Anaru
Inkling Cadet
You're not supposed to guess a letter twice in succession. I'll let it slide this time. There's no R.
_ _ A _ (1/4)
O R E (3/10)
_ _ A _ (1/4)
O R E (3/10)
Hangsquid!
- Thread starter Cuttleshock
- Start date
Dual
Inkling Cadet
Sorry, no one else is up. If you're still accepting, my last letter is S.
Cuttleshock
Inkling Commander
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- Apr 1, 2016
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- 459
I'll guess S in Dual's stead! Or, uh, scratch that - T is more my style.
Right, yes, that's what I thought. I know I say that all the time but I only say it if it's true.
Anaru
Inkling Cadet
Cuttleshock
Inkling Commander
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Or you could just spell 'otter', English for my second-favourite mammal. It took me a few tries to get it to recognise my α but hey, I'm bookmarking that website! Maybe it'll replace always having to open Word for capital umlauts and stuff.
... wait, no, it doesn't seem to have Latin letters, or else just refuses to recognise them.
And it only has 110 characters, Anaru. Where'd you get that figure from?
... wait, no, it doesn't seem to have Latin letters, or else just refuses to recognise them.
And it only has 110 characters, Anaru. Where'd you get that figure from?
Anaru
Inkling Cadet
Hmm. That seemed inaccurate because I was testing it and it didn't seem to have very many. I'm not sure how I missed that page containing the letters, I see it now. I'm very confused where I got over 700 from.
Oh, here's a better one, Shapecatcher. It claims to have over 11,000. So it's much better, Mausr is now obsolete.
It seems like some of them don't have copyable text along with the image, though. That's sort of strange.
Oh, here's a better one, Shapecatcher. It claims to have over 11,000. So it's much better, Mausr is now obsolete.
It seems like some of them don't have copyable text along with the image, though. That's sort of strange.
Cuttleshock
Inkling Commander
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- 459
Anyone wanna guess?
Ikaheishi
Bobble-chan
Anaru
Inkling Cadet
More people guessing! And a correct guess! But how'd you forget the thread? I posted in the Signature Description thread, and my signature has a link to this thread in it!
S _ A _ (1/4)
O T T E R (English for a semiaquatic fish-eating mammal of the weasel family, with an elongated body, dense fur, and webbed feet) (4/10)
I just finished a fun probability problem, now I can finally do my actual Math homework. After checking all my alerts.
S _ A _ (1/4)
O T T E R (English for a semiaquatic fish-eating mammal of the weasel family, with an elongated body, dense fur, and webbed feet) (4/10)
I just finished a fun probability problem, now I can finally do my actual Math homework. After checking all my alerts.
Cuttleshock
Inkling Commander
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- 459
In that case, I'll get a P.
A few months back, I came up with and solved a probability-related problem which I really loved. So... you have two inputs connected to an output through an OR gate: if either or both inputs are ON, the output is ON. You wish to switch the output OFF. It is ON to start with. By pressing an input, you randomise its state, regardless of what it was before. In attempting to turn the output OFF, you decide to press inputs 1 and 2 alternately until it goes off. What is the expected number of button-presses required to turn everything OFF?
Minor spoiler, but I'm pretty sure you need to know the general term for the Fibonacci sequence to solve this; I couldn't come up with any way without using it. That term is:
F(n) = c(a^n - b^n), where:
a and b are the two roots of x^2 - x - 1 = 0 and a > b
c = 1/sqrt(5)
One logical follow-up to this question would be to ask if there's any more efficient method to switch everything OFF (like, say, always pressing whatever's most likely to be ON, but that might just be the method that we used). I don't know!
A few months back, I came up with and solved a probability-related problem which I really loved. So... you have two inputs connected to an output through an OR gate: if either or both inputs are ON, the output is ON. You wish to switch the output OFF. It is ON to start with. By pressing an input, you randomise its state, regardless of what it was before. In attempting to turn the output OFF, you decide to press inputs 1 and 2 alternately until it goes off. What is the expected number of button-presses required to turn everything OFF?
Minor spoiler, but I'm pretty sure you need to know the general term for the Fibonacci sequence to solve this; I couldn't come up with any way without using it. That term is:
F(n) = c(a^n - b^n), where:
a and b are the two roots of x^2 - x - 1 = 0 and a > b
c = 1/sqrt(5)
One logical follow-up to this question would be to ask if there's any more efficient method to switch everything OFF (like, say, always pressing whatever's most likely to be ON, but that might just be the method that we used). I don't know!
Dual
Inkling Cadet
Is there an N?
Anaru
Inkling Cadet
That “minor spoiler”, unless my logic is extraordinary flawed, is one of the most misleading “hints” I’ve ever been given. How did you solve it? I'm pretty sure what I did works just fine without having to use a sequence of Fibonacci Numbers, that seems unnecessarily confusing. The problem seems fairly simple.
Two of the starting situations seem to basically be switching to OFF twice in a row, when the starting value is (OFF,ON) or (ON,ON), which has a 2/3 chance, I can expect 6 presses to get me to (OFF,OFF), because there is a 50% chance to press ON, and a 50% chance to switch to OFF, which leads to a 25% chance to get ON then OFF, and a 25% chance to get OFF then OFF.
N = expected amount of flips needed to get to OFF twice in a row. I add the amount of presses used to fail in that particular way to N and multiply it by the probability of doing that and then add them all up.
N = (.5)(N+1)+(.25)(N+2)+(.25)(2)
N = (.5N+.5)+(.25N+.5)+(.5)
N = .75N + 1.5
To solve for N, subtract .75N from each side
.25N = 1.5
Then multiply each side by 4
N = 6
So in two of 3 cases, the expected amount is 6
For the other case, (ON,OFF), it would be similar, but on the first press, there would be a 50% chance for there to already be two OFF’s. That’s sort of a strange situation, because there’s only one opportunity for that to happen. It’s a 50% chance, so I think I could just divide the last equation by 2 and get the answer, 3 alternating presses.
N = 6/2
N = 3
Then I’d average all of those N values
6+6+3 = 15
15/3 = 5
So I’d expect to need to alternatively press the switches 5 times before if turns OFF. I’m fairly confident that this is the most efficient way to turn it OFF, as if you press a switch twice in a row, the first of those presses pretty much becomes obsolete, and you’ve wasted a press.
Back on topic! None of those guesses are in this word.
S _ A _ (1/4)
R E P T O N sounds like it could be a word, I'll search for it. Hmm... it seems to be a village in the Southern district of Derbyshire, England (6/10)
くコ:
Ikaheishi
Bobble-chan
Is there an L?
Anaru
Inkling Cadet
Yes, there is! Just one letter left.
S _ A L
R E P T O N (6/10)
くコ:
S _ A L
R E P T O N (6/10)
くコ:
Dual
Inkling Cadet
I'll guess sail. Can someone else guess saul?
Anaru
Inkling Cadet
I thought you said you were good at spelling! And isn't Saul a Proper Noun? I don't think those are allowed.
S _ A L
R E P T O N SAIL (7/10)
くコ:ー
S _ A L
R E P T O N SAIL (7/10)
くコ:ー
Cuttleshock
Inkling Commander
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Plausible letters are C, H, K, M, U, V, W, Y and Z... none of which make a word I know. A few others are also pronounceable but just don't seem realistic. I'll ask for H, anyway.
Very interesting! I found the problem a lot more complicated, taking a few hours of calculation and ending on the answer 16/3. My father ran a computer simulation of the set-up and got an average result equal to that within a few significant figures (after 1,000,000 trials, I think), so I'm not backing down! First thing I did was make a somewhat extensive probability tree, and the pattern of recurrence in that brought about Fibonacci numbers.
G1ng3rGar1
Inkling Fleet Admiral
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- Inkopolis, born and raised!
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- SW-7732-8500-1233
So I stopped getting notifs for this…I'll see what we have and take a guess
I've got nothing. And Dual's previous guess had the A in the wrong place :P
How about an U just to get rid of the definite vowels
I've got nothing. And Dual's previous guess had the A in the wrong place :P
How about an U just to get rid of the definite vowels
Last edited:
Dual
Inkling Cadet
Yeah I meant to type in sial but I misclicked. Don't know why I put Saul in there.
Anaru
Inkling Cadet
S _ A L
E H N O P R T U SAIL (9/10)
くコ:テ
I was sure that I was correct about the first part, but I was also pretty sure that dividing by 2 wasn't quite what I needed to be doing near the end. I had a few other ideas, but none of them seemed to make much sense. One thing would have led me to the solution of 16/3, if I had calculated the expected amount of switches if that problem contained only one switch, then averaged that with the problem containing 2 switches starting in neutral positions (If they started in neutral positions, there would be an expected amount of 6 switches, just like in the cases of (ON,OFF) and (ON,ON)). And I suppose it does seem somewhat logical. I think when I tried that the first time, I had thought that it wouldn't work because doing the second expected press for the problem with one switch wouldn't be possible, but also, I'm now thinking that if the first press did work, then the second part of it with 6 expected presses wouldn't be possible to start, so I guess those situations cancel each other out? It doesn't make a whole lot of sense to me but it's the only thing I can think of leading to the solution you got.
For one switch:
N = (0.5)(N+1)+(0.5)(1)
N = .5N+.5+.5
.5N = 1
N = 2
(2+6)/2 = 4
(6+6+4)/3 = 16/3
And I tried using this logic to solve for a problem with three switches, where you press switch one, then two, then three, then repeat that until they all turn OFF.
For three switches:
N = (2/3)(N+1)+(2/9)(N+2)+(2/27)(N+3)+(1/27)(3)
N = (26N/27)+(13/9)
N = 27(13/9) = 39
4 situations (ON,ON,ON), (OFF,OFF,ON), (ON,OFF,ON), (OFF,ON,ON) where you must get 3 presses of OFF in a row.
2 situations (OFF,ON,OFF), (ON,ON,OFF) where if the first 2 presses are OFF, then it outputs OFF, otherwise you need 3 in a row.
1 situation (ON,OFF,OFF) where if the first press is OFF, then it outputs OFF, otherwise you need 3 in a row. I think you just have to average 39 with 2 for this, rather than averaging 39, 6, and 2, because a situation where you only need to press OFF twice in a row is absolutely impossible to reach here.
(39+39+39+39+((39+6)/2)+((39+6)/2)+((39+2)/2))/7 = 443/14 ≈ 31.64
And that does seem like a plausible answer, so doing the problem this way could work. But I don't know how I'd be able to accurately check if it does, unless you're able find the answer to it using something similar to what you did originally and seeing if we still have the same answer for a problem with 3 switches.
E H N O P R T U SAIL (9/10)
くコ:テ
I was sure that I was correct about the first part, but I was also pretty sure that dividing by 2 wasn't quite what I needed to be doing near the end. I had a few other ideas, but none of them seemed to make much sense. One thing would have led me to the solution of 16/3, if I had calculated the expected amount of switches if that problem contained only one switch, then averaged that with the problem containing 2 switches starting in neutral positions (If they started in neutral positions, there would be an expected amount of 6 switches, just like in the cases of (ON,OFF) and (ON,ON)). And I suppose it does seem somewhat logical. I think when I tried that the first time, I had thought that it wouldn't work because doing the second expected press for the problem with one switch wouldn't be possible, but also, I'm now thinking that if the first press did work, then the second part of it with 6 expected presses wouldn't be possible to start, so I guess those situations cancel each other out? It doesn't make a whole lot of sense to me but it's the only thing I can think of leading to the solution you got.
For one switch:
N = (0.5)(N+1)+(0.5)(1)
N = .5N+.5+.5
.5N = 1
N = 2
(2+6)/2 = 4
(6+6+4)/3 = 16/3
And I tried using this logic to solve for a problem with three switches, where you press switch one, then two, then three, then repeat that until they all turn OFF.
For three switches:
N = (2/3)(N+1)+(2/9)(N+2)+(2/27)(N+3)+(1/27)(3)
N = (26N/27)+(13/9)
N = 27(13/9) = 39
4 situations (ON,ON,ON), (OFF,OFF,ON), (ON,OFF,ON), (OFF,ON,ON) where you must get 3 presses of OFF in a row.
2 situations (OFF,ON,OFF), (ON,ON,OFF) where if the first 2 presses are OFF, then it outputs OFF, otherwise you need 3 in a row.
1 situation (ON,OFF,OFF) where if the first press is OFF, then it outputs OFF, otherwise you need 3 in a row. I think you just have to average 39 with 2 for this, rather than averaging 39, 6, and 2, because a situation where you only need to press OFF twice in a row is absolutely impossible to reach here.
(39+39+39+39+((39+6)/2)+((39+6)/2)+((39+2)/2))/7 = 443/14 ≈ 31.64
And that does seem like a plausible answer, so doing the problem this way could work. But I don't know how I'd be able to accurately check if it does, unless you're able find the answer to it using something similar to what you did originally and seeing if we still have the same answer for a problem with 3 switches.