Train Puzzle

Factual Questions
1

Hi all. I received much to chew on last time I posted a James Randi puzzle on this board. So, here we go again - below is the “Train Puzzle”, which seems so straightforward (see my answer below). Is it as simple as I think, or am I being deceived? I would go nowhere else but the SDMB. Thanks in advance for your insight(s).

Here, I quote from James Randi’s Website <www.randi.org>

[Situation: Stan, a young chap living in Manhattan right near a subway stop that goes north-and-south (uptown-and-downtown) is fortunate enough to have two equally attractive and single girlfriends, one who lives in the Bronx (north on the subway) named Nora, and the other in Brooklyn (south on the subway) named Sophie. (Names have been changed to protect the innocent). Stan really can’t make up his mind which of these two lovelies he will ask to marry him. And, when he visits one of them, he never announces that he’s going to be there, because he’s part of this puzzle, and it wouldn’t work if he didn’t behave that way.

Trains in each direction arrive once an hour. Trains bound for the Bronx always arrive at 10 minutes past the hour, and Brooklyn trains always arrive on the hour. (Obviously, this is a fictional situation, because I know from long experience that you’d never encounter such regularity on the NYC subway system.) But this schedule hasn’t dawned on Stan, who just goes into the subway station at any old time, waits on the upper platform until he hears a train (Bronx- or Brooklyn-bound, he doesn’t care) pulling into the station. At that point he goes to the train that’s arrived, boards it, and goes to see the lady who resides in the direction the train is traveling.

We introduce here the wise opinion of our consulting psychologist, who opines that the lady Stan sees most often will undoubtedly be the one he ends up marrying. We will take that as a valid point.

Who will Stan marry? And, part two, what are the exact mathematical chances that he will marry that girl?]

I’m inclined to think that he will marry the Brooklyn girl (I did, and it worked out great!), because for any given random arrival time, he has an 83.33% (5/6) chance of going to brooklyn, verses 16.66% (1/6) Bronx (there is only a ten-minute-out-of-sixty window for Bronx trips). So in the long run he will go to Brooklyn 83.33% of the time, and seeing her more means he will marry her (the faulty reasoning of which I will not go into…)

Am I right? Could it be so simple?

2

Yes.

Yes.

3

Well, I’ve heard from Randi. The puzzle must be more complicated than I thought (whic is what I thought).

To quote Randi’s e-mail to me:

“…that’s not the right answer… Re-read.”

So, what is it? Maybe he does marry the Brooklyn girl, but my percentages are wrong… any takers??

4

Punoqllads - I wish it were so simple!! Seems like I’m (we’re?) missing something. See my above post…

5

Nope, you’re right. Unless you accidentally said “Bronx” instead of “Brooklyn”. I’d recheck.

6

The girl is right. The odds are not.

7

Nope. I checked my e-mail, I definitely said Brooklyn.

Actually, my first answer (ooo, I hope this doesn’t relegate us to the BBQ Pit…) was, “He marries the one with the biggest tits.” Then I got serious. I think Randi has a loophole up his sleeve…

8

Punoqllads, if you have an answer, please share! I’m dying to know! (Or are you just guessing that that’s the case?)

9

I may have answered my own question. There may only be NINE possible bronx-bound minutes (after Brooklyn train leaves, til Bronx comes) which shifts the percentages to 85%/15%.

Those sound neater, I’m thinking maybe that’s what he’s aiming at…

10

If that’s the case, it’s a very poorly worded question, and anyway, it’s still 83.33/16.67. If you give the Brooklyn train from H:00 to H:01 to sit there, you have to give the Bronx train the same extra minute from H:10 to H:11.

Assuming that the likelihood he will marry either girl is the same as the amount he sees her, the proper answer to the question as asked is 5/6, 1/6. Or, 5-1 odds. You were right.

11

The answer is the Brooklin girl, with a probability of 85%. The reason is that there is only a nine-minute time window where he can go to the Bronx. If he arrives at the top of the hour, he can still catch the Brooklin train. Thus the probability is 51/60, or 85%. Am I right?

12

RickJay quote:

If that’s the case, it’s a very poorly worded question,

RickJay - unfortunately, “poor wording” has been a problem for Randi’s puzzles before (see our “Red Hats Puzzle” Thread, sorry I don’t know how to set up a link…)

I don’t know, Diceman, I think I agree with RickJay that there IS a ten minute window because the same rules have to apply to both trains’ arrival/rest times! Let’s assume that the minute of arrival is the only minute during which he can board the train. If he arrives at :11 - :00, he takes the Brooklyn; if he arrives at 01 - 10, he takes Bronx. That’s 50/10!)

But paradoxically, I think that the answer Randi is looking for is 85/10, which in my mind is arguable at best.

13

The odds are damn near close to 100%. I haven’t yet worked out the right limit equation to give the correct value, so I’m not exactly sure what the number is, but for merely 5 subway trips, the odds are over 96%.

For 2N + 1 trips, the odds are:


Sum(i = 0..N; Choose(2N + 1, i) * ([sup]1[/sup]/[sub]6[/sub])[sup]i[/sup] * ([sup]5[/sup]/[sub]6[/sub])[sup]2N + 1 - i[/sup])

where Choose(a,b) = a! / b!(a-b)! and the factorial operator ! is defined for any positive integer k as k * (k-1) * … * 1, and 0! = 1. So, for 5 trips, the probability he’ll marry Sophie is


Choose(5,0) * ([sup]1[/sup]/[sub]6[/sub])[sup]0[/sup] ([sup]5[/sup]/[sub]6[/sub])[sup]5[/sup] + Choose(5,1) * ([sup]1[/sup]/[sub]6[/sub])[sup]1[/sup] ([sup]5[/sup]/[sub]6[/sub])[sup]4[/sup] +
 Choose(5,2) * ([sup]1[/sup]/[sub]6[/sub])[sup]2[/sup] ([sup]5[/sup]/[sub]6[/sub])[sup]3[/sup]
= (5[sup]5[/sup] + 5 * 5[sup]4[/sup] + 10 * 5[sup]3[/sup]) / 6[sup]5[/sup]
~= .9645
14

I think Punoqllads has it. The probability of 5/6 only applies to one trip. The question assumes multiple trips, but doesn’t tell us how many. So we can’t really determine the probablility.

I’m not really capable of doing the math, but the chance Stan will marry Sophie seems to approach 1:1 as the number of trips approaches infinity. But the exact answer is dependent on the actual number of trips.

Then again, why would Sophie want to marry a guy who can’t figure out a train schedule? Brooklyn girls are known for their practical bent. Maybe she wouldn’t consent, so the chance that he’d marry her is not the same as the chance he’ll ask her.

15

Punoqllads, according to the Central Limit Theorem the limit of your expression as N goes to infinity is the same as the limit of the probability that a unit normal random variable assumes a value greater than


((N+0.5)-(2*N+1)*5/6)/(sqrt(2*N+1)*5/36)

And since that expression goes to -infinity as N goes to infinity, the probability that a unit normal random variable assumes a value bigger than that goes to 100% as N goes to infinity.

Of course, there’s probably a simpler answer than that; that’s just the first answer I dug up from my probability textbook…

16

Okay, I have heard more from Randi. He suggests I read carefully what the psychologist said. I take this to mean that, regardless of the exact odds of going to Brooklyn/Bronx in one trip (85/15, 83/17, whatever) he definitely will see the Broooklyn girl more often; he therefore has a 100% chance of marrying (if we take the psychologist’s evaluation at face value). If he saw the Brooklyn girl even 51% of the time, by the psychologists reasoning he will still have a 100% chance of marrying her.

I think this is Randi’s “trick answer”.

17

Welllllll…if he wants to get technical…

So get technical on him. There is still a 15% chance that he will see the Bronx girl more often.

Unless all those neat numbers above are correct. Then he has a 0% chance. Personally, I’d move to Jersey.

18

Drop, posting these questions in their entirety may or may not be within “fair use” for copyright purposes, but they are certainly outside the spirit of fair use.

In the future, please just supply the link if you want to enlist Dopers’ help on these.

19

As a related question, as far as I can tell, there are only three subway stations in Manhattan that one could do this. What are they?

20

If that’s the answer it’s the worst puzzle I have ever read.