## User Tag List

1. G-Virus:

For simplicity, let's imagine that Adam lives exactly 20 minutes from the west terminal and 80 minutes from the east terminal. Let's also imagine that one train is run going in either direction every 100 minutes, both at precisely the same time. If Adam misses the eastbound train by a nanosecond (again, for simplicity), the westbound will arrive in just under 60 minutes. On the other hand, if Adam misses the westbound train by a nanosecond, the next eastbound train will arrive in just under 40 minutes.

So here, while the trains are running as frequently and passing through Adam's station the equal number of times per hour, we still wind up with Adam being, by the luck of the draw, about 1.5 times more likely to end up heading out east.

Is this sounding about right? I'll try to hammer out the math if it seems I'm on the right track. Kudos, though, this is a tough one!

2. You are on the right track you just gotta work a little with the numbers and I am seriously impressed. Great job.

3. Originally Posted by G-Virus
You are on the right track you just gotta work a little with the numbers and I am seriously impressed. Great job.
Damn... I've always had the quirk of being good with logic and horrible with math! All right, I'll try to work out the numbers after getting some rest tonight. (Assuming someone else doesn't beat me to it.)

Edit: my progress thus far:

My thinking was a bit reversed, so let's switch Adam's station to the 10-minute mark from the east terminal. Now, so long as the distance from Adam's station to the west terminal is 90 minutes, for a 9:1 ratio, if Adam misses the westbound, the eastbound will arrive in 80 minutes. If he misses the eastbound, the next westbound will arrive in 20 minutes, giving us the 4:1 ratio. My thinking yesterday was reversed: this being the case, Adam is 4 times more likely to stroll into the station in that 80-minute lull than the 20-minute lull, meaning he will wind up on the eastbound train 4 times more often.

However! Let's assume that some forward-minded train line administrator has come to the revelation that less people would be beating one another up and committing suicide if they weren't packed in like cattle and arranges to have 2 trains run in either direction every 50 minutes. Now, should Adam miss the westbound, the next eastbound will arrive in 30 minutes. If Adam misses the eastbound, the next westbound will arrive in 20 minutes. Adam is still more likely to wind up in the 30-minute lull than the 20-minute lull, but only 1.5 times moreso. Hence there is still more math to be done...

4. Is Mycroft the only genius on here. how about solving this one or giving us some new riddles guys.

5. Originally Posted by G-Virus
Adam is a man-slut (or playa for some) that has two girlfriends he likes equally. Each Saturday afternoon he visits one of them, and in hopes of not having to decide, he used his logic and knowledge of limits to infinity. Since both of them live on the same train line but in opposite directions from his train station, he simply strolls right into the station and hops on the first train to come. If the eastbound train arrives first, he visits Jackie, and if the westbound train comes first, he visits Donna. Since there is an equal number of eastbound and wesbound trains, Adam reasons that in the long run he will visit them with more or less equal frequency. After a few months, he realizes the he is visiting Jackie four times more often than he is visiting Donna. What is Adam' mistake?

(Hint, think of the P and Q values of the random walk and remember that the probability in a gaussian distribution should always equal 1).
Presumably the answer lies in the fact that they live different distances from the station?

6. ED111: Presumably the answer lies in the fact that they live different distances from the station?

Is that your answer? If yes, how would distance acount for the discrepancy?

7. ED111: Presumably the answer lies in the fact that they live different distances from the station?

Is that your answer? If yes, how would distance acount for the discrepancy?

Girl A === Station === === === Girl B

Whilst waiting at the station it is three times more likely that the train will be between the station and Girl B than it will be between the station and Girl A. Therefore it is three times more likely that he will catch the train as it is travelling in Girl A's direction.

8. Originally Posted by G-Virus
how about solving this one or giving us some new riddles guys.
I'll have a new riddle shortly! In the midst of my most busy section of the week presently.

9. Originally Posted by G-Virus
Adam is a man-slut (or playa for some) that has two girlfriends he likes equally. Each Saturday afternoon he visits one of them, and in hopes of not having to decide, he used his logic and knowledge of limits to infinity. Since both of them live on the same train line but in opposite directions from his train station, he simply strolls right into the station and hops on the first train to come. If the eastbound train arrives first, he visits Jackie, and if the westbound train comes first, he visits Donna. Since there is an equal number of eastbound and wesbound trains, Adam reasons that in the long run he will visit them with more or less equal frequency. After a few months, he realizes the he is visiting Jackie four times more often than he is visiting Donna. What is Adam' mistake?

(Hint, think of the P and Q values of the random walk and remember that the probability in a gaussian distribution should always equal 1).
Here is my take on this "riddle":
Since you are calling this a riddle I doubt this is what you are looking for, but this is how I see it:

He only visits them on Saturday and after a few weeks he's visited Jackie four times more often than Donna. So for say twenty weeks he's visited Jackie 16 times and Donna 4. There is actually nothing unreasonable about this at all. Now the expected value (mean) for his trips would be that he visits them each ten times, but when talking about probability you have no guarantee that the end result will be the mean. The mean is only the most likely result, but there are plenty of other possible results. It's even possible that he visits Jackie every time even though there is only a .5 probability of visiting her each time.

So if we take my example of using 20 visits total, then the probability of visiting one 16 times and the other 4 is 20C16*.5^20 = .00462, but one must keep in mind that the probability of visiting them both equally is only 20C10*.5^20 = .176. Therefore with only a roughly 18% chance of visiting them both equally after 20 weeks we can see that Adam's strategy is a poor one if he intends to see them both equally.

Now if he intends to visit them an infinite number of times then his strategy is a good one since this strategy will converge to the mean at infinity, but assuming all three people are mortal he will never be able to manage an infinite number of visits if he is only taking a train on Saturdays.

10. I love all the solutions, I personally took an approach more like that of mycroft. I'll post my answer up soon.

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