/sci/ - Science & Math

how is the answer not 50%

Why the fuck would you post that picture?

Now I have to fap tonight

Nice going jackass

>>3800760

You can tie.

>>3800760
Because A's wins have to EXCEED B's

>>3800756
Don't remember exact answer.
It was something close to 47.56%

>>3800760
Because there is a 1/3 chance of a draw.

However the chance of an overall drawn is not 1/3 since you can have 99 draws and 1 victory still count as a victory (or loss) to Person A. Therefore, it must be a bit less than 50% because of the slight chance that number of wins are even.

>>3800765
>>3800768
It's so obvious

>>3800770
I mean, seriously. So obvious. It IS 50%. You stupid rednecks.

>>3800777
Whats the chance the moon explodes tomorrow night for no reason?

50% it either happens or it doesnt hru uururduruurrrrrrrr

>>3800780
Holy fuck, I didn't realise it was so high until now.

>>3800780
>50% chance that the moon will explode

>>3800780
Someone post this on /x/

Posted this MATLAB code yesterday too.

Answer consistently ends up being around 1/3

c=0;
choices=[1 2 3];
for j=1:1000000
A=floor(rand*3)+1;
B=floor(rand*3)+1;
if choices(A)==choices(B)
c=c;
elseif choices(A)==1 &&choices(B)==2
c=c;
elseif choices(A)==1 &&choices(B)==3
c=c+1;
elseif choices(A)==2 &&choices(B)==3
c=c;
elseif choices(A)==2 &&choices(B)==1
c=c+1;
elseif choices(A)==3 &&choices(B)==1
c=c;
elseif choices(A)==3 &&choices(B)==2
c=c+1;
end
end
disp(c/1000000)

>>3800777
Are you implying that my trip wasn't open before I posted.

What if there are 100 draws?

>pic related

>>3800806
what if god exists?

then god is also in your ass

>>3800797
That assumes that a higher number of draws results in the set being a draw. It isn't.

99 draws and 1 win is still a win to Person A.

sauce on the girl GIVE ME

100 games can be divided into three categories: win, lose, draw... so if you enumerate all of them:

0 draw, 0 lose, 100 win
0 draw, 1 lose, 99 win
etc.

then you have 49/100 winning possibilities for 0 draws, 49/99 winning possibilities for 1 draw, 48/98 winning possibilities for 2 draws, 48/97 winning possibilities for 3 draws, and so on and so forth

the total number of winning games is: 49 + 49 + 48 + 48 + 47 + ... + 1 + 1
the total number of games is: 100 + 99 + 98 + ... + 2 + 1 = 101 * 100 / 2 = 5050

the number of winning games is 2 * (49 + 48 + 47 ... + 1) = 2 * 50 * 49 / 2 = 2450

so the winning ratio is 2450/5050 = 48.5%

>>3800843
sorry, I made a mistake, there are actually 101 + 100 + 99 + ... + 1 total games, so rather than 5050 that should be 5151
similarly, there are actually 50 + 49 + 49 + ... + 1 + 1 winning games so 2450 should be 2500

ratio win/loss = 48.5%

sauce on OP's pic ???

>>3800866
>>>/r/9003952

>>3800843
This is fucking my brain over... Can anyone explain this in even more depth?

>>3800863
wait I guess I forgot that a random sampling forms a normal distribution so this might be more complex than I thought

something else that is interesting to notice about this is that there are only 51 kinds of draw, which happens when you have an even number of wins and losses... this can only happen on games where you have an even number of draws, i.e. 0/50/50, 2/49/49, 4/48/48, ... 100/0/0

but of course there is 100 choose 50 ways to draw in the first case and 100 choose 0 ways to draw in the last case

this is so complicated

>>3800912
Bro... I'm a fucking dumbass, how'd you get so smert anon

>>3800928
I love discrete math

Not enough information. We are only told that two people play the game, not how they will determine their moves. For all we know, A could choose scissors every time and B chooses rock.

>>3800912
Had this problem last night. Many people took a shot thinking it was simple but problems cropped up with every possible solution. Someone should propose the question to a math professor.

>>3800980
Assume it's 100% random you pedantic fuck.

It's really tedious. For 10 games the probability of winning is 25048/59049 = ~42.42%

>>3800991
I have a symbolic solution but computing it is difficult. It involves very large numbers and would probably take a very long time for a computer to handle.

Basically, you count the total number of games (3^100) and remove the draws (big complicated expression) and half of what's remaining is the number of winning games

>>3800991
>accuses others of being "pedantic" for needing information necessary to make an accurate mathematical calculation

you fail at science and math

Guys I did it:
If you have a 33% chance of winning a single hand, then that could also be used to calculate 100 hands. Therefore, 100^33% = 1E66. Your chance of winning 100 hands is 1E66.

>>3801006
actually based on how quickly my calculator can compute 100! it is probably feasible

so essentially the total number of games is 3^100 (like tossing a 3-sided die 100 times), that's the easy part

the total number of draws is more complicated, as I demonstrated before... it's equal to 100c0*100c50 + 100c2*98c49 + 100c4*96c48 + ... 100c0*0c0

so in other words
# draws is the sum from k=0 to 50 of 100c(k)*(100-2k)c(50-k)

calculating now

>>3801016
Except the question is obviously a theoretical question with the actual situation involving purely random outcomes.

When someone asks you if you have $50 and want to buy a $20 item, how much will you have left" you don't say "I can't answer that, HE MIGHT GET DISCOUNTS"

Notice how noone else brought it up? Trolling or not, you're an idiot.

>>3801016
You're that guy that thinks his smart because he points out the 'flaws' with a purely theoretical question.

Everyone hates you.

Here's some sauce on OP.

>>3801028
>actually based on how quickly my calculator can compute 100!
Calm down there, buddy.

There are 3 possibilities for each game, so there are 3^100 total outcomes.

We can get the draws using the multinomial theorem.
For each even number n, the number of outcomes that it ends in a draw with exactly n draws is 100!/(n! *((100-n)/2)!((100-n)/2)!). So you just have add the number when n is even from 0 to 100.

Subtract the draws from 3^100, divide by 2.

>>3801047
What were you like in high school

>>3801047
http://www.wolframalpha.com/input/?i=%28%283^100+-+sum+i+%3D+0+to+i+%3D+50%2C+100%21%2F+%28+%282*i%2
9%21+*+%2850-i%29%21+*+%2850-i%29%21%29%29%2F2%29%2F%283^100%29

There we go.

>>3801057
Why 3^100?
If you generalizate to n number of games, it would be 3^n
Taking n=2 => 3^2 = 9
But the possibilities:
W L D
2 0 0
0 2 0
0 0 2
1 1 0
1 0 1
0 1 1

n(possibilities) = 6
Am I worng?

There are three outcomes which, with random drawing, each have a probability of one third:
Win
Lose
Draw

There are a vast number of possible ways person A can win (99 draws, one win; 100 wins; 34 wins, 33 losses, 33 draws, etc). However, this is exactly equal to the number of ways person B can win. Each possible outcome has a probability of (1/3)^100. In addition, there is a smaller number of possible ways that they can draw (51 - up to 50 wins each, down to zero). And so you have 1 = 2*(probability of A wins) + (probability of drawing). The probability of drawing is 51*(1/3)^100 = 9.7*10^-47, and so the probability of A winning the series is (50 - 9.7*10^-45) percent.

>MFW you guys will never work it out

Who sucks now?

-Biofag

OH GOD IT'S MONTY HALL ALL OVER AGAIN!!!

>>3801057
I mistyped the expression, it should be:
sum from k=0 to 50 of 100c(2k)*(100-2k)c(50-k)

I got the same answer in my lisp program, roughly 47.6%

>>3801080
yes, order matters... for those games where you have 2 wins, 2 losses, 2 draws, each of those can come in two different forms, which makes 9 with the other three possibilities

>>3801086
> posts reasonable argument
> screws up subtraction

>>3801100
d'oh I'm an idiot, order matters for the OTHER 3, the ones that aren't 2 of the same

>>3801080
That doesn't account for how many times each happens.

Let 0 represent a win, 1 represent a loss, and 2 a draw.

Then the total number of ways the game can go is as follows.
00, 2 wins
01, 1 win 1 loss.
02, 1 win 1 draw
10, 1 win 1 loss.
11, 2 losses
12, 1 loss 1 draw
20, 1 win 1 draw
21, 1 draw 1 loss
22, 2 draws.

>>3801100
>>3801107
Thanks for explaining that.
High Schooler here, and still forget when orders matter.

Not a math question.
These are human beings we are talking about. One will be beaten, punked, pwned, flustered and probably throw a chair.
My informed opinion (I have been human for some time now) it approaches certainty.

OK this is what i got so far. Check it out.
Don't be too hard on me ;)

NOTE: This is ONLY right if we assume that none of the players has a preference for rock or papper or sis, their choice is aleatory.

Tell me what you guys think.

Also if you could give me your opinion on this i would appreciate it. :)

>>3800553

http://www.wolframalpha.com/input/?i=%28%283^n+-+sum+i+%3D+0+to+i+%3D+n%2F2%2C+n%21%2F+%28+%282*i%29
%21+*+%28n%2F2-i%29%21+*+%28n%2F2-i%29%21%29%29%2F2%29%2F%283^n%29%2C+n+%3D+10

Here's a more general equation. Works for even n's. You'll notice it gives the same exact answer for n = 10 games, as what my program simulated earlier in this thread.

>>3801125
lrn2latex

>>3801125
I'm fairly certain that the answer is 47.56% based on my train of thought beginning here >>3801028
(Please note that I make a minor typo in the expression for # of draws)

The idea is that you count the total number of possible games, subtract the ones that result in draws, and divide the remainder by 2.

>>3801125
>posible
>favouroble

It's impossible to determine statistically. There are too many unknown variables, and Rock, Paper, Scissors is not a straight-up game of statistics. Go play your nearest four-year-old in ten games of RPS. You'll win more than he will.

>>3801125
It's wrong.

We want to look at the number of trinary bitstrings of length 100 that have more 0's than 1's.

The total number of strings is 3^100.
Instead of calculating the number of wins, it's easier to calculate the number of draws. And use symmetry to find the number of strings that count as a win.

A draw can only happen if there are an even amount of draws. We use the multinomial theorem to compute the number of strings with k draws and (100-k)/2 wins and loses. For example. A draw can occur with 30 draws, and (100-30)/2 = 35 wins, and 35 loses. The total number of ways this can happen is 100!/(30!*35!*35!).

We sum the number of draws. Subtract that from the total number of possibilities. That number is the number of ways it doesn't end in a draw. Through symmetry we see that half of those strings are wins, and half are loses.

>>3801166
It is so easy to idealize, that your comment instantly reduces to idiocy.

>>3801166
Yeah, you're a retard.

this question is trivially easy.

its p(not draw)/2

or (1-p(draw))/2

pdraw is bin expansion of probability of draw (1/3) with 100 trials and 50 successes i.e. comb(100,50)*(1/3)^50*(2/3)^50 or about 0.00022 or 0.022% so p(somebody wins) = 0.499889795 (as accurate as my calculator can)

you can also get exact fractions but i dont really like typing large numbers

>>3801238
That's wrong. I'll check my cache from last nights thread. Brb

>>3801238

sorry thats the probability that a wins and prob that b wins. the prob that either wins is double that.

>>3801107
I got the same solution:

ways to tie (w-l-d) look like 0-0-100, 1-1-98, 2-2-96... 50-50-0.
so find the probability of each of those outcomes and add them up to find probability of a tie.

for the first one it is simply (1/3)^100, which is the chance of getting any unique set. for the next one, it is (1/3)^100 * the number of different sets of 98 ties and 1 win each, which is 100 p 2.
For 2-2-96, it is 1/3^100 * 100 p 4/(2!*2!), again 100p4/2!2! is simply the number of ways to distribute two wins each throughout the 100 games. continue the pattern all the way up to 50-50-0 which is 1/3^100 * 100p100/50!50!

subtract the sum from 1 and divide by 2:

http://www.wolframalpha.com/input/?i=%281+-+%281%2F3%29%5E100+*+%28the+sum+from+n+%3D+0+to+50+of+100
%21%2F%28n%21*n%21*%28100-2n%29%21%29%29%29%2F2

= 47.56%

>>3801238
This is wrong. You don't draw if you have 50 draws.

You could have 30 wins, 10 loses and 60 draws. You still have more wins than loses.

>>3801238
no

50 draws and 25 wins each is not the only way to draw.

0 draws and 50 wins each is a draw
2 draws and 49 wins each is a draw
4 draws and 48 wins each is a draw
etc.

Rock Paper Scissors is not a game of chance, there is some skill to it, just like poker or other casino games.

http://www.worldrps.com/

>>3801238
>pdraw is bin expansion of probability of draw (1/3) with 100 trials and 50 successes i.e. comb(100,50)*(1/3)^50*(2/3)^50 or about 0.00022 or 0.022% so p(somebody wins) = 0.499889795 (as accurate as my calculator can)

This calculates the probability that out of the 100 games exactly 50 were draws. It tells use nothing about what the other outcomes were. It could be 50 wins, 50 draws. Or 40 wins 50 draws and 10 losses.

<span class="math">\frac{245121627770311584039909305322618634402319806940}{51537752073201133103646112976562127270210752
2001}[/spoiler]

>>3801337
>>3801309
>>3801337
>>3801337


>It could be 50 wins, 50 draws. Or 40 wins 50 draws and 10 losses.

either of those case somebody won. so a chance A wins is half the total.

Why are people complicating this so much, Just work out the probability of a draw and take it from 1.

This will give the probability of A or B winning, and since they have the same probability of winning dividing this by 2 will give the odds of A winning.

>>3801331
Way to contribute nothing to this thread. In questions like these it's common to assume people play perfectly fair and random, regardless that it's impossible.

Besides we can rephrase the question to be, what's the probability that a randomly generated trinary bitstring of length 100 has more 0's than 1's?

In before ASPIES saying "nothing is truly random."

>>3801358
pic related

already worked on this before the "obvious" symmetry argument oh well

>>3801366
You said that that was the probability of a draw. I showed you it wasn't.

Besides 50 draws, and 25 wins a piece is still a possibility. Which means someone didn't win in that case.

All you did was calculate what the probability is that you have exactly 50 draws. If you have 50 draws you can still either win, lost or have drawn.

>>3801367
This has been done.
Here
>>3801283
and here
>>3801057

>>3801390
http://xkcd.com/297/

There is no difference between the chances of playing 2 games or 100 games, so all you have to do is find the chances of winning playing 2 games.

There are 9 possible outcomes
Draw + Lose
Draw + Win
Draw + Draw
Win + Win
Win + Lose
Win + Draw
Lose + Lose
Lose + Win
Win + Draw

3 possibilities lead to winning
3 to losing
3 to drawing

the chances are 1/3

>>3801283
the wolfram alpha answers overcount 1 case or I undercount one case. Examining the output of the winning partitions my racket code generates, I suspect the draw calculation overcounts 1 case, though I can't figure out why.

>>3801419
>xkcd

>>3801476
Try running it for 10 games, and 2.

Fairly sure your program is wrong, not the actual math.

>>3801464
not true, doesn't even hold up for 4 games:
LLLL - L
LLLW - L
LLWL- L
LLWW - D
LWLL -L
LWLW - D
LWWL - D
LWWW - W
WLLL - L
WLLW - D
WLWL - D
WLWW - W
WWLL- D
WWLW - W
WWWL - W
WWWW - W

probability of win is 5 / 16

>>3801520
roger that, it was wrong, I was undercounting because the stupid "in-range" stream is "up to but not including". Ahh, good old off-by-one errors.

>>3801464
Try it for 3 games then.
WWW win
WWL win
WWD win
WLW win
WLD
WLL
WDW win
WDL
WDD win
LWW win
LWL
LWD
LLW
LLD
LLL
LDW
LDL
LDW
DWW win
DWL
DWD win
DLW
DLL
DLD
DDW win
DDL
DDD

10/31, not the same as for 2 games.

>>3801546
forgot draws

>>3801564
has to be even number of games to be comparable

>>3801591
>has to be even number of games to be comparable
I do not believe you. Where is your reasoning for this? In the odd case, comparing 1 game and 3 games gives you different odds of winning.

>>3801564
> list 27 cases, 10 of which are wins
> 10/31

This is the same as asking what are the odds that 100d3 will be a sum higher than 200.
The answer is .4756

>>3801649
clever

>>3801591
Here's all the results for 4 games. it was too long to post here.
http://pastebin.com/xRbNxci5
>>3801648
Correct, 10/27. It's late, not sure what I was thinking.

In the above a 0 is a win, a 1 is a loss, 2 is a draw.

31/81 is not 1/3.

>>3801649
a winrar is you

>>3801649
*Tries to imagine a fair 3 sided dice, Head explods*

>>3801649
This is nice, but how is wolfram computing this?

>>3801759

>>3801759
3-sided die is just a 6-sided die where 4, 5, and 6 are another set of 1, 2, and 3.

>>3801759
I have a 6 sided die that has 1,2 and 3 on two faces each. Not 3 sided but it would work.

>>3801803
probably with a trinomial distribution or some shit

but it could also be just a large number of simulated trials

>>3801803
I think you could use the normal distribution, but that would only be an estimation right?

>>3801852
Not sure about a normal distribution but a multinomial distribution should be exact.

Disregard my comment, I am homosexual.

>>3801930
True.