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.
Proud supporter of Plan 2525. Everything you think, do and say will be in the pill you took today.
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I was thinking you could model it as a lattice path of n=94, but only after the first two votes come in. But I haven’t figured out how to work the probabilities in.
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@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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@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 -
DON’T let them eat cake.
- RFK Jr
