Puzzle Time - Election Edition
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You can treat the votes as a weighted random walk, right?. I studies this many years ago, but can't remember a damn thing about the analysis. I Wiki'd it briefly, and then closed the page in horror at how much I've forgotten.
@Doctor-Phibes said in Puzzle Time - Election Edition:
You can treat the votes as a weighted random walk, right?. I studies this many years ago, but can't remember a damn thing about the analysis. I Wiki'd it briefly, and then closed the page in horror at how much I've forgotten.
Yep, thought about that, too. What makes this problem difficult is that it's not a Markov chain: the probabilities change based on previous outcomes.
A standard example in stochastic processes is that of a drunkard who either takes steps towards a cliff with probability p or the other way with probability 1-p, and then to compute the probability that he will eventually fall over the cliff. But that's simpler because it is a Markov chain, I think.
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Here's why:
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Let's call a sequence of votes a path.There's a one-to-one correspondence between the paths that start with H and the paths that start with K but lead to a tie.
Meaning there are just as many of the former as of the latter.
The "successful" paths are the remaining ones.
So the probability of being on a successful path is
1 - 2*(probability of starting with H).
The probability of starting with H is 95/200. Hence
1-2*95/200 = (105-95)/(105-95) = 0.05.
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That was way easier than I thought. Which probably means I screwed up
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Here's how to construct the 1:1 correspondence.
Assume a path that leads to a tie, say
KKHH...
which yields a tie after 4 votes.
Now take every vote until the tie and flip K with H and vice versa.
The remainder stays the same.HHKK...
That's the corresponding path starting with H.
That correspondence works both ways because every path starting with H must eventually be tied at some point (because K has more votes).
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No, there are plenty of cases where K has the lead, loses the lead for a while, and gains it back.
@jon-nyc said in Puzzle Time - Election Edition:
No, there are plenty of cases where K has the lead, loses the lead for a while, and gains it back.
Exactly. Those cases shouldn't count as successful. And I don't count them, since they are among the paths where there is at least one tie in between.
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Let me illustrate that my solution works with a simpler case:
Let's say that K wins with 3 votes against 2 votes for H.
According to my solution, the probability would be (3-2)/(3+2) = 20%.
Let's consider all 10 possible sequences:
HKKKH
HKKHK
HKHKK
HHKKK
KHKKH
KHKHK
KHHKK
KKHKH
KKHHK
KKKHHOnly two of these are successful, namely:
KKHKH
KKKHH2 out of 10; exactly the 20% my formula predicted.
You can also see the 1:1 correspondence of the remaining 8 ones: There's an equal number of paths starting with H and unsuccessful paths starting with K, namely 4 each. Flip at the first tie and you get the corresponding other one. Here are the four pairs of the correspondence.
HKKKH - KHKKH
HKKHK - KHKKH
HKHKK - KHHKK
HHKKK - KKHHK -
Ah, I thought you meant those that ended in a tie. Not those that tied at all. Let me look at it again after Iām done with lunch
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By the way, there's an interesting pattern in the puzzles you post. They seem to be extremely complicated and involve all kinds of advanced maths, but then it turns out there's some kind of trick that only applies in the very specific situation that suddenly makes all the complexity go away and there's a very simple solution. I assume one could also come up with all the formulas to compute the number of distinct successful paths (which would involve Catalan numbers and stuff), divide it by the total number of paths, and, after a lot of algebraic manipulation, end up with the same formula. So the actual puzzle is to find a shortcut to the formula, which, in this case, turns out to be the identification of the path correspondence.
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Your solution is neat.
Another is similar to a cycle lemma proof.
Imagine randomly arranging the votes in a circle and trying to decide where to start in order to satisfy the goal of K always being ahead. You can remove any adjacent pairs KH as they will not lead to an increase in H votes. You are left with 10 Ks at the end of this process, corresponding to starting points that would have resulted in K always being ahead.
10/200=5%
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Your solution is neat.
Another is similar to a cycle lemma proof.
Imagine randomly arranging the votes in a circle and trying to decide where to start in order to satisfy the goal of K always being ahead. You can remove any adjacent pairs KH as they will not lead to an increase in H votes. You are left with 10 Ks at the end of this process, corresponding to starting points that would have resulted in K always being ahead.
10/200=5%
@jon-nyc said in Puzzle Time - Election Edition:
Imagine randomly arranging the votes in a circle and trying to decide where to start in order to satisfy the goal of K always being ahead. You can remove any adjacent pairs KH as they will not lead to an increase in H votes. You are left with 10 Ks at the end of this process, corresponding to starting points that would have resulted in K always being ahead.
You lost me at "You are left with 10 Ks...". Can you elaborate?
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You would finish the process and there would be 10 āKā votes remaining. Had you started counting from any of those 10 spots on the circle than you would have always had a positive number because every H vote you came across would have been preceded (not necessarily immediately) by a canceling K vote.
