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02-23-2018 , 03:20 PM
#526
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Quote:
Originally Posted by eyebooger
Depends on the coin.
Quote:
Originally Posted by e_holle
You need to take that into consideration for the answer.
Also depends on the flipper
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02-23-2018 , 03:30 PM
#527
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Quote:
Originally Posted by xander biscuits
Spoiler:


Starting with n = 1

5n-3,
(5n+2)2
((5n+2)2+5)2
(((5n+2)2+5)2+5)2
...
Basically what I got as well, not sure how to write it more cleanly

Spoiler:

for n = 0 to inf
5n+2
5n+(5n+2)^2
5n+(5n+(5n+2)^2)^2
etc.

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02-23-2018 , 03:34 PM
#528
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Quote:
Originally Posted by ibavly
this is a bad riddle because it completely depends on your priors.
Any answer that can be supported with no a priori knowledge of the coin is a good answer. To get an actual number may require a bit of extra math knowledge, but I think this is very tractable for the people here.
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02-23-2018 , 04:05 PM
#529
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Quote:
Originally Posted by e_holle
Any answer that can be supported with no a priori knowledge of the coin is a good answer. To get an actual number may require a bit of extra math knowledge, but I think this is very tractable for the people here.
If you can 'solve' this I'd expect you to win a nobel prize for disproving Bayes' theorem!
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02-23-2018 , 04:43 PM
#530
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Quote:
Originally Posted by ibavly
Basically what I got as well, not sure how to write it more cleanly

Spoiler:

for n = 0 to inf
5n+2
5n+(5n+2)^2
5n+(5n+(5n+2)^2)^2
etc.
The actual answer is pretty interesting, I'm not sure how you would get here without having an idea of the answer to begin with:

Spoiler:

1 and all multiples of 5 are not in S. Makes sense because no multiple of 5 can fulfill the criteria starting from 2, only 5s interact with other 5 in each of the criteria. Same rationale for 1.

Then you can show manually that you can get to 3,4,6 using the two rules, so by the +5 rules all other numbers are in S

2 ⇒ 49 ⇒ 54^2 ⇒ 56^2 ⇒ 56 ⇒ 121 ⇒ 11 ⇒ 16 ⇒ 4 ⇒ 9 ⇒ 3 ⇒ 36 ⇒ 6
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02-23-2018 , 09:31 PM
#531
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Yep, you can take other lines to get to the same answer, but still a cool answer. I didn't get it either when I saw the problem. My first thought was the answer you gave as well.

Hint if you don't want the answer

Spoiler:
Easier question...

Is 13 in S?
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02-25-2018 , 12:38 AM
#532
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Quote:
Originally Posted by e_holle
If you flip a coin and get Heads four times in a row, what are the odds that you will get Heads the fifth time?
Easy answer based on the style of their other questions is 100%.

Either it's a 2-sided head coin, that you simply flip over once repeatedly, so the next one will be heads.

Or, it's a standard coin, and each flip is a full rotation, meaning it always lands on heads.

That's the only reasonable answer to this question without additional details about the fairness of the coin and fairness of the flipper.
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02-25-2018 , 01:11 AM
#533
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Quote:
Originally Posted by housenuts
Easy answer based on the style of their other questions is 100%.

Either it's a 2-sided head coin, that you simply flip over once repeatedly, so the next one will be heads.

Or, it's a standard coin, and each flip is a full rotation, meaning it always lands on heads.

That's the only reasonable answer to this question without additional details about the fairness of the coin and fairness of the flipper.
That may be. I'm curious about how they approach the problem when they reveal their solution. I think the puzzle is most interesting in how people repond to it. It's been unanimous, I think, in that people are avoiding/evading, rather than trying to come up with an answer. At least you have provided a solution with a rationale in your post just now.

I'm going to say the answer is 5/6, and give my thinking when I have the puzzle setter's solution.

By the way, housenuts, it's good to hear from you. I, for one, have missed your participation.
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02-25-2018 , 01:20 AM
#534
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How about if the problem were put differently?

Suppose a computer program uses a good RNG to produce a number between 0 and 1. If the number is larger than some preset value (unknown to you), the program prints 1, otherwise it prints 0. Now, if the program has printed four consecutive 1s, what is the chance that the next number printed will be 1?
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02-25-2018 , 01:26 AM
#535
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Quote:
Originally Posted by e_holle
How about if the problem were put differently?

Suppose a computer program uses a good RNG to produce a number between 0 and 1. If the number is larger than some preset value (unknown to you), the program prints 1, otherwise it prints 0. Now, if the program has printed four consecutive 1s, what is the chance that the next number printed will be 1?
If the intent of that question was to be equivalent to the question you posed here, the writers failed miserably.
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02-25-2018 , 01:51 AM
#536
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02-25-2018 , 03:26 AM
#537
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Quote:
Originally Posted by e_holle
It's been unanimous, I think, in that people are avoiding/evading, rather than trying to come up with an answer.
Look at some of the difficult problems tackled itt. No one is avoiding it, its just a bad question. The best case scenario is that its a wordplay type question, but I don't think thats the sort of thing most people here enjoy.

Quote:
Originally Posted by e_holle
How about if the problem were put differently?

Suppose a computer program uses a good RNG to produce a number between 0 and 1. If the number is larger than some preset value (unknown to you), the program prints 1, otherwise it prints 0. Now, if the program has printed four consecutive 1s, what is the chance that the next number printed will be 1?
Thats still completely unsolvable. I'm guessing you are assuming the preset value is uniformly distributed between 0 and 1, but there's nothing in the problem to imply this.

Here's an answer: if we have zero knowledge of the prior state of the coin, the MLE is that the next flip will be heads with 100% probability. Accurate, but completely uninteresting.
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02-25-2018 , 03:46 AM
#538
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Here's a similar question I was asked in an interview once, try solving it in your head

you have 2 coins in front of you, one is fair, one lands on heads with 75% probability

you don't know which is which

you pick one up and flip it 3 times, landing HHT

What is the probability that you are holding the fair coin?
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02-25-2018 , 04:32 AM
#539
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done entirely in my head so I'm likely to be wrong but:

Spoiler:
16/39?
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02-25-2018 , 10:28 AM
#540
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My head hurts.

No idea if this is right:
Spoiler:
8/17
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02-25-2018 , 10:50 AM
#541
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I'll go with eb.
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02-25-2018 , 11:45 AM
#542
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Quote:
Originally Posted by eyebooger
My head hurts.

No idea if this is right:
Spoiler:
8/17


That's correct
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02-25-2018 , 12:56 PM
#543
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Spoiler:
(0.53)/(0.752x0.25+0.53)=8/17

I had the right calculation, but obviously did something wrong in my head
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02-26-2018 , 01:33 PM
#544
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Quote:
Originally Posted by ibavly
Look at some of the difficult problems tackled itt. No one is avoiding it, its just a bad question. The best case scenario is that its a wordplay type question, but I don't think thats the sort of thing most people here enjoy.
When do they post answer to this lol question.
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02-26-2018 , 01:59 PM
#545
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Quote:
Originally Posted by housenuts
When do they post answer to this lol question.
Friday, I think.
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02-26-2018 , 03:06 PM
#546
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Maths Help Needed

x3 - xyz = 2
y3 - xyz = 6
z3 - xyz = 20

x, y, z are real

what are the possible values for x3 + y3 + z3?

I've got some answers from Wolfram Alpha, but I don't know how to justify them

https://www.wolframalpha.com/input/?...z%5E3-xyz%3D20

Thanks for any help in advance
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02-26-2018 , 07:39 PM
#547
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Quote:
Originally Posted by xander biscuits
Maths Help Needed

x3 - xyz = 2
y3 - xyz = 6
z3 - xyz = 20

x, y, z are real

what are the possible values for x3 + y3 + z3?

I've got some answers from Wolfram Alpha, but I don't know how to justify them

https://www.wolframalpha.com/input/?...z%5E3-xyz%3D20

Thanks for any help in advance
Spoiler:
Just solve it.
Spoiler:
A series of hints follows.
Spoiler:
hint: set w = x^3
Spoiler:
hint2: get y^3 in terms of w.
Spoiler:
hint3: get z in terms of w,x and y from the first equation
Spoiler:
Q: why is xy not 0?
Spoiler:
A: the first two equations would then give a contradiction
Spoiler:
substitute for y and z into the third equation.
Spoiler:
magic happens and you get an equation in w alone
Spoiler:
which if you are attentive should be a quadratic
Spoiler:
which factorizes as ...
Spoiler:
only open this if you want to see the answer...
Spoiler:
(7w+1)(w+2)=0.
Spoiler:
Now take real cube roots of the solutions for w to obtain the two real solutions for x ... if you want to see x, y and z explicitly. (I'll leave you to see how to get y and z.)
Spoiler:
Or just use w to get y^3 and z^3 and add them all together.
Spoiler:
.. or use w to get xyz & add the three original equations together, giving the sum of the cubes as a real.
Last edited by tchaz; 02-26-2018 at 08:05 PM. Reason: "real"
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02-27-2018 , 02:49 PM
#548
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thank you

it is very much appreciated
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02-27-2018 , 02:51 PM
#549
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Except I didn't really follow this step

Quote:
substitute for y and z into the third equation.
I substituted for z only since all the ys cancelled.

Worked out to the same result thought, so thanks again

edit, unless you mean substitute for y3. If that's the case then yeah, I did that
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02-27-2018 , 03:49 PM
#550
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Quote:
Originally Posted by xander biscuits
Except I didn't really follow this step



I substituted for z only since all the ys cancelled.

Worked out to the same result thought, so thanks again

edit, unless you mean substitute for y. If that's the case then yeah, I did that
Yes - sorry - that's what I meant.

Spoiler:
Let w = x3. Number the equations (1), (2) and (3).
From (1) and (2) we have y3 = w + 4.
By (1) we have -xyz = 2 - w.
Since we can't have xy = 0 (as if xy = 0 then multiplying (1) and (2) would give 0 = 12),
we thus have z = (w-2)/xy.
Hence z3 = (w-2)3/x3y3 = (w-2)3/w(w+4).
Sticking this together, (3) gives us (w-2)3 - (w-2)w(w+4) = 20w(w+4).
The left hand side is (w-2)(w2 -4w +4 - w2 - 4w) = -4(w-2)(2w+1)
and rearranging the equation gives, after some multiplying out adding and factoring ..
(7w+1)(w+2)=0. Hence w = -1/7 or w = -2.
finally
As z3 = 20 + xyz = w + 18 and y3 = w + 4
we get x3 + y3 + z3 = 3w + 22
which is 151/7 or 16
Last edited by tchaz; 02-27-2018 at 04:03 PM.
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