Shouldn't it just be 0.5? t. regards /g/ brainlet

I think i speak for everyone on /sci/ when i say;

>>8700077
If you rolled N, the odds of you getting a bigger number is pretty much

(100 - N)/(100)

So solve for

(100 - N)/(100) > 0.5
100 - N > 50
-N > -50
N < 50

pretty easy.

of course it's 0.5

>>8700077
Without knowing the mean and standard deviation, it's not really possible to tell. I mean, I'd recommend going with anything that's above average in that situation, hopefully one that's a StdDev above.

>>8700096
>Gee i wonder what the mean is of a Uniform Distribution

It's .67 (2/3 rounded up)

File: plot.png (979KB, 1360x1186px) Image search: [Google]

979KB, 1360x1186px

>>8700077
Yeah, it's 0.5.
At most, this can increase the chance of winning to 5/8 (assuming the opponent is dumb and either never discards or always discards).

assuming both players numbers are picked from the same pool of numbers, then this gets interesting. then you have to consider each case for each round on their own.
round 1
case 1
you get a GOOD number
chances they get a good number: (NofGOOD-1) / NofBAD
result 2
you get a BAD number
chances they get a GOOD number
(NofGOOD) / (NofBAD-1)

these chances affect the second round also, you are picking from the same pool after all.
you have to consider all the 16 possible outcomes, for each round you may get:
GG, GB, BG, BB

>>8702083
How many numbers are there between 0 and 1? infinite?

>>8700077
Fuck off dumb anime poster

>>8700077
It's phi-1

>>8702126
uncountably infinite, so even if you have an infinite amount of players and they all get good numbers your chances are no worse than if there were no other players

>>8702126
Sure, but an infinite number can not be displayed on the screen. We do not know how many, but we know there is a specific amount, that is why we can use the amount of numbers as a variable. The number may be 0.00...1.00
or 0.01...0.99 or 0.0000...1.0000 or 0.001....0.999
does not matter, there still is a set amount of numbers. Be it 100, 98, 10000 or 998 does not matter. There is an N. The problem calls for a group of numbers 0 < N < 1 or 0 <= N <= 1 that was not clear, but what was clear is that the numbers are displayed on a screen, thus they are limited in length.

File: plot.png (7KB, 427x238px) Image search: [Google]

7KB, 427x238px

>>8700077
>>8702335 is correct. If you just want to get the highest possible number on average, you should switch if you get below [math]\frac{1}{2}[/math]. However, you are not trying for the highest average, you are trying to beat the other person and it doesn't matter by how much you win or lose. If the enemy sticks with [math]\frac{1}{2}[/math], your odds of winning depending on when you choose to switch follow the graph in pic related.
By instead switching when below [math]\frac{7}{12}[/math] you can actually increase your chances of winning to [math]\frac{97}{192}[/math]. Of course, your enemy can adjust to this and switch if below [math]\frac{277}{456}[/math], to which you can adjust again etc.
The limit of this is [math]\frac{1}{2} \left(\sqrt{5}-1\right)[/math].

>>8702479
is there no number that has higher chance of winning than phi1 and why

>i.e. what if i knew that my opponent chose phi1

>>8702479
>Of course, your enemy can adjust to this and switch if below 277/456
How would this be in any way advantageous to my enemy? It will lower his average outcome compared to switching at .5.

File: plot2.png (6KB, 374x234px) Image search: [Google]

6KB, 374x234px

>>8702668
Depending on what your opponent does, [math]\phi -1[/math] may not be the perfect choice (like the above plot shows, the best option in case that are sure that his number is [math]\frac{1}{2}[/math], using [math]\frac{7}{12}[/math] would give an even higher advantage, for example). However, [math]\phi -1[/math] is the best strategy in so far that it is the only number that guarantees a chance of [math]\geq \frac{1}{2}[/math] to win no matter what the opponent chooses. This also means that if your opponent uses [math]\phi -1[/math], the best choice is to use it yourself as well, giving you a [math]\frac{1}{2}[/math] chance to win.
Pic related shows the chance to win using[math]\phi -1[/math] as a function of the enemy's choice: as you can see, the curve always stays above [math]\frac{1}{2}[/math].

>>8702697
Im not whom you ask, but i guess that maybe knowing nonsmooth distribution of your opponent's numbers you can take small advantage of it by switching a bit above his switching point. But im too lazy to calculate it myself

>>8702748
That first sentence should read "Depending on what your opponent does, [math]\phi -1[/math] may not be the perfect choice (like the above plot shows, in case you are sure his number is [math]\frac{1}{2}[/math], [math]\frac{7}{12}[/math] gives an even higher advantage)"

>>8702748

Could you elaborate on how you're making these graphs, or provide a more thorough mathematical basis for the problem? I'm finding it hard to get my head around the answer.

>>8702753
>>8702758

nvm, here is a nice explanation:

http://mindyourdecisions.com/blog/2016/04/19/538s-hot-game-show-riddler-explained-game-theory-tuesdays/

>>8701548
Well, I take this back.
Looking more carefully at the plot, the answer does indeed look to be phi-1.