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I see what you're saying, but have no idea how to solve for it. If the dog does in fact backtrack, the duck could just make another 90 degree turn, this time "north" and put the dog in another quandary. If the dog again backtracks, the duck can do it again....and eventually get to the shore with the dog still somewhere on the far side. Maybe I'm changing my answer....
My initial thought is that it seems like there's a pretty simple, non-mathy way to see what the answer is. Not 100% sure if my answer is correct though. I'll work on it a little later this morning to see if I can prove it, but won't post it for now so others can discuss their ideas freely.
Pardon my French, but their answer to last week (rather it's explanation) sucks. Seems to me they just tell you it's 4.6 and leave it at that. Even the little video (the first one) has the duck swimming in his little circle, then making a beeline for the shore closest to him (which was my example which gave me pi +1), not your example where the duck would head on a tangent to the shore (is that the way to describe his path?) which ended up being 4.6.
I’m with you on that one... To ensure the calculations were ok, I’ve looked at a two places plane, three places plane (same answer) and when looking at all the four places scenarios – an easy conclusion is drawn for any size plane.
Bah, I got 10 wrong because I forgot to double the number of triangles in the half a hexagon I drew out7/10But I did them all quickly in my head because I was impatient. If I'd actually used the 24 minutes allotted or whatever I definitely would've gotten at least 9, probably 10 of them. I'm the opposite of you I think, I'd definitely have gotten 8 but might've still gotten 10 wrong just by being careless.
Hmm, I went in all mathy and I'm stuck on the third guy. Maybe I'm missing something.
Not sure if this helps, but by choosing a random seat, the first passenger is opening a "loop" so to speak. You'll get your seat if and only if that loop gets closed before someone takes your seat.
Right, but he's got a 1% chance of that loop never opening if he picks his own seat. That first guy also has a 1% chance of picking your seat from the start - a situation which leaves only 1 unhappy passenger, you.
This is correct - and there is an easy way to show the generic n-seats plane probability recursively.
You're correct that right off the bat, there's a 1% chance that the first passenger immediately closes the loop (in which case you get your seat) and also a 1% chance that he takes your seat (in which case you don't). The other 98% of the time he just kicks the can down the road to another passenger. Let's assume that last one is what happens.
Ok, that's spot on. The one thing I didn't see right off the bat is that there would never be "two cans being kicked". Meaning, if the first guy picked the 3rd guys seat and then the 2nd guy gets on and picks the 4th guys. That wouldn't happen as the the 2nd guy would simply take his own seat. At any given time there can only be a max of 1 person being displaced. Once I saw that light, your explanation makes perfect sense.
My wild guess is that the probability of the last guy getting his own spot is 1/(n+1) with n random sitters (i.e. number of 'important seats', all other sitting at their respective spot).
For the sake of simplicity, yes - but I'm not sure that changes the probability outcome if they're the n first to board or placed randomly in the queue.
First one is: I have successfully solved the FiveThirtyEight Riddle!... just by guessing the A is I and translation of other letters, pretty sure the others won't be as fun to play with
Played around a bit and assigned the a-holes randomly among the passengers
I had the same thought. I have no idea where to start on #3.
That one was basically as easy as #1, no idea why the author thought that one would be toughest to crack.
How about a clue on the second? Am I correct in that it's not simply a "letter for letter swap" like the first one? There has to be some other step in the process.
Correct. The first and fourth puzzles are simple substitution ciphers. The second one is a polyalphabetic cipher. The third one is a digram cipher.
And you'd win that bet.I bet you can guess it.
Another hint: It appeared in a previous answer
I thought this would be the ONE THREAD where we wouldn't discuss politics!
Yeah, I thought of that and at first concluded that it didn't matter. But now I'm having second thoughts. Maybe it's not as easy as it seems.
I thought of that initially too - interaction with the other guy... But aren't we maximizing our chance of winning by determining our own optimal cutoff point regardless of the strategy the other guy uses?...
So here's what I've seen via simulation that I'm trying to make sense of:
Top of my head and trying to play devil's advocate here (no empirical validations)... We agree that the optimal 'stand alone' cutoff is at 0.5 - which gives an expected value of 0.625.... If we suppose P2 believes that P1 uses that strategy and he goes for a 0.625 cutoff on his own - we know that it yields an expected value lower than 0.625 (since it is optimal at 0.5) - meaning that P1-0.5 strategy versus P2-0.625 strategy... Results in P1 winning more often than P2 (the opposite of the conclusion above).So here's what I've seen via simulation that I'm trying to make sense of:
A player's expected value is highest when he uses 0.5 as a cutoff. This is intuitive, since if his first number is less than 0.5 he's more likely than not to improve by trying again, and vice versa. Using a cutoff of 0.5 yields an expected value of 0.625.
On the one hand, it would then seem that using any other cutoff would be suboptimal. However, if Player 2 assumes Player 1 is using this simple strategy, then he knows on average he's going to have to beat a score of 0.625 to win. So perhaps he should use that as a cutoff instead (like on the Price is Right, it normally wouldn't make sense to spin the big wheel again if your first spin was $0.90, but if the guy before you already got $0.95 you have to spin again).
And it looks like using a cutoff of 0.625 does indeed result in a lower expected value, BUT it results in Player 2 winning more frequently!
So the obvious response is, if Player 1 knows that this is what Player 2 would do, then he will also adjust to use a cutoff of Player 2's expected value (which looks to be something like 0.617). And then Player 2 would do the same, etc. And this rapidly converges to some value around 0.617 where the two players have roughly a 50/50 shot of winning the prize.
No idea if that's actually correct, I usually start with simulations to get an idea of the answer and then figure out how to prove it analytically, so that's the next step.
I'm guessing the apparent paradox is resolved sort of like a presidential candidate winning the popular vote but losing the election - the magnitude of Player 2's losses end up being greater than the magnitude of his wins, so overall he has a lower average score but ends up with more wins. Something like that, probably.
But Player 2's goal isn't to beat Player 1's average value, right? It's to beat his "naive strategy" of staying on anything above 0.5 (which we already know yields a 0.625 AV).So here's what I've seen via simulation that I'm trying to make sense of:
A player's expected value is highest when he uses 0.5 as a cutoff. This is intuitive, since if his first number is less than 0.5 he's more likely than not to improve by trying again, and vice versa. Using a cutoff of 0.5 yields an expected value of 0.625.
On the one hand, it would then seem that using any other cutoff would be suboptimal. However, if Player 2 assumes Player 1 is using this simple strategy, then he knows on average he's going to have to beat a score of 0.625 to win. So perhaps he should use that as a cutoff instead (like on the Price is Right, it normally wouldn't make sense to spin the big wheel again if your first spin was $0.90, but if the guy before you already got $0.95 you have to spin again).
And it looks like using a cutoff of 0.625 does indeed result in a lower expected value, BUT it results in Player 2 winning more frequently!
So the obvious response is, if Player 1 knows that this is what Player 2 would do, then he will also adjust to use a cutoff of Player 2's expected value (which looks to be something like 0.617). And then Player 2 would do the same, etc. And this rapidly converges to some value around 0.617 where the two players have roughly a 50/50 shot of winning the prize.
No idea if that's actually correct, I usually start with simulations to get an idea of the answer and then figure out how to prove it analytically, so that's the next step.
Expected value doesn't really matter though. What matters is how often your number is higher than your opponents.
I think this will be the key to figuring out the answer mathematically. Whatever the optimum number is, you should have a 50% chance of beating it if you respin on anything below.
