/sci/ - Science & Math

>>9649459
I'm not reading all that.

>Reminder: /sci/ is for discussing topics pertaining to science and mathematics, not for helping you with your homework or helping you figure out your career path.


>If you want help with your homework, go to /wsr/ - Worksafe Requests.

If the number is greater than or equal to .5 then you keep it. If its less than .5 you risk it for the biscuit.

>>9649543
false

>>9649547
prove me wrong.

>>9649459
There's some kind of recursive solution - knowing the other player sets 0.5 as his cutoff... something like that

>>9649459

P(2nd number > 1st number | 1st number) = P(2nd number > 1st number, 1st number) / P(1st number). But P(1st number) = 0. So you should always re-roll, even if you got a 1 in your first roll.

Brainlets.

>>9649459
0.75

>>9649549
0.5 gives you the advantage, but only before the switch. You'd be giving the opponent two chances to pass 0.5, while you stayed at your value.

Its like playing Blackjack, if ur hand is under 11 then you HAVE to hit or you risk losing. But if you hit and only get sometbing like a 2, hit again. Unless you get 16 then stop or its like gambling with your life on the line.

Man I suck at this kind of math

It's root 2 divided by 2

>>9649459
>2018
>probability
>>>/his/

>>9649459
0.61803

>>9649459
Chance of winning given your cutoff is a and your opponent's is b is

0.5 a^2 b - 0.25 a b^2 + 0.5 a b - 0.75 a - 0.5 b^2 + 0.5 b + 0.5 when b > a

and

-0.5 a^2 b + 0.25 a b^2 - 0.5 a b + 0.75 a + 0.5 b^2 - 0.5 b + 0.5 when b < a

This is maximized when a = (b^2 - 2 b + 3)/(4 b)

The minmax is when b = sqrt(3)

sqrt(3) is the optimal cutoff.

>>9649749
Too much effort for a troll. You're just retarded.

>>9649758
More like too little effort, sqrt(3) is greater than 1

>>9649775
By effort, I meant length, but yeah

>>9649459
Monte Carlo sim with 500,000 samples for different cuttoffs shows no significant differences in win rates with respect to changing the cutoff all close to 50% for both players and no obvious pattern. The greatest difference in win rates was 0.3%. So if there is a dominant strategy it doesn't add that much of an advantage even if the other player isn't being rational

>>9649786
Following up and trying to find an analytical solution suggests that this problem is similar to the unexpected hanging paradox.
https://en.wikipedia.org/wiki/Unexpected_hanging_paradox

So the perfectly rational players recursively increase there cuttoffs toward 1 until they reach 1 which is the equivalent of just having 1 chance since the first chance is guaranteed to be rejected.

>>9649800
It's not about trying to win every time, but giving yourself the best chance to win given the first number. You can use the most optimal strategy and still lose.

there literally is no optimal number. the probablillity you will get any number on the line is equal. so you could get 0.1 and roll again and get 0.05, so you fucked yourself. this will fuck u up as often as it will give u mad gainz, so there really is no strategy to implement. this is the nature of a uniform distribution. derp.

>>9649812
Monte Carlo doesn't lie. There is no strategy that gives a statistically significant effect on your chance of winning.

phi-1

>>9649645
Its not about the number of chances, its about your own chances to increase vs decrease your number.

If you only have one potential shot, you should not attempt to get a higher number unless your chances of doing so are better than half.

>>9649459
vaguely remember seeing something like this in "mathematics of poker". nash equilibrium is like 0.632 or something

at first glance you might think "anything above 0.5 is a favorite to win"
but the intuition is that opponent never rerolls above a certain threshold but always rerolls below some other threshold
thus there is an increased likelihood that you will face a number higher than 0.5.

>>9649845
lel fair play for coding a little experiment, but that really was not necessary as I stated here: >>9649842

you can use the analytic definitions of a uniform distribution to prove this rigorously without having to run any trials.

lets put it another way. you get to flip a coin either 1 or 2 times every heads is +1 every tails is -1. so now you can see that it doesnt matter if you flip once or twice, your average score over many rounds is 0

>>9649645
let a be the chance point where switching is optimal
let b be the chance that your opponent's number is larger
b = (1 - b) + a * b
2*b = 1 + a * b
both players have the same strategy, so b = .5
1 = 1 + a * .5
0 = a

You should always switch

>>9649842
>the probablillity you will get any number on the line is equal. so you could get 0.1 and roll again and get 0.05, so you fucked yourself
>either you lose or you don't so there's no optimal strategy
I've never posted a wojak with caved-in skull before, had to go find this just for you

>>9649870
its made to be irrelevant whether you switch or not

>>9649871
>if a coin flip is heads 10 times its more likely to be tails

>pic related is you being a pseud who should stick to nintendo switch

>>9649459
If you get a lower than average number you have above a 0.5 chance of increasing your number by switching.

If you get a higher than average number you have less than a 0.5 chance of increasing your number by switching

So is your goal to maximize your number, or to maximize the rate at which your number exceeds that of your opponent? Do these two problems have the same optimal strategy?

>>9649870
Wait. Shit.

let a be the point where switching is optimal
let b be the chance that your opponent's number is larger
b = (1 - a) + a * b
b = 1 - a + a * b
both players have the same strategy, so b = .5
.5 = 1 - a + a * .5
- .5 = - .5 * a
1 = a

You should never switch

>>9649459
this question is such dogshit ofc there is no number that will give one player an advantage because they would both use that number

You must pick last

>>9649881
>rolls a 0.01

>>9649881
>>9649870
The question asks for what number you should switch at
>Always switch
Rolls 1. Switch?!
>Never switch.
Rolls 0. Switch?!

>>9649876
>in a uniform distribution from 0 to 1, the chance of drawing a number higher than 0.1 is 50%, either it's higher or it's not
>n-nintendo switch
back to /v/ nigger

>>9649882
Is it really this simple? That sounds like an obviously correct answer

>>9649889
>rolls again
>gets 0.001 this time

derp

>>9649882
If you had to play this game for your life, you wouldn't at least favor switching for some numbers rather than others? If some cutoffs are more optimal than others, than there has to be a most optimal cutoff, even if multiple cutoffs tie for the most optimal.

>>9649881
>both players have the same strategy, so b = .5
Why isn't b based on your roll?

>>9649895
if its a uniform distribution, there is an equal probability to draw any number from the set.

>what is a real number line

>>9649896
it's not that simple
the correct answer is [math]\frac{\sqrt 5-1}{2}\approx0.618034[/math]

>>9649895
this is you

>BUT THERE IS MORE NUMBERS BETWEEN 0.1 AND 1 THAN 0 AND 0.1 REEEEEEEE MUH PEBBLES IN A BAG.

>>9649900
>>9649896
Assume the optimal cutoff (the number when chosen gives a statistically significant higher chance to win) is x
player 1 picks x
player 2 picks y
player 1 wins more often
both players are rational
player 2 uses x from now on
But this situation contradicts our definition of x.

>>9649921
Okay, maybe there is no way to win more than the other person once an equilibrium is reached

But there is a way to win less than the other person: by choosing a non-optimal cutoff

What is the equilibrium? It surely doesn't involve switching if you get 1.0, and it surely involves switching if you get 0.0. And it can't hurt to switch at 0.5. And it can't hurt to switch if you get 0.1 either. And it can't hurt to switch if you get 0.4 either. So is the equilibrium 0.5?

>>9649910
>>9649905
>the length of the interval doesn't matter
Letting negroes use the internet was a mistake

>>9649921
https://en.wikipedia.org/wiki/Nash_equilibrium

>>9649921
(cont)
if this is single player going for the highest number then there is an optimal number to switch but because the OP is two player no such number exists.

>>9649921
>optimal
I don't think you know what that word means

>>9649935
>When someone thinks that they learned something

>>9649938
So why doesn't it help to switch if you get 0.001?

>>9649940
I literally defined it (the number when chosen gives a statistically significant higher chance to win)

ok lets restate it AGAIN for you dumb fucks who cant understand.
>if you roll a 1 on a dice, whats the probabillity you will roll a 1 next time?
re rolls are not giving you any exta probabillity of rolling higher than one other than by giving you more chances to roll a 6.

so to answer. if you got zero then roll again, if you got one, then dont. everything else is pointless.

>>9649944
Because if you and you opponent both decide to switch if you get .001 or lower the outcome after that switch is random

>>9649960
>>9649959
Wrong, if you have a 6 sided die and you roll a 1, you should always re-roll. If you roll a 2 you should also re-roll. So there is an optimal cutoff number, this doesn't imply it's a number which will give you an advantage over the other player, but it does mean it will avoid you letting them take an advantage over you

extended:
the only way to make a decision is to try to guess their threshold at which they will not re roll.
if your first roll is lower than the guess, then roll again for another chance to go better.

Let x be your cutoff point. If your initial roll is below x, you re roll. Your expected Value is:
(x/2)*P(<x)+(x+x/2)*P(>x)
P(<x)=x^2
P(>x)=(1-x)+(1-x)x
Therefore your expected value is 1/2(x+1-x^2)
You maximize your score at x=.5 (EV=.625)
However your opponent is thinking the same way. Therefore your x is now .625 so on and so forth. This a roundabout way of looking at a 1-d iterated function where f(x)=1/2(x+1-x^2).

Fortunately there is a fixed point (where f(x)=x) in (0,1) at a=(sqrt(5)-1)/2 furthermore |f'(a)|<1 therefore a is a sink *

therefore assuming both players are rational they will both choose a cutoff point of around .618
*Chaos an Introduction to Dynamical Systems by Kathleen T. Alligood et al. pg 10

>>9649960
Think geometrically. 0.001 is much closer to 0 than 1. There's a greater probability that a point will lie to the right of 0.001 than to the left. First roll is bad luck, try again.

>>9649947
It gives the highest possible chance to win, given y's strategy
If y cuts x's absolute winrate down by also playing optimally, as long as x's strategy is winning more than any other strategy he could play, it's still the optimal strategy

>>9649944
Obviously it does
And obviously you shouldn't switch if you get 1 (a mind-numbingly stupid conclusion unironically suggested by some anons ITT)

>>9649959
>dumb fucks
But you're wrong too.

>>9649963
>this doesn't imply it's a number which will give you an advantage over the other player, but it does mean it will avoid you letting them take an advantage over you
This.
by extension, if they fail to play that strategy, your strategy automatically exploits theirs

>>9649966
Did you post this >>9649906 ?

>>9649965
But if their cutoff is 0.99 you don't have to try to surpass it, because there's a small chance they'll get 0.99 on the first number. I can't believe I have to explain this

>>9649972
No I saw the post early on and wanted to work it out by myself. But I am glad someone else got the same answer.

>>9649972

>>9649947 (You)
I am saying that there exists no way for one player to obtain the prize more often than the other player

ok i was being a brainlet. the average roll is 0.5. so if you are lower, your likely to do better to have another go, if you are above then take your chances.

im not a brainlet now because this the the only safe way to play.

>>9649459
TLDR so its 5

>>9649966 is right below is the python code
import random, math


def game(cutoff):
score=random.uniform(0,1) #original roll
if score >= cutoff:
return score #Don't reroll if you like your score
else:
return random.uniform(0,1) #only one Reroll as per rules

#Everybodyelse's Guess
Brainlet_Cutoff=.5
#Chaos Theory Anon's Guess
Chad_Cutoff=(math.sqrt(5)-1)/2

#Initial Scores
Brainlet_Score=0
Chad_Score=0

#How many games you want to try
trials=10000000

for i in range(trials):
Brainlet_Num=game(Brainlet_Cutoff)
Chad_Num=game(Chad_Cutoff)

if Brainlet_Num > Chad_Num:
Brainlet_Score +=1
elif Chad_Num > Brainlet_Num:
Chad_Score +=1
#if happen to tie, then it dosen't score.

#print(str(trials-Brainlet_Score-Chad_Score))#should be 0
print("Brainlet's Score: " + str(Brainlet_Score)) #4,956,477
print("Chad's Score: " + str(Chad_Score)) #5,043,523
Chad_Ratio=Chad_Score/trials
print( str(Chad_Ratio))
stdev=math.sqrt(trials*Chad_Ratio*(1-Chad_Ratio)) #1,581.078927456533
print("Standard Deviations Over: " + str((Chad_Score-Brainlet_Score)/stdev))#55.05481003407597

>>9652388
After running the game for 10,000,000 times, the Chad that uses [math]\frac{\sqrt{5}-1}{2}[/math] as his cutoff point will beat the Brainlet who uses .5 as his cutoff point 50.4% of the time. Furthermore we are over 27 standard Deviations over the null hypothesis of a 50/50 chance. Therefore we can reject the null hypothesis.

It's looking like 0.5 from here.

>>9652475
why can't opponent re-roll?

>>9649845
What criterion are you using to test statistical significance?

>>9652591
Don't know. Tried again; I need to debug this but it's late so I'll do that tomorrow. Preliminary glance shows that there does seem to be a sweet spot around ~0.61

>>9653104
Also since both players are rational it feels like there's going to be some kind of infinite regress when choosing their thresholds if they aren't openly disclosed ahead of time.

>>9652411
Sure, you can reject any null hypothesis based on a p-value with a sample of 10,000,000 you brainlet

This isn't as much of a probability game as it is a psychological one. If you were to participate in such a game the real cut-off "threshold" isn't strict constant numerical value eg: 0.5,0.61 etc but it should be something like > estimated opponent's threshold.

>>9649459
0.999...

>>9649459
Anything below 0.5?

Mash dat button!

>>9649618
No, 1 would be the optimal number

There's a lot bigger chance you'd get < 1 if you press the button again that that you'd end up back to that optimal number '1'

Getting a higher number than your opponents unknown number basically means you want a as high as possible number

So don't press again when there's a 99% chance of going down, that's not a smart thing to do

>>9649855
Assuming the other person isn't as smart as me, because most people aren't,

There's a 50/50 chance they'd press again when they got < 0.5
Meaning that their chances slightly improve over 0.5

I don't know how to calculate that number but that would be my optimal cutoff

I would have to account for the fact that making my cuttoff higher I increase my chances of getting a lower number on the second press

Using my Intuition, cuz I don't know how to do the math here, I'd say 0.585

@9653533

>>9649971
>There's a greater probability that a point will lie to the right of 0.001 than to the left.
No there isnt. There is a infinite amount of numbers on the left and right (as you put it) of 0.001, the chance is equal you would get a smaller number as a bigger number.

>>9653521
intuitively yes, but you did not take into account that you are in a fight with another player.
See 0.5 as a heuristic lowerbound?

>>9649459
A random number between zero and one l, eh?

Given that there an infinite number of numbers between and two figures no matter what those figures are, it is approaching certain that I will be given a number between 0 and 10^-(10^(10^(10)))...

I lose.

>>9654014
You can represent the numbers between 0 and 1 with a number line. Say that you draw the line 10 inches long. Common sense tells you a random number could be anywhere on the line, and that the average should be around 0.5, not near the ends. Yet you say that it's certain the point will be hardly distinguishable from 0, neglecting that it could be close to 1, or 0.5, or 0.3256, or 0.716 just as likely.

Fact: if the machine is able to pick a number, then the pool of possible numbers to pick from is finite.
You can't pick a truly random number between 0 and 1 the same way you can't pick a random number between 0 and infinity.
That is assuming a standard uniform distribution.

Thus, if your number is 0.1, there are more numbers after 0.1 than before it, so you should switch.

>>9654144
That is, of course, if the numbers are regularly spaced between 0 and 1.

>>9654039
It is equally certain that the number chosen will be infinitely long, but that's the point, you cannot two outcomes be certain, never mind an infinite number of outcomes be certain.

Normally this would be fine, but as we are asked to choose a single figure to be a cut-off point, it generates problems with the question itself.

>>9649749
i thought it was the sqrt of 42?

hint: look at the distribution of random numbers 0 to 1. what does ye old graph look like captain?

>pick random elements of an infinite set with standard distribution

do brainlets really do this?

>>9653533
>There's a 50/50 chance they'd press again when they got <0.5
not quite accurate but at least you show some intuition

Correct answers here
>>9652411
>>9649966
>>9649906

this thread is proof that /sci/ is really shit, i'm out

>>9649459
Thanks OP, saving this for when I teach Monte Carlo simulation in R