Time's AWasting
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.
A digital clock tells the time in hours and minutes. During what fraction of the day does it display at least one '1'?
Answer
720/1440 or 1/2
Here's the systematic approach to getting the answer:
Convert all the times of the day into simple numbers: 1:00 = 100, etc...
From 100159, there are 60 times that there is at least one '1'.
This is the same for the ranges of 10001059, 11001159, and 12001259.
From 200259, there are 15 times that there is at least one '1'
This is the same for ranges of 300359, 400459, 500559, 600659, 700759, 800859, and 900959.
That means 8 ranges have 15 times, 4 ranges have 60 times. We'll double this to convert to a 24hour day:
2[(15)(8) + (60)(4)] = 720 total times that there is at least one '1'.
To figure out the total number of times that any number is shown at least once, we know there are 60 times between any range listed above, so we simply multiply by 24 to get the total number of times in 24 hours.
(24)(60) = 1440 total times.
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Comments
tpg76
Feb 16, 2008
 Interesting teaser, and thanks for the explanation. I think, however, there are two problems with your reasoning:
1. You don't properly account for the times when more than 1 appears  in other words, you count a time like "4:11" twice by your method, and it only should be counted once.
2. You don't take into account that on a standard 12hour clock (which is logically assumed)  EVERY time between 10 and 12 o'clock meets the criterion, based on its unchanging leading '1'.
The best way to calculate such a thing and avoid the pitfall (1) is to calculate the probability that it WON'T happen, and then take its difference from one. Let me offer this alternative:
1. Put aside the 2 (of 12) hours with the leading '1'  for now. For the next few steps only consider the other 10 hours....
2. The probability (given the above) of otherwise NOT having a '1' in the first digit is 9/10.
3. The probability of not having a '1' in either of the tensofminutes digit is 5/6.
4. The probability of not having '1' in the last digit is 9/10.
5. Combining 24, the probability of not having '1' in any of these three is 9/10*5/6*9/10, or 0.675.
6. The probability of having AT LEAST ONE '1' is thus 1.675 = .325.
7. Reentering the other two hours (from point 1), simply take the weighted average of these two answers:
[10(.325) + 2(1.0)]/12 = 0.4375.
I believe this is the correct answer: 0.4375. 
kratsg
Feb 16, 2008
 What method did I use that says I double count? Read the teaser:
"During what fraction of the day does it display at least one '1'?"
So, at least one '1' means that if it shows at least one '1', it's included, otherwise, it isn't. 
tpg76
Feb 16, 2008
 But the numbers don't work out. Maybe I'll put it into an excel sheet to demonstrate. Thanks. 
kratsg
Feb 16, 2008
 Oh, also, that method above doesn't work. Having a one in any digit is influenced by the other digits, hence they are not independent (cannot directly multiply) but they are also not disjoint (since ones can appear in both places at the same time). Therefore, the above method is wrong when you multiply (since they are not independent) and you cannot subtract from 1 (since they are not disjoint, their total probabilities would not add up to one.) 
tpg76
Feb 16, 2008
 I did the spreadsheet and YOU WERE RIGHT. My reasoning was off in that there are three hours with the leading '1', not '2'.
For 4 hours (leading in 01, 10, 11, 12) you always have a '1'; of the other 4 you have it 15/60 minutes. The net is 360/720, as you said.
My apologies for the confusion. I'll try to delete the earlier post but don't know if I can!
It was fun to check. Thanks. 
kratsg
Feb 16, 2008
 No problem :P 
tpg76
Feb 16, 2008
 On your comment about nonindependence, that's what I had tried to eliminate by taking the first digit separately. I think by doing that the other three are independently variable, that is, over the nine hours between 10 and 9:59.
Notice now that this makes the (new) first digit effectively base nine (no zero); the second digit remains base six; the third is base 10. So my methodology can work with the appropriate modification:
Over the NINE hour period
 p(first digit NOT 1) = 8/9
 p(sec. digit not 1) = 5/6
 p(third digit not 1) = 9/10
Simultaneous probability from 10 through 9:59, inclusive, of having at least one '1':
= 1  (8/9)(5/6)(9/10) = 1/3.
Proability from 10012:59 of having at least one '1' = 1.0 (3 hrs)
Now do the proper weighted average over the whole clock:
P(net) = [3(1.0) + 9(1/3)]/12 = 0.5.
Again, you were right and your method may be easier. But I couldn't figure out why mine didn't work (first) until I realized there were three hours with the leading '1', and also relegates the next digit to base nine, not ten.
Whew. Fun stuff, but I'd better sign off and do some useful work today! Thanks again. 
bbbz
Jul 31, 2008
 tpg76. After you fixed it, your solution is actually very elegant. Good job. Having to work that one digit as base nine would have tripped me up too. I never realized until now that every digit on a digital clock is effectively a different base. How strange some of the simplest things in our world are. 
bbbz
Jul 31, 2008
 Actually the first "digit" being the pair 01 thru 12 is base 12. Is that right? I think that makes it base 12 if its 112 as opposed to 012 which would be base 13. Second is base 6 and third is base 10 of course. just to clarify what I was getting at with the above comment. 
unklemyke
Aug 13, 2008
 Wow, guys way to overthink a problem! You can actually do it on your fingerswithout considering ANY exotic bases. The hours of 100, 110, 120 and 10 OBVIOUSLY have at least one "1" in each of their 60 minutes. That's 240 minutes, and since those hours recur in the PM, we have a total of 480 minutes each day.
The hours that do NOT have a "1" in their hours digits display "1's" at the 1, :10, :11, :12, :13, :14, :15, :16, :17, :18, :19, :21, :31, :41, and :51  (that's 15 in all) in each of 16 hours (24  for a total of 240 more minutes.
Adding 240 to 480 gives us a total of 720 minutes in which our trusty timepiece displays at least one "1"  no bases required. 
unklemyke
Aug 13, 2008
 Be VERY, VERY thankful that our timepiece did not display seconds!
I'd have to take my shoes off.
Nice Teaser, kratsq! 
unklemyke
Aug 13, 2008
 Interesting.... When I type a colon followed by a zero, it posts as an
 OH, well, DUH of course it does, you dimbulb! (Don't mind me folks  just talkin' to myself...) 
bbbz
Aug 14, 2008
 unklemyke. The solution you showed is exactly the way I initially solved the problem. I was simply complimenting another user for coming up with a different and original solution, which also brought to light something I never realized about clocks and their use of different bases. But yes, you are right. The soution you show is more straight forward and solving this problem does require use or knowledge of other bases. 
bbbz
Aug 14, 2008
 sorry. quick edit. Meant to say: Problem does NOT require use of different bases. 
javaguru
Dec 09, 2008
 Even unklemyke overworked the problem a bit. 1/3 of the hours have a 1 and 1/4 of the remaining 2/3 of the hours have a 1. Thus:
1/3 + (1/4 x 2/3)
= 1/3 + 1/6
= 1/2
This took, oh, like 10 seconds to figure out. 
Juanito
Jun 25, 2010
 javaguru you are soooo smart 
spikethru4
Jan 11, 2011
 The probability that the last digit is NOT a 1 is 9/10.
The probability that the penultimate digit (i.e. first digit of the minutes portion) is NOT a 1 is 5/6.
The probability that the hour portion does not contain a 1 is 8/12, for 12hour displays, or 12/24, for 24hour displays. (The teaser needs to state whether it is a 12 or 24hour display.)
Thus, for 12hour displays, the probability of at least one 1 is:
1  (9/10 * 5/6 * 8/12) = 1/2
and for 24hour displays, is:
1  (9/10 * 5/6 * 1/2) = 3/8 
spikethru4
Jan 11, 2011
 Sorry, 5/8 for 24hour displays, not 3/8. 
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