Poll

No votes (0%)
2 votes (13.33%)
2 votes (13.33%)
1 vote (6.66%)
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3 votes (20%)
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5 votes (33.33%)
5 votes (33.33%)

15 members have voted

Ayecarumba
Ayecarumba
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Thanks for this post from:
beachbumbabs
January 10th, 2019 at 4:16:18 PM permalink
Quote: beachbumbabs

...What if I'm seated at exactly 12 on your clock face?

Hehe... Then hope it is/isn't a cuckoo? Hey Yoooooooo...
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TomG
TomG
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January 10th, 2019 at 4:17:54 PM permalink

Making a big messy graph of 105 minutes (3 x 5 x 7), I end up with probabilities of 53% for Amy, 28% for Betty, and 19% for Cathy.

I may have made some counting / arithmetic mistakes. In that case, I guess 50, 30 and 20 percents, because it feels right
Wizard
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Wizard
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January 10th, 2019 at 4:33:53 PM permalink
Quote: beachbumbabs

I think the clock analogy is a good one for demonstrating there CAN be ties. The minute hand and the hour hand will be tied 12 times in 12 hours. The minute hand and second hand will be tied every just-over-a-minute. As will the second hand and the hour hand. And all 3 will be together 12 times in every 12 hours. What if I'm seated at exactly 12 on your clock face?



If you're at a random point on the clock, the probability of both hands passing you at exactly the same time is zero -- but it could happen. Like my random number from 0 to 10 analogy, which was discussed in a math paradox thread a while ago.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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January 10th, 2019 at 4:34:46 PM permalink
Quote: TomG


Making a big messy graph of 105 minutes (3 x 5 x 7), I end up with probabilities of 53% for Amy, 28% for Betty, and 19% for Cathy.

I may have made some counting / arithmetic mistakes. In that case, I guess 50, 30 and 20 percents, because it feels right



The first answer was pretty close, but not quite.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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January 10th, 2019 at 4:38:33 PM permalink
Let me reword the problem, to help avoid the confusion about ties.

There are three ants on a circle, both going around the circumference in the same direction, and constant but different rates. It takes ant A three minutes to make a revolution. It takes ant B five minutes and ant C seven minutes to do the same. All three ants started at random points on the circle before you arrived.

You pick a random point on the circumference of the circle too and stay there. What is the probability each ant reaches you first?

Extra credit: Does the answer change if the any given ant goes the opposite direction as the other two?
It's not whether you win or lose; it's whether or not you had a good bet.
beachbumbabs
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beachbumbabs
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January 10th, 2019 at 4:41:46 PM permalink
Quote: Wizard

If you're at a random point on the clock, the probability of both hands passing you at exactly the same time is zero -- but it could happen. Like my random number from 0 to 10 analogy, which was discussed in a math paradox thread a while ago.



Ok, thanks. Sorry to belabor the point.
"If the house lost every hand, they wouldn't deal the game."
Wizard
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Wizard
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January 10th, 2019 at 4:44:30 PM permalink
Yes, congratulations, you're right! I like your solution too, much more intuitive than mine, which involved a triple integral.

I owe you a beer for being first.
It's not whether you win or lose; it's whether or not you had a good bet.
TomG
TomG
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January 10th, 2019 at 4:46:08 PM permalink
Quote: Wizard

The first answer was pretty close, but not quite.



Hopefully I was on the right track, just getting too sloppy on a first try. Took some extra time, made it neater. Amy = 74/138 = 54%; Betty = 38/138 = 28%; and Kathy = 26/138 = 19%

edit: see someone else got it better using an actual math method.
djtehch34t
djtehch34t
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January 10th, 2019 at 4:52:34 PM permalink
Quote: Wizard

Yes, congratulations, you're right! I like your solution too, much more intuitive than mine, which involved a triple integral.

I owe you a beer for being first.



Cool, thanks! My first thought was the triple integral, but I definitely wanted to avoid it :p.
Wizard
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Wizard
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January 10th, 2019 at 4:53:23 PM permalink
Quote: TomG

Hopefully I was on the right track, just getting too sloppy on a first try. Took some extra time, made it neater. Amy = 74/138 = 54%; Betty = 38/138 = 28%; and Kathy = 26/138 = 19%



Again, close but no cigar
It's not whether you win or lose; it's whether or not you had a good bet.

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