Otherwise, don't we need the added requirement that midnight would cut short a viewing session, or that A and B don't do this on a daily basis - just once? But then we'd need the probability that they chose the same day.
   
   Edit - OK, read your last post, see what you intended.
   
   
   
3. How did you get discrete start times?
   

[Prof. Templeton]

1. I didn't read the OP as requiring A or B to leave their windows at midnight if the 1/N duration had not expired. I interpreted the OP as saying they went to their windows, daily, at 6 random times, staying there for the 1/N duration. That would permit A's last viewing time to line up with [part of] B's first, or v.v.
   
   
   
2. Otherwise, don't we need the added requirement that midnight would cut short a viewing session, or that A and B don't do this on a daily basis - just once? But then we'd need the probability that they chose the same day.
   
   Edit - OK, read your last post, see what you intended.
   
   
   
3. How did you get discrete start times?
   

There has to be an arbitrary cut-off time wheather it is 12AM or some other time as long as the "day" ends 24 hours later and the ladies get their window viewing in that 24 hour frame. Since the minimum chance involves solid blocks of time some of the start times are excluded so the block does not extend outside of the time frame. I think as we move away from solid blocks of time and into smaller segments that the ladies spend in front of their windows your analysis is sound.