The puzzle doesn’t rule out the possibility of either of the ladies visiting Stan. And if they, also, arrived at their stations at random times, how does that affect the odds? And you have to factor in the odds of one of them not being home, because they’re in transit, assuming each visit is x number of minutes long? And then you also have to factor in the time it takes to get home.
And does he really marry the lady he sees “most often”? I would think it would be “for the longest accumulative duration.” What if he visits the Bronx for 2 hours each time, but only stays in Brooklyn 1 minute?
Punoqllads method is right on.
For 9 visits there is a 99.105% chance that Stan will visit Sophie more often. For 11 visits there is a 99.5391% chance that Sophie will be visited more frequently. Since Randi doesn’t give us the exact number of visits we have to make an assumption of how many visits occur. If I were Stan, I’d have to see a girl about a million times before I proposed, but I’m afraid of commitment. Let’s assume Stan would need to see a girl at least 20 times before he proposed to her. 39 trips would guarantee that the girl he visited most often was visited at least 20 times. At 39 trips the chance that Sophie would be visited more often is, for all practical purposes, 100%. (The odds are about 100% for any number of trips over 11)
My answer would be that Stan would marry Sophie, and that the exact odds of this happening are 100%.
Actually, with 39 visits (which is an admittedly arbitrary number), the odds of Nora being visited more frequently is about 1 chance in 1,394,000. So even if Nora is a 1 in a million girl, the odds are against her.
So did Randi ever say what his answer was?
Nope, not yet. But he DID tell me that 85%/15% and 83.33%/16.66% were wrong. I wrote back with my guess of 100%, based on the psychiatrist’s observation (which he told me to consider) that he would DEFINITELY marry whoever he saw more often. He did not respond - which he may also have done to a third wrong answer, but I think that’s what he was looking for.
I’ll update this thread on Monday when he posts his answer.
I still say there’s way too much data missing to reach any sort of “accurate” answer. Any conclusion has to rest on too many assumptions (and even 1 is too many.).
My guess is that the lady he sees MOST often would be the lady selling the subway tickets; therefore there’s a 100% chance that he will marry her!
The rest is all just a smoke screen…
Well, James Randi has posted his answer on his website (scroll up for link, it’s http://www.randi.org).
Turns out my most recent guess was right - Randi says there is a 100% chance that Stan will marry the Brooklyn girl, based on what the psychiatrist said. (Of course, I only guessed it because he said the other percentages were wrong…) He did concede that he should have said that Stan makes at least a dozen trips, to make it clearer…
He posted a new puzzle that I’m curious to get people’s opinions on. It involves a “coin-flip” bet that the flipper can’t lose. If you read it, please post your suggested answers!
Since Manhattan objected about posting the entire puzzle, here’s a summary of the new one (I’ll try to ensure I get all the valid points). The original URL is here.
Stan has dinner with friends. Makes the following offer:
Flip a coin. Heads, we each pay our share, tails, I pay the whole bill. Asks a guy in the group to call it, then tosses the coin.
Stan never has to pay the whole bill, only his own. The guy who’s asked to call it is not a confederate. And Stan never tries the trick twice with the same group of friends. What’s the secret?
I believe the key to the puzzle is the fact that Stan asks someone to call the toss after he’s already stated what he’s going to do (heads=we all pay, tails=I pay). However, I haven’t figure out how this translates into never losing. Here are the following possibilities:
Guy Calls/Toss Lands:
Heads/Heads - this permutation would seem to guarantee that he pays
Heads/Tails or Tails/Heads - conceivably these two could be a cop-out
Tails/Tails - Obviously never pays for all
He also says “Think deviously, think tricky. Be bold. Our Stan can save the situation.” Not sure if that’s part of the puzzle or not. Thoughts?
Whoops, I made a mistake in my table. Should have said:
tails/tails would seem to guarantee that he always pays;
heads/heads he wouldn’t have to pay.
At any rate, for any given dinner, there would still seem to be a 25% chance that he’d have to pay the entire bill.
Frogstein - actually, I started a whole new thread to deal with the new puzzle -
http://boards.straightdope.com/sdmb/showthread.php?threadid=43312
see you there!
I have to comment on Randi’s answer to this.
It is a load of crap. He asked for the exact odds. A 100% chance of marrying Sophie means that there is no chance of marrying Nora. There is always a small chance that he will marry Nora no matter how many visits he makes. I stated above that Nora had about a 1 in 1.4 million chance if there were 39 visits. Sophie’s odds are real close to 100% but the “exact” odds are not 100%. Randi seems to have a problem with poor wording. He admitted to it in the three hats puzzle, said he should have specified more than 12 visits for this puzzle, and apparently rewrote the coin puzzle after posting it to the internet. These puzzles aren’t any fun if the guy posting them (Randi not DropOfaHat) keeps screwing up.