"Musical chair" for clock hands

Question

A colleague and me usually make bets on the time when a simulation is finished. Next time I would suggest the following: At the time the simulation is finished we take a look at the analog clock on the wall:

If the hour hand is closer to the 12 (in terms of the enclosed angle $\color{red}{\alpha}$) I win. If the minute hand is closer to 12 (in terms of $\color{red}{\beta}$) you win.

If all simulation finish times are equally likely, what is the chance for me winning this bet?

Both clock hands move continuously, not in discrete steps.

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_{If anybody wonders about the title: This is like the children's game Musical chairs with just 1 chair (at the position of the 12) and 2 players (hour and minute hand) left. When the music (or simulation) stops, the player who is closer to the chair is most likely to win.}

Answer 1

The chances of you winning the bet are

just over half, at $\frac{72}{143}$.

I figured this puzzle out by going through all the steps the long way. The idea is:

1. Find all the times where the hands are equally far from the 12, and calculate the intervals between each one
   
2. Sum up the intervals according to who would win the bet at this time interval
   

Let's call the angle of the hour hand $a$ and the angle of the minute hand $b$. The angles will be expressed in fractions of one revolution.
For a given time (hour, minute or $h$:$m$) we can calcluate the hand angles:
\begin{align} b &= \frac{m}{60} \\ a &= \frac{h}{12} + \frac{\frac{m}{60}}{12} \\ &= \frac{h}{12} + \frac{b}{12} \end{align} eg. for the time $11$:$15$, $h=11$ and $m=15$ so $b=\frac{1}{4}$ and $a=$ some fraction getting close to $1$.

There are two ways for the hands to be equally far from $12$. Either on the same side, or opposite sides (like the OP's picture). These times are represented by $a = b$ and $1 - a = b$.

First, $a = b$, by substitution

\begin{align} h / 12 + b / 12 &= b \\ h &= 11b \\ \text{and} \\ m &= 60b \end{align}
So, for any given $h$, we can find the $b$ then find $m$ for it. Knowing that the hands cross at most once per hour we can find for $h=0..12$, these are the times when the hands cross.

      h  b
     (0, 0)      12:00
     (1, 1/11)    1:(60/11)
     (2, 2/11)    2:(120/11)
     (3, 3/11)    3:(180/11)
     (4, 4/11)    4:(240/11)
     (5, 5/11)    5:(300/11)
     (6, 6/11)    6:(360/11)
     (7, 7/11)    7:(420/11)
     (8, 8/11)    8:(480/11)
     (9, 9/11)    9:(540/11)
     (10, 10/11) 10:(600/11)
     (11, 1)     12:00
 

Now, do the same for when $1 - a = b$,

\begin{align} 1 - \frac{h}{12} - \frac{b}{12} &= b \\ 12 - h &= 13b \\ -h &= 13b - 12 \\ h &= -13b + 12 \\ b &= \frac{12 - h}{13} \end{align}
So again we have a formula relating $b$ to $h$, and we use $h$ to find $b$. Then use $b$ to find $m$. This gives the time in $h$:$m$ where the hands are in an 'iscoscoles' shape.

      h  b
     (0, 12/13)  12:(60*12/13)
     (1, 11/13)   1:(60*11/13)
     (2, 10/13)   2:(60*10/13)
     (3, 9/13)    3:(60*9/13)
     (4, 9/13)    4:(60*8/13)
     (5, 7/13)    5:(60*7/13)
     (6, 6/13)    6:(60*6/13)
     (7, 5/13)    7:(60*5/13)
     (8, 4/13)    8:(60*4/13)
     (9, 3/13)    9:(60*3/13)
     (10, 2/13)  10:(60*2/13)
     (11, 1/13)  11:(60*1/13)
     (12, 0)     12:00
 

Now let's change both of those sets of times into proper fractions with a common denominator,

     Crossing 'x':
     (0, 0)      12:00
     (1, 1/11)    1:(780/143)
     (2, 2/11)    2:(1560/143)
     (3, 3/11)    3:(2340/143)
     (4, 4/11)    4:(3120/143)
     (5, 5/11)    5:(3900/143)
     (6, 6/11)    6:(4680/143)
     (7, 7/11)    7:(5460/143)
     (8, 8/11)    8:(6240/143)
     (9, 9/11)    9:(7020/143)
     (10, 10/11) 10:(7800/143)
     (11, 1)     12:00
 

     Isosceles shape 'v':
     (0, 12/13)   12:(7920/143)
     (1, 11/13)    1:(7260/143)
     (2, 10/13)    2:(6600/143)
     (3, 9/13)     3:(5940/143)
     (4, 9/13)     4:(5280/143)
     (5, 7/13)     5:(4620/143)
     (6, 6/13)     6:(3960/143)
     (7, 5/13)     7:(3300/143)
     (8, 4/13)     8:(2640/143)
     (9, 3/13)     9:(1980/143)
     (10, 2/13)   10:(1320/143)
     (11, 1/13)   11:(660/143)
     (12, 0)      12:00
 

Now let's actually put the times into the order they occur and label them with x or v. Also label whether the interval is 'winning' for the short hand or the long hand, which alternates although note the two v's which switch it up in the middle.

     (0, 0)       12:00           x      +7920                   short
     (0, 12/13)   12:(7920/143)  v               +660  +780      long
     (1, 1/11)     1:(780/143)    x      +6480                   short
     (1, 11/13)    1:(7260/143)  v               +1320 +1560     long
     (2, 2/11)     2:(1560/143)   x      +5040                   short
     (2, 10/13)    2:(6600/143)  v               +1980 +2340     long
     (3, 3/11)     3:(2340/143)   x      +3600                   short
     (3, 9/13)     3:(5940/143)  v               +2640 +3120     long
     (4, 4/11)     4:(3120/143)   x      +2160                   short
     (4, 9/13)     4:(5280/143)  v               +3300 +3900     long
     (5, 5/11)     5:(3900/143)   x      +720                    short
     (5, 7/13)     5:(4620/143)  v               +3960 +3960     long
     (6, 6/13)     6:(3960/143)  v       +720                    short
     (6, 6/11)     6:(4680/143)   x              +3900 +3300
     (7, 5/13)     7:(3300/143)  v       +2160
     (7, 7/11)     7:(5460/143)   x              +3120 +2640
     (8, 4/13)     8:(2640/143)  v       +3600
     (8, 8/11)     8:(6240/143)   x              +2340 +1980
     (9, 3/13)     9:(1980/143)  v       +5040
     (9, 9/11)     9:(7020/143)   x              +1560 +1320
     (10, 2/13)   10:(1320/143)  v       +6480
     (10, 10/11)  10:(7800/143)   x              +780  +660
     (11, 1/13)   11:(660/143)   v       +7920
     (11, 1)      12:00           x
 

Now it's a task of actually summing the numbers and that will give you a value $n$, where $\frac{n}{143}$ is the amount of minutes in a $12$ hour period. To calculate this I went through and calculated the minutes, each time an hour ticked over, I added the amount of minutes left in that hour. It's tedious and I didn't realise a nice way to do this yet.

The total amount of time that the short hand leads is $\frac{51840}{143}$ minutes every $12$ hours and the long hand is $\frac{51120}{143}$ minutes. $51840:51120$ reduces to $72:71$ giving the chances of each outcome at $\frac{72}{143}$ and $\frac{71}{143}$.

Answer 2

@Jay's answer is correct and complete, but OP has said "anyone finding a shorter way of calculating this is welcome to add another answer to this question"

As said in that answer,

find all the times where the hands are equally far from the 12. These are of two types: a conjunction, where the minute and hour hands coincide, and a reflected conjunction, where the minute and hour hands are mirrored in the vertical.

But we can simplify things by

just dividing the circle into 143 equal slices. The minute hand moves 12 times as fast as the hour hand.
When the hour has moved 11 whole slices ahead of 12 o'clock, the minute hand has moved 132 = 143-11 slices, so this is a reflected conjunction. At 13 slices the minute hand has moved 156=143+13 slices, so this is a conjunction. In fact every multiple of 11 whole slices is a reflected conjunction, and every multiple of 13 whole slices is a conjunction.
Write out the multiples of 11 and 13 up to 143. Starts at 0, then the hour hand is closer for 11 slices, then the minute hand takes over up to 13 (2 slices), then the hour hand up to 22 (9 slices), then the minute hand up to 26 (4 slices), and so on.
Except around six o'clock, where you get two reflected conjunctions in a row.
Here is a table: [spent time fighting with SE markup then gave up]
You can do the sums to see the minute hand is short one slice.

So that's all, the odds are

72:71

and you can stop reading now, the rest is just me rambling.

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You might have noticed a pattern in the sums, with an obvious 'glitch' in the middle number: 11+9+7+5+3+1+1+3+5+7+9+11 versus 2+4+6+8+10+11+10+8+6+4+2.
And, if that 'glitch' were 'fixed' in the obvious way, making the middle 11 into 12, the sums would be equal.
Here is my half-baked explanation [EDIT: now 60% baked] as to what happened to the 'missing slice'. Let's do this:
- Subtract 2 from each number in the sum.
- Remove the two 0s and the two -1s.
- The left sum loses 12*2, but then we delete (-1)*2.
- The right sum loses 11*2.
- So this is "fair", and the result is 9+7+5+3+1+1+3+5+7+9 versus 2+4+6+8+9+8+6+4+2.
Repeat to get:
- 7+5+3+1+1+3+5+7 versus 2+4+6+7+6+4+2.
- 5+3+1+1+3+5 versus 2+4+5+4+2.
- 3+1+1+3 versus 2+3+2.
- 1+1 versus 1.
What is the point of that? Well, you can check that these sums represent the corresponding answer for a 10-hour, 8-hour, 6-hour, 4-hour and finally 2-hour clock.
Let's look at the last case: imagine there were only two hours on the clock, ie. the minute hand moves twice as fast as the hour hand. Divide the circle into 3 equal slices. Start from the top and follow the hour hand through one rotation. The first slice belongs to the hour hand, the second to the minute hand, and the third to the hour hand, and then we're back at the top. The odds are 2:1.
So, for an even number of hours: divide the circle into $(N+1)(N-1) = N^2-1$ slices, which is an odd number. The two slices adjacent to the top of the circle belong to the hour hand, the single slice at the bottom belongs to the minute hand, and the $N^2-4$ other slices are divided fairly.
On the other hand, for three hours, the hour and minute hands meet at the bottom of the circle, which is a conjunction and a reflected conjunction, just like at the top. This is a fair bet: divide the circle into four, the two slices adjacent to the top of the circle belong to the hour hand, and the two slices adjacent to the bottom of the circle belong to the minute hand. You can check that this remains fair for any odd number of hours.
It seems like it's the asymmetry between top and bottom that shortchanges the minute hand out of one slice.

So I am left with the feeling that I am still missing something, and there's a simpler trick that lets you see why the answer is that number.

Answer 3

After some time of inactivity I'd like to post another answer. I hope it's understandable despite its brevity.

The chances are

$\frac{72}{143} \approx 50.34965 \, \%$

that the hour hand is closer to the 12 at a random time.

Proof:

The angle distance between the 12 and the $\color{blue}{\text{hour}}$/$\color{orange}{\text{minute}}$ hands varies with a period of 12 h.

But the time between 6 - 12 is just a mirror image of the time between 0 - 6. So the chances in one of these invervals are the same as for the whole day. Let's zoom into the first 6 hours:

Using the units [h] and [# rotations] for the $x$- and $y$-axis respectively, we can describe the angular distance between the 12 and the $\color{blue}{\text{hour}}$ hand as $$\color{blue}{\frac{x}{12}}$$ and the distance between the 12 and the $\color{orange}{\text{minute}}$ hand as $$\color{orange}{x - k}$$ for the first half of the $k$th hour and as $$\color{orange}{k + 1 - x}$$ for the second half of it.

The $\color{blue}{\text{hour}}$ hand is closer to the 12 when the $\color{blue}{\text{blue}}$ graph is below the $\color{orange}{\text{orange}}$ one. To sum up the length of these intervals we first need to calculate all the intersections. For the first half of the $k$th hour the intersection is at $$\begin{align} \color{blue}{\frac{x_{k \uparrow}}{12}} &= \color{orange}{x_{k \uparrow} - k} \\ \Leftrightarrow \quad x_{k \uparrow} &= \frac{12}{11} k \end{align}$$ and for the second half at $$\begin{align} \color{blue}{\frac{x_{k \downarrow}}{12}} &= \color{orange}{k + 1 - x_{k \downarrow}} \\ \Leftrightarrow \quad x_{k \downarrow} &= \frac{12}{13} (k+1) \text{.} \end{align}$$ The summed length of these intervals is $$\sum_{k=0}^{5} \left( x_{k \downarrow} - x_{k \uparrow} \right) = \sum_{k=0}^{5} \left( \frac{12}{13} (k+1) - \frac{12}{11} k \right) = \frac{432}{143} \text{.}$$ Divided by the overall time of 6 h we get $$\frac{432 / 6}{143} = \frac{72}{143} \text{.}$$