Posted March 14, 2019
![Lifthrasil](https://images.gog.com/e2259851c6e86609b166b01345176fdc7b1fed6147789a7fb8042dc24095c8da_forum_avatar.jpg)
Lifthrasil
Bring the GOG-Downloader back!
Registered: Apr 2011
From Germany
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JoeSapphire
Consultant Liar
Registered: Jun 2011
From United Kingdom
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Lifthrasil
Bring the GOG-Downloader back!
Registered: Apr 2011
From Germany
Posted March 14, 2019
I neglected one case in my above formula, also it contains a mathematical error. There are actually two special seats one has to consider. The chance that 'Vitek' sits in his own seat was covered. But there is also the chance that he sits in the seat of the last person. In which case the chance to get it correct is 0.
So the correct formula would be: P(n) = (1/n)*1 + (1/n)*0 + ((n-2)/n)*P(n-1). Sure, the (1/n)*0 falls out of the equation since it's 0. But the -2 instead of the -1 makes a big difference. It allows to solve by forward induction instead of working backwards over sums (which I neglected to do formally correct).
We know that P(2)=1/2
Now let's take the above equation while assuming that we know that P(n-1)=1/2
P(n) = 1/n + ((n-2)/n)*(1/2)
= 2/2n + (n-2)/2n = (2+n-2)/2n = n/2n = 1/2
So, if any P(n) is 1/2, then all higher P(n+m) are also 1/2 and we know that for n=2 the assumption P(n)=1/2 is true.
So the correct formula would be: P(n) = (1/n)*1 + (1/n)*0 + ((n-2)/n)*P(n-1). Sure, the (1/n)*0 falls out of the equation since it's 0. But the -2 instead of the -1 makes a big difference. It allows to solve by forward induction instead of working backwards over sums (which I neglected to do formally correct).
We know that P(2)=1/2
Now let's take the above equation while assuming that we know that P(n-1)=1/2
P(n) = 1/n + ((n-2)/n)*(1/2)
= 2/2n + (n-2)/2n = (2+n-2)/2n = n/2n = 1/2
So, if any P(n) is 1/2, then all higher P(n+m) are also 1/2 and we know that for n=2 the assumption P(n)=1/2 is true.
![JoeSapphire](https://images.gog.com/a490d1df3c59d6cfc9f3621bc26316c4a94e1db0cc982d64629b643a8766acd3_forum_avatar.jpg)
JoeSapphire
Consultant Liar
Registered: Jun 2011
From United Kingdom
Posted March 14, 2019
![avatar](http://images.gog.com/49a6dca56c974f870f544ec04ded133f39b1e20ef314499dfe21e91abda65914_avm.jpg)
So the correct formula would be: P(n) = (1/n)*1 + (1/n)*0 + ((n-2)/n)*P(n-1). Sure, the (1/n)*0 falls out of the equation since it's 0. But the -2 instead of the -1 makes a big difference. It allows to solve by forward induction instead of working backwards over sums (which I neglected to do formally correct).
We know that P(2)=1/2
Now let's take the above equation while assuming that we know that P(n-1)=1/2
P(n) = 1/n + ((n-2)/n)*(1/2)
= 2/2n + (n-2)/2n = (2+n-2)/2n = n/2n = 1/2
So, if any P(n) is 1/2, then all higher P(n+m) are also 1/2 and we know that for n=2 the assumption P(n)=1/2 is true.
glad I could help. ;)
![ZFR](https://images.gog.com/8d7a9f712d0a0e0f7c0ba66102e81933a36652ad0a2a17b9bc4734d717abd952_forum_avatar.jpg)
ZFR
I love gold!
Registered: Jan 2010
From Ireland
Posted March 14, 2019
And now I'm embarassed for notchecking your solution thoroughly...
I looked at the final answer, saw the word "induction" and skipped the maths.
But yeah, in general the first drunk either
sits in his own seat. Probability 1/n. Problem solved.
sits in last guy's seat. Probability same 1/n. Problem unsolvable.
sits in someone else's seat. Say kth person.
If the last option happens, then all people up to k sit in their own seats and k becomes... the new drunk. Again, he either sits in the original drunk's seat. Or in the last seat. Or in someone else's who becomes the new new drunk... etc
So for any n there is an equal chance of solving it, making it unsolvable or reducing it to an equivalent problem with a smaller number of people.
I looked at the final answer, saw the word "induction" and skipped the maths.
But yeah, in general the first drunk either
sits in his own seat. Probability 1/n. Problem solved.
sits in last guy's seat. Probability same 1/n. Problem unsolvable.
sits in someone else's seat. Say kth person.
If the last option happens, then all people up to k sit in their own seats and k becomes... the new drunk. Again, he either sits in the original drunk's seat. Or in the last seat. Or in someone else's who becomes the new new drunk... etc
So for any n there is an equal chance of solving it, making it unsolvable or reducing it to an equivalent problem with a smaller number of people.
![JoeSapphire](https://images.gog.com/a490d1df3c59d6cfc9f3621bc26316c4a94e1db0cc982d64629b643a8766acd3_forum_avatar.jpg)
JoeSapphire
Consultant Liar
Registered: Jun 2011
From United Kingdom
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ZFR
I love gold!
Registered: Jan 2010
From Ireland
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yogsloth
GRAAAAAAH!!!!!
Registered: Dec 2013
From United States
Posted March 14, 2019
It's legitimately distressing to me that I can no longer follow the math. I feel like less of a man.
I brute-forced n=2, 3, and 4, and they all came up 50%.
I brute-forced n=2, 3, and 4, and they all came up 50%.
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JoeSapphire
Consultant Liar
Registered: Jun 2011
From United Kingdom
Posted March 14, 2019
(you can make yourself feel better by claiming to have checked someone's working and telling them it all adds up. Makes you look very smartbrains.)
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Bookwyrm627
ADD Jumping Bean
Registered: Nov 2013
From United States
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my name is catte
i touch your foods
Registered: Mar 2010
From United Kingdom
![ZFR](https://images.gog.com/8d7a9f712d0a0e0f7c0ba66102e81933a36652ad0a2a17b9bc4734d717abd952_forum_avatar.jpg)
ZFR
I love gold!
Registered: Jan 2010
From Ireland
Posted March 14, 2019
Viiiiiiiiiiiiiiiiiteeeeeeeeeeeeeeeeeeeek
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my name is catte
i touch your foods
Registered: Mar 2010
From United Kingdom
Posted March 15, 2019
Good news everybody! I've found the trail of our missing mod.
Someone just needs to go and threaten the seller until they tell us who the buyer was.
Someone just needs to go and threaten the seller until they tell us who the buyer was.
![JoeSapphire](https://images.gog.com/a490d1df3c59d6cfc9f3621bc26316c4a94e1db0cc982d64629b643a8766acd3_forum_avatar.jpg)
JoeSapphire
Consultant Liar
Registered: Jun 2011
From United Kingdom
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JoeSapphire
Consultant Liar
Registered: Jun 2011
From United Kingdom
Posted March 15, 2019
if we can just post in this thread 3000 more times we can get the reply count to spell 'boob'
Well, 2999 now.
Go!
Well, 2999 now.
Go!