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I am wondering how to calculate the probability that $3$ hands of a clock are in the same semi-circle ? I know a similar question is that if we randomly choose n points in a circle, then the probability that all of them in the same semi-circle will be $n/2^{(n-1)}$ with $n = 3$.

But when it comes to the dial hands, where I guess there are correlations between them, I am not sure whether my question is equivalent to the previous one. If they are not equivalent, and this question will be hard to calculate, then whether we can determinate this probability will be greater or smaller than $3/4$?
Anyone can help? Thanks.

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  • $\begingroup$ Yes $3$ points on a circle in the same half is your question. Think this way. If two hands are making angle $ \theta $, the third can be anywhere in $ (2 \pi - \theta) $ of the circle. The only area that does not work is angle $ \theta $ just opposite of it. Here is a link - math.stackexchange.com/questions/342293/… $\endgroup$ – Math Lover Nov 22 '20 at 20:55
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    $\begingroup$ I'm not sure if this is helpful, but I checked this computationally, and the probability appears to be approaching $3/4$ as you suspected and @MathLover claimed. However, I'm not entirely convinced because as you pointed out, the hands are correlated (in fact, assuming exact clocks, the hour hand determines the place of the other hands), so this argument about "the third [hand] can be anywhere in $(2\pi - \theta)$" needs a little more to make it precise, possibly something about small perturbations. $\endgroup$ – Math Helper Nov 23 '20 at 2:52
  • $\begingroup$ @DavidC.Ullrich yes but it is a good approximation. If we take hour and minute hands at a given time, the movement of seconds hand has to be within a certain angle. If we do this over $12$ hours, it is a good approximation. $\endgroup$ – Math Lover Nov 25 '20 at 15:35
  • $\begingroup$ @DavidC.Ullrich If you take a certain time, the positions of all three hands are fixed. The probability of them being in the same semicircle is really based on what time you are looking at at it. I said $12$ hours because in $12$ hours window, you would have covered all possible positions and angles between $3$ hands (including angles between hour and minute hands). $\endgroup$ – Math Lover Nov 25 '20 at 15:56
  • $\begingroup$ The answer would depend on what time of day it is $\endgroup$ – wolfies Nov 26 '20 at 3:42
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The hands in fact move deterministically; it is just the time when we view them that can be random. The movement repeats ever twelve hours, so assume we view the hands at some moment uniformly distributed between noon and midnight. Then the probability they will be within a semicircle is the number of hours they spend in a semicircle, divided by $12.$

Starting at noon, the hands are within one semicircle until the second hand is opposite the hour hand. The second hand makes $12\times 60 = 720$ revolutions in $12$ hours while the hour hand makes one revolution in the same time, so the time until the second and hour hands first align after noon is $12/719$ hour, which is $720/719$ minutes; but the time from noon until the hands are exactly opposite is half that time, $360/719$ minute.

For a short time after the first $360/719$ minute, the hands are not within a semicircle. They are in a semicircle again starting at the instant when the second hand is opposite the minute hand. The second hand aligns with the minute hand for the first time after noon at $60/59$ minutes after noon, so the first time they are opposite is at $30/59$ minute after noon.

Let $t_h = 360/719$ and $t_m = 30/59,$ and let's take the convention that "at time $t$" means at $t$ minutes past noon. So the first time interval when the hands are not in a semicircle starts at $t_h$ and ends at $t_m.$ The length of this interval is $t_m - t_h$ minutes; specifically, the time during which the hands are continuously not in a semicircle, in minutes, is $$ \frac{30}{59} - \frac{360}{719} = \frac{330}{42421}. $$

The second such time interval starts at $3t_h$ and ends at $3t_m$ for $3(t_m - t_h)$ minutes not in a semicircle, the third starts at $5t_h$ and ends at $5t_m$, and so forth as long as the minute hand remains less than $180$ degrees clockwise from the hour hand.

This is a promising pattern: after $n$ intervals like this, we will have spent a total of $n^2(t_m - t_h)$ minutes during which the hands do not fit in a semicircle.

Alas, the full solution is not that simple. The intervals increase in length until they are almost half a minute long, and then this simple pattern is broken.

The minute hand first becomes exactly opposite the hour hand at $6/11$ hour after noon, that is, at $360/11$ minutes after noon. Shortly before that, at $65t_h$ minutes, the second hand passes opposite to the hour hand, making the hands not within a semicircle; but at $360/11$ minutes after noon the second hand is suddenly finds itself between the minute and hour hands so that the hands are all within a semicircle again, even though $65t_m$ minutes have not yet passed.

After this abbreviated interval of not-in-a-semicircle time, the next interval is almost half a minute long, and the next several intervals decrease in length. But once they have decreased enough, they start increasing again, and the pattern repeats but with different intervals each time as the relative position of the second hand is different each time the minute hand overtakes the hour hand or comes opposite to it.

At about this point in the analysis I decided I would rather have a computer count the time for me. The events where the hands start fitting in a semicircle or stop fitting in a semicircle all occur at exact integer multiples of either $1/11,$ $1/59,$ or $1/719$ minute after noon. Since those fractions have a common denominator of $11\times 59\times 719 = 466631,$ every event of interest (where the hands start or stop being within a semicircle) occurs at some multiple of $1/466631$ minute after noon. By taking $720\times 466631$ equal timesteps of that length we cover twelve hours. At the start of each timestep we know from the relative positions of the three hands whether the hands will fit in a semicircle or not during that timestep. (It is one way for the entire duration of the timestep.)

I wrote the analysis in the python script shown below. I had the script count the first six hours, take a checkpoint, and then do the last six hours. Everything after $6$ o'clock is a mirror image of what happened before, in reverse order, so the same sequence of lengths of intervals occurs in reverse order after $6$ o'clock. I had the script record the start, end, and duration of each interval to verify this. Note that since each interval starts about a minute after the previous one, there are almost $360$ intervals in six hours, but not quite $360$ due to the way the hands "catch up" with each other.

According to my script, it takes $720\times 11\times 59\times 719 = 335974320$ timesteps to complete twelve hours, and during $83982960$ of those timesteps the three hands do not fit within a semicircle. The proportion of time not in a semicircle is therefore $$ \frac{83982960}{335974320} = \frac{1977}{7909} \approx 0.2499683904412694. $$ So the probability that the hands are not within a semicircle is almost, but not quite, $1/4$; the probability that they are within a semicircle is therefore slightly over $3/4.$


To relate these calculations to G Cab's method, let $x$ be the number of units of $12/11$ hour since noon (that is, $x$ is $0$ at noon and $11$ at midnight). If $k = \lfloor x\rfloor$ (an integer), then for each $k$ (that is, for each period of $12/11$ hour) there are two sequences of intervals during which the three hands are not within a semicircle. During the first half of this period, while the angle from the hour hand to the minute hand is between $0$ and $180$ degrees clockwise, the three hands are not in a semicircle whenever $$ m + \frac12 \leq \frac{15(2k)}{22} + \frac{719}{11} \tau < m + \frac12 + \tau $$ for some integer $m,$ where $\tau = x - k,$ and during the second half of this period, while the angle from the minute hand to the hour hand is between $0$ and $180$ degrees clockwise, the three hands are not in a semicircle when $$ n + \tau \leq \frac{15(2k + 1)}{22} + \frac{719}{11} \tau < n + \frac12 $$ for some integer $n,$ where $\tau = x - k - \frac12.$ In either case we also have the restriction that $0 \leq \tau < \frac12.$

In the typical case, the beginning or end of an interval occurs when one of the inequalities becomes an equation; the exceptions occur when the equation is true only for $\tau < 0$ or $\tau > \frac12$. Solving these equations for $\tau,$ we find $$\tau = \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag1$$ at the start and $$\tau = \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right)\tag2$$ at the end of a typical interval in the first sequence, but $$\tau = \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right)\tag3$$ at the start and $$\tau = \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right)\tag4$$ at the end of one in the second sequence. The length of the first interval (in the same units as $\tau$) is therefore \begin{align} \Delta\tau_1 &= \frac{11}{708} \left(m + \frac12 - \frac{15(2k)}{22}\right) - \frac{11}{719} \left(m + \frac12 - \frac{15(2k)}{22}\right) \\ &= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right) \end{align} and the length of the second interval is \begin{align} \Delta\tau_2 &= \frac{11}{719} \left(n + \frac12 - \frac{15(2k+1)}{22}\right) - \frac{11}{708} \left(n - \frac{15(2k+1)}{22}\right) \\ &= \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right) \end{align}

The combined length of two intervals, one from each sequence, is \begin{align} \Delta\tau_1 + \Delta\tau_2 &= \frac{121}{708(719)} \left(m - \frac{30k - 11}{22}\right) + \frac{121}{708(719)} \left(\frac{30k + 723}{22} - n\right) \\ &= \frac{11}{708(719)} \left(367 + 11(m - n)\right) \\ \end{align}

If we hold $m - n$ constant, $\Delta\tau_1 + \Delta\tau_2$ is independent of the value of $m$ as long as all of the $\tau$ values from which we computed $\Delta\tau_1$ and $\Delta\tau_2$ are between $0$ and $\frac12.$ For example, if we set $n = m$, we get a pairing of intervals in which each pair of intervals has total length $\frac{11(367)}{708(719)}$ except in some cases where an interval starts at $\tau=0$ or ends at $\tau=\frac12$ or where an interval has no corresponding interval in the other sequence. Trial and error shows that if we set $n=m+1$ we minimize the number of intervals that are unmatched or that are in a pair where one interval ends at $\tau=0$ or $\tau=\frac12$ instead of where the equations above say it will (there are $14$ such intervals); each matched pair then has total length $\frac{11(367-11)}{708(719)} = \frac{11(89)}{3(59)(719)}$ in the units of $\tau.$ This is equal to $\frac{356}{59(719)}$ hour, which is $234960$ timesteps of the brute-force method.

According to my spreadsheet calculations, if $n = m+1$ there are $32$ matched pairs of intervals for each value of $k.$ since there are $11$ values of $k,$ the total time in intervals that belong to typical matched pairs is $352 \times 234960$ timesteps. It then remains to account for the $14$ intervals that are not in typical matched pairs. Those intervals can be identified for each $k$ by finding the maximum $m$ such that right-hand side of Equation $(1)$ is negative and the minimum $n$ such that the right-hand side of Equation $(4)$ is greater than $\frac12.$ A relatively quick hand calculation should therefore be possible, though it would still be longer than I would like to write in detail in this answer.

Update: The calculations are not as long as I feared after all, as shown in G Cab's edited answer.


Here is the script for the brute-force method:

# Program to figure out what proportion of the time the three hands
# of a clock are not within a semicircle.

# Assume there are 59*719*720 equally-spaced positions around the clock face.
# Note: 59*719 == 42421; 42421*11 = 466631.

# Each timestep of this program is 1/42421 of a minute.
# In one timestep, the second hand moves 720 positions, the minute hand
# moves 12 positions, and the hour hand moves 1 position.
# In 42421 timesteps, one minute has passed and the second hand returns
# to the starting position,


numPositions = 11*59*719*720


# Returns the new position after having moved the given amount forward
# from the given position. That is, the return value is the sum of
# the two input values, modulo numPositions.
# It is assumed the input values are in the range 0 to numPositions - 1.
def newPosition(fromPosition, movement):
    toPosition = fromPosition + movement
    if toPosition >= numPositions:
        toPosition -= numPositions
    return toPosition


# Returns the number of steps clockwise to get from position1 to position2.
# positions measured in number of steps from the start.
def stepsClockwise(position1, position2):
    distance = position2 - position1
    if distance < 0:
        distance += numPositions
    return distance


# Returns 1 if the hands are at the start of a timestep in which the
# smallest angle encompassing all three hands will be greater than 180 degrees,
# 0 if the angle will be less than 180 degrees.
def isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
    # A 180 degree angle
    halfCircle = numPositions // 2
    # A position exactly opposite the seconds hand
    antiSecondsPosition = newPosition(secondsPosition, halfCircle)
    minutesOffset = stepsClockwise(hoursPosition, minutesPosition)
    if minutesOffset < halfCircle:
        # The minutes hand is < 180 degrees clockwise from the hour hand
        antiOffset = stepsClockwise(hoursPosition, antiSecondsPosition)
        if (antiOffset >= 0 and antiOffset < minutesOffset):
            # The anti-seconds hand will be between the hours and minutes
            # during this timestep, so the seconds hand will be more than
            # 180 degrees past the hour hand but less than 180 degrees
            # before the minute hand.
            return 1
        else:
            return 0
    else:
        # The hour hand is <= 180 degrees clockwise from the minute hand.
        # During the next timestep the short arc between the hands will be
        # clockwise from the minute hand to the hour hand.
        hoursOffset = stepsClockwise(minutesPosition, hoursPosition)
        antiOffset = stepsClockwise(minutesPosition, antiSecondsPosition)
        if (antiOffset >= 0 and antiOffset < hoursOffset):
            # The anti-seconds hand will be between the minutes and hours
            # during this timestep, so the seconds hand will be more than
            # 180 degrees past the minute hand but less than 180 degrees
            # before the hour hand.
            return 1
        else:
            return 0


# Runs the given number of timesteps from the given starting position
# and returns the number of times the hands spanned more than 180 degrees.
def countLargeAngles(hoursPosition,
                     minutesPosition,
                     secondsPosition,
                     startingTimestep,
                     numTimesteps):
    numLargeAngles = 0
    wasLargeAngle = 0
    largeAngleStarted = 0
    numIntervals = 0
    for i in range(numTimesteps):
        if isLargeAngle(hoursPosition, minutesPosition, secondsPosition):
            numLargeAngles += 1
            if wasLargeAngle == 0:
                largeAngleStarted = startingTimestep + i
                wasLargeAngle = 1
        else:
            if wasLargeAngle == 1:
                largeAngleStopped = startingTimestep + i
                numIntervals += 1
                print(">180 degrees interval ", numIntervals,
                      " from ", largeAngleStarted,
                      " to ", largeAngleStopped, " for ",
                      (largeAngleStopped - largeAngleStarted), " timesteps")
                wasLargeAngle = 0
        hoursPosition   = newPosition(hoursPosition, 1)
        minutesPosition = newPosition(minutesPosition, 12)
        secondsPosition = newPosition(secondsPosition, 720)
    print("After ", numTimesteps, " timesteps")
    print("hour hand at ", hoursPosition)
    print("minute hand at ", minutesPosition)
    print("second hand at ", secondsPosition)
    print(numLargeAngles, " positions with > 180 degrees")
    return numLargeAngles, hoursPosition, minutesPosition, secondsPosition


# Run 360*466631 timesteps to simulate 6 hours,
# then another 360*466631 timesteps to simulate the next 6 hours.
halfTotalTimesteps = numPositions // 2
hoursPosition = 0
minutesPosition = 0
secondsPosition = 0

(numLargeAngles,
 hoursPosition,
 minutesPosition,
 secondsPosition) = countLargeAngles(hoursPosition,
                                     minutesPosition,
                                     secondsPosition,
                                     0,
                                     halfTotalTimesteps)
(additionalLargeAngles,
 hoursPosition,
 minutesPosition,
 secondsPosition) = countLargeAngles(hoursPosition,
                                     minutesPosition,
                                     secondsPosition,
                                     halfTotalTimesteps,
                                     halfTotalTimesteps)
numLargeAngles += additionalLargeAngles

print("Total number of timesteps was ", numPositions)
print("Total number of > 180 degrees was ", numLargeAngles)
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  • $\begingroup$ Your clock seems to be digital rather than analog. Wouldn’t your computation have been a little easier with an analog clock? $\endgroup$ – Lubin Nov 23 '20 at 4:04
  • $\begingroup$ @Lubin It's an analog clock, but instead of using floating-point arithmetic, I recognized that all events on the clock face (at least, all events that are interesting for this problem) occur at rational numbers of minutes past noon, and I found the common denominator for all of those numbers. This lets me do a straightforward timestep-and-count. With floating-point numbers, I could have calculated the start and and of each interval without the common denominator, but I think there would have been more cases to worry about, and the results would not be exact. $\endgroup$ – David K Nov 23 '20 at 4:29
  • $\begingroup$ Well, since using my own mathematical common sense, I got a wrong answer, I can’t complain too much. But it does seem to me that a more conceptual approach would have involved a lot less talk. $\endgroup$ – Lubin Nov 23 '20 at 4:50
  • $\begingroup$ @Lubin I could have saved a lot of talk by starting at the paragraph that begins, "At about this point in the analysis ...," but I thought I needed a justification for taking a brute-force approach. I have the vague recollection of having done a problem somewhat like this without resorting to computer programming, but it was a frightful mess then too IIRC. (Too bad I haven't been able to find my former solution yet.) $\endgroup$ – David K Nov 23 '20 at 4:54
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Let's denote by $\omega$ the angular speeds, so $$ \left\{ \matrix{ \omega _{\,h} = {{2\pi } \over {12h}} = {{2\pi } \over {12 \cdot 3600}}s^{\, - 1} \hfill \cr \omega _{\,m} = {{2\pi } \over h} = {{2\pi } \over {3600}}s^{\, - 1} \hfill \cr \omega _{\,s} = {{2\pi } \over m} = {{2\pi } \over {60}}s^{\, - 1} \hfill \cr} \right. \quad \Rightarrow \quad \left\{ \matrix{ \omega _{\,m} - \omega _{\,h} = 11\;\omega _{\,h} \hfill \cr \omega _{\,s} - \omega _{\,h} = 719\;\omega _{\,h} \hfill \cr} \right. $$

Clearly we are interested in the relative angles $$ \left\{ \matrix{ \alpha _{\,m} = \left( {\omega _{\,m} - \omega _{\,h} } \right)t\, \quad \left( {\bmod 2\pi } \right)\quad \Rightarrow \quad \rho _{\,m} = {{\alpha _{\,m} } \over {2\pi }} = \left\{ {11{{\omega _{\,h} \,t} \over {2\pi }}} \right\}\, \hfill \cr \alpha _{\,s} = \left( {\omega _{\,s} - \omega _{\,h} } \right)t\, \quad \left( {\bmod 2\pi } \right)\quad \Rightarrow \quad \rho _{\,s} = {{\alpha _{\,s} } \over {2\pi }} = \left\{ {719{{\omega _{\,h} \,t} \over {2\pi }}} \right\} \hfill \cr} \right. $$

The hands will be on the same half circle when $$ \left\{ \matrix{ 0 \le \alpha _{\,m} < \pi \hfill \cr 0 \le \alpha _{\,s} < \pi \; \vee \hfill \cr \;\pi + \alpha _{\,m} \le \alpha _{\,s} < 2\pi \hfill \cr} \right.\;\; \vee \;\;\left\{ \matrix{ \pi \le \alpha _{\,m} < 2\pi \hfill \cr \pi \le \alpha _{\,s} < 2\pi \; \vee \; \hfill \cr 0 \le \alpha _{\,s} < \alpha _{\,m} - \pi \hfill \cr} \right. $$ or more simply they will not be so when $$ \left\{ \matrix{ 0 \le \alpha _{\,m} < \pi \hfill \cr \pi \le \alpha _{\,s} < \pi + \alpha _{\,m} \; \hfill \cr} \right.\;\; \vee \;\;\left\{ \matrix{ \pi \le \alpha _{\,m} < 2\pi \hfill \cr \alpha _{\,m} - \pi \le \alpha _{\,s} < \pi \; \hfill \cr} \right. $$

The cycle will endure $12$ h, that is $$ 0 \le {{\omega _{\,h} \,t} \over {2\pi }} < 1 $$ and let's put $$ 11{{\omega _{\,h} \,t} \over {2\pi }} = k + x\quad \left| \matrix{ \;0 \le k \in \mathbb Z \le 10 \hfill \cr \;0 \le x \in \mathbb R < 1 \hfill \cr} \right. $$

Then using $[ ]$ to denote the Iverson bracket and $\overline K (k,x)$ to denote the indicator function of the "forbidden" area, we get $$ \eqalign{ & \overline K (k,x)\quad \left| \matrix{ \;0 \le k \le 10 \hfill \cr \;0 \le x < 1 \hfill \cr} \right.\quad = \cr & = \left[ {0 \le x < {1 \over 2}} \right]\left[ {{1 \over 2} \le \left\{ {{{719} \over {11}}k + {{719} \over {11}}x} \right\} < {1 \over 2} + x} \right] + \cr & + \left[ {{1 \over 2} \le x < 1} \right]\left[ {x - {1 \over 2} \le \left\{ {{{719} \over {11}}k + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] = \cr & = \left[ {0 \le x < {1 \over 2}} \right]\left( \matrix{ \left[ {{1 \over 2} \le \left\{ {{{719\left( {2k} \right)} \over {22}} + {{719} \over {11}}t} \right\} < {1 \over 2} + x} \right] + \hfill \cr + \left[ {x \le \left\{ {{{719\left( {2k + 1} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] \hfill \cr} \right) = \cr & = \left[ {0 \le x < {1 \over 2}} \right]\left( \matrix{ \left[ {{1 \over 2} \le \left\{ {{{15\left( {2k} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2} + x} \right] + \hfill \cr + \left[ {x \le \left\{ {{{15\left( {2k + 1} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] \hfill \cr} \right) \cr} $$

Therefore the probability $\overline P$ of not having the hands on the same half circle will be $$ \eqalign{ & \overline P = {1 \over {11}}\sum\limits_{k = 0}^{10} {\int_{\,x\, = \,0}^{\,1} {\overline K (k,x)dx} } = \cr & = {1 \over {11}}\sum\limits_{k = 0}^{10} {\int_{\,x\, = \,0}^{\,1/2} {\,\left( \matrix{ \left[ {{1 \over 2} \le \left\{ {{{15\left( {2k} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2} + x} \right] + \hfill \cr + \left[ {x \le \left\{ {{{15\left( {2k + 1} \right)} \over {22}} + {{719} \over {11}}x} \right\} < {1 \over 2}} \right] \hfill \cr} \right)dx} } \cr} $$

$\overline K (k,x)dx $ is represented by the segments of the two sawtooth waves intercepted
inside the respective forbidden areas like those shown in the following sketch. lanc_orol_2

Note that for better clarity the inclination $m$ of the lines is much lower than $719/11$.

The integral translate into the sum of the $\Delta x$ of each segment.
To compute them let's start with computing the intercepts with $y=1/2$.
The minimum intercept $s_{\,0}$ will be $$ \eqalign{ & y = \left\{ {q + m\,x} \right\} = \left\{ {\left\{ q \right\} + m\,x} \right\} = {1 \over 2}\quad \Rightarrow \cr & \Rightarrow \quad \left\{ q \right\} + m\,x = n + {1 \over 2}\quad \Rightarrow \cr & \Rightarrow \quad m\,x = n - \left( {\left\{ q \right\} - {1 \over 2}} \right) \cr & \Rightarrow \quad m \, s_{\,0} = \left\lceil {\left\{ q \right\} - {1 \over 2}} \right\rceil - \left( {\left\{ q \right\} - {1 \over 2}} \right) = \left( {{1 \over 2} - \left\{ q \right\}} \right) - \left\lfloor {{1 \over 2} - \left\{ q \right\}} \right\rfloor = \cr & = \left\{ {{1 \over 2} - q} \right\} \cr} $$ For the lower triangle we can simply replace $q$ with the negative of the intercept with $x=1/2$ , i.e.: $$ m\,s_{\,0} = \left\{ {{1 \over 2} + {m \over 2} + q} \right\} $$

Next we take the upper triangle and slant it as shown lanc_orol_3

where $$ \Delta y(x) = \left\{ {\matrix{ {{m \over {m - 1}}x} & {\left| {\;0 \le x < {1 \over 2}\left( {1 - {1 \over m}} \right)} \right.} \cr {m\left( {{1 \over 2} - x} \right)} & {\left| {\;{1 \over 2}\left( {1 - {1 \over m}} \right) \le x < {1 \over 2}} \right.} \cr } } \right. $$ so that we will compute $\Delta x$ as $ 1/m \, \Delta y$.

Finally, we are sampling the slanted triangle at points starting from $s_{\, 0}$ (which depends on $q$ and thus on $k$) and progressing with $\Delta s = 1/m$. $$ s(k,j) = s_{\,0} + {j \over m} = \left\{ {\matrix{ \matrix{ {1 \over m}\left\{ {{1 \over 2} - q} \right\} + {j \over m} = {{11} \over {719}}\left( {\left\{ {{1 \over 2} - {{15\left( {2k} \right)} \over {22}}} \right\} + j} \right) = \hfill \cr = {{11} \over {719}}\left( {\left\{ {{{11 - 8k} \over {22}}} \right\} + j} \right) = s_{\,u} (k,j) \hfill \cr} \hfill & {{\rm upper}\;{\rm Tr}{\rm .}} \hfill \cr \matrix{ {1 \over m}\left\{ {{1 \over 2} + {m \over 2} + q} \right\} + {j \over m} = {{11} \over {719}}\left( {\left\{ {{1 \over 2} + {{719} \over {22}} + {{15\left( {2k + 1} \right)} \over {22}}} \right\} + j} \right) = \hfill \cr = {{11} \over {719}}\left( {\left\{ {{{19 + 8k} \over {22}}} \right\} + j} \right) = s_{\,d} (k,j) \hfill \cr} \hfill & {{\rm lower}\;{\rm Tr}{\rm .}} \hfill \cr } } \right. $$

Since $$ \eqalign{ & {{11} \over {719}}\left( {\left\{ {{{11 - 8k} \over {22}}} \right\} + j} \right) = {1 \over {2 \cdot 719}}\left( {22\left\{ {{{11 + 14k} \over {22}}} \right\} + 22j} \right) \cr & 22\left\{ {{{11 + 14k} \over {22}}} \right\}\quad \Rightarrow \quad 11 + 14k = 2n + 1\quad \left( {\bmod 22} \right)\quad \Rightarrow \cr & \Rightarrow \quad 5 + 7k = n\quad \left( {\bmod 11} \right)\quad \Rightarrow \quad 7k \equiv n + 5\quad \left( {\bmod 11} \right) \cr} $$
and $$ \eqalign{ & {{11} \over {719}}\left( {\left\{ {{{19 + 8k} \over {22}}} \right\} + j} \right) = {1 \over {2 \cdot 719}}\left( {22\left\{ {{{19 + 8k} \over {22}}} \right\} + 22j} \right) \cr & 22\left\{ {{{19 + 8k} \over {22}}} \right\}\quad \Rightarrow \quad 19 + 8k = 2n + 1\quad \left( {\bmod 22} \right)\quad \Rightarrow \cr & \Rightarrow \quad 9 + 4k = n\quad \left( {\bmod 11} \right)\quad \Rightarrow \quad 4k \equiv n + 2\quad \left( {\bmod 11} \right) \cr} $$
then the {Independent Residues theorem](https://en.wikipedia.org/wiki/Residue_number_system)
assures us that the the two relations above are bijections between
$$\left[ {0 \le k \le 10} \right] \Leftrightarrow \left[ {0 \le n \le 10} \right]$$

We have that the set of the sampling points are the same for both the triangles, and we can simply put
$$ \left\{ \matrix{ s_u (n) = s_d (n) = {{2n + 1} \over {2 \cdot 719}} \hfill \cr 0 \le s(n) < {1 \over 2}\left( {1 - {1 \over m}} \right) = {1 \over 2}\left( {{{708} \over {719}}} \right)\quad \Rightarrow \quad 0 \le n \le 353 \hfill \cr {1 \over 2}\left( {1 - {1 \over m}} \right) \le s(n) < {1 \over 2}\quad \Rightarrow \quad 354 \le 2n \le 358 \hfill \cr} \right. $$

And finally obtain
$$ \eqalign{ & \overline P = {1 \over {11}}\sum\limits_s {\Delta x(s)} = {1 \over {11m}}\sum\limits_s {\Delta y(s)} = {2 \over {11m}}\sum\limits_{\,0 \le n \le 358} {\Delta y(s_u (n))} = \cr & = 2\left( {{1 \over {708}}\sum\limits_{\,0 \le n \le 353} {{{2n + 1} \over {2 \cdot 719}}} + {1 \over {11}}\sum\limits_{\,354 \le n \le 358} {\left( {{1 \over 2} - {{2n + 1} \over {2 \cdot 719}}} \right)} } \right) = \cr & = {1 \over {719}}\left( {{1 \over {708}}\sum\limits_{\,0 \le n \le 353} {\left( {2n + 1} \right)} + {1 \over {11}}\sum\limits_{\,354 \le n \le 358} {\left( {718 - 2n} \right)} } \right) = \cr & = {1 \over {719}}\left( {{{354 + 354 \cdot 353} \over {708}} + {{5 \cdot 718 - 2 \cdot 5 \cdot 356} \over {11}}} \right) = \cr & = {1 \over {719}}\left( {{{354} \over 2} + {{30} \over {11}}} \right) = {{1977} \over {7909}} = {1 \over 4}\left( {1 - {1 \over {7909}}} \right) \cr} $$

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  • $\begingroup$ I worked the formula out in Excel (on 363 lines of a spreadsheet) and came up with exactly the same answer I got in by brute-force counting the intervals in python. (After I corrected the script, that is; while trying to translate the Excel figures into brute-force intervals I found a non-integer value on the 33rd line, which tipped me off that my interval was too large by a factor of 11.) $\endgroup$ – David K Nov 26 '20 at 4:32
  • $\begingroup$ Each Iverson bracket in the final integral, treated as a function of $t,$ is $1$ over a finite set of intervals. It turns out that most of the intervals can be paired off in such a way that the sum of lengths of two consecutive intervals is $21360/42421$ minute. There are maybe a dozen intervals that I did not manage to pair up this way, all starting at $t=0$ or ending at $t=1/2,$ suggesting that there may be a way to collect most of the terms of the calculation in a relatively simple expression and just have to handle these exceptions separately. $\endgroup$ – David K Nov 26 '20 at 4:48
  • $\begingroup$ @DavidK: yes there is somehow the possibility to group up the intervals. I will work on that also $\endgroup$ – G Cab Nov 26 '20 at 10:25
  • $\begingroup$ @DavidK: in fact it is possible to convert the integral in a sum. I recasted my answer accordingly $\endgroup$ – G Cab Nov 30 '20 at 21:51
  • $\begingroup$ At one point $\left\{\frac12 - q\right\}$ turns into $\left\{1 - q\right\}$ in the upper-triangle formulas. The formula with $\frac12 - q$ matches my results for the upper triangle. But by symmetry (comparing a backward-running clock to a regular clock), the intervals in the lower triangles should be the upper-triangle intervals in reverse order. I have not been able to reproduce that effect with the latest formulas in this answer. $\endgroup$ – David K Dec 1 '20 at 3:38
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Take two of the hands. On average they will be a quarter of a circle apart. Distance varies uniformly from zero to half a circle apart. Turn the dial so that one hand is at 12 and the other hand at 3. Now when the third hand is in the interval from 9 to 6 all three hands are on the same semicircle. When the third hand is between 6 and 9 they are not on the same semicircle. So on average the three hands will be on the same semicircle 3/4 of the time.

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