/sci/ - Science & Math

>>2313743
1 in 100
10*10=100

Die in a fire.

>>2313743
Are the coins re-flipped if they can't fit the statement? If so, is it after examining all of the coins at each step?

7

This thread is on every single board on 4chan right now.

This proves 4chan is full of GED retards

Amazing that you can post multiple copies of the same troll thread on a single board and still manage to troll people

>>2313763
It sounds like all the flips happen at the beginning, then it's just me and my friend grabbing coins from eachother. I'm having trouble with the first part of the question... what are the odds of at least three coins being heads in ten coin tosses?

1/5

1 in 20

>>2313796
Figuring that out is just too much work for me to bother as I have a headache. The problem is interesting though, if only because it's so complicated, I'd like to see a solution if somebody does work it out.

Assuming a previous prepared statement and reflips at any juncture where the coins don't match the statement:
I don't fucking know.

I was going to write all the permutations of the first 10 here, but there are I 2^10 possibilities or 1024 permutations.

So fuck that.

82/100

fuck you.

The question is:
>What are the probability that both of your friend's two remaining coins are heads?
Therefore, the only thing that matters is that my friend has pick 5 coins at random and at least 2 of them are heads; and then, I take 3 of them and at least 1 is head. (I don't think that what happened before really matters, at least not if we just want a good approximation)
So out of 32 possible combinations, My friend could have:
1x HHHHH
5x HHHHF
10x HHHFF
10x HHFFF
5x HFFFF (no 2 H)
1x FFFFF (no 2 H)

I'll cont. in my next post

Ignoring the fact that this is a troll, did anybody seriously solve it?
I got ~16.18%

>>2313947
1/26... all H
5/26... chances my friend has 2H left: 3/5 out of 5/26
10/26... chances my friend has 2H left: 3/10 out of 10/26
10/26... (my friend doesn't have 2H left)

So... I get about 7/26 or 27%, but that's with making abstraction of the beginning, so it is probably not be correct, but it sould be a good approximation.

fuck you troll
this problem is elementary with matlab
solution in picrelated
result is about 16% btw

# Bruteforcing it, hope there\'s not any bugs...

from fractions import Fraction

def outcomes(n):
    if n == 0:
        yield []
    else:
        for x in outcomes(n-1):
            yield [False] + x
            yield [True] + x

num = 0
den = 0
for coins1 in outcomes(10):
    if sum(coins1) >= 3: # at least 3 heads
        for choice2 in outcomes(10):
            if sum(choice2) == 5: # he chose 5 coins
                coins2 = [coins1[i] for i in range(10) if choice2[i]]
                if sum(coins2) >= 2: # at least 2 were heads
                    for choice3 in outcomes(5):
                        if sum(choice3) == 3: # you chose 3 coins
                            coins3 = [coins2[i] for i in range(5) 
if choice3[i]]
                            if sum(coins3) >= 1: # at least on
e is heads
                                den += 1
                                if sum(coins2) - sum(coins3) == 
2:
                                    num += 1

print(Fraction(num, den))

# I get 32/113 ~ 0.2831858407079646

>>2314006
The program reflips if there aren't the specified number of heads in the coins, correct?

The question OP is making fun of is an effective troll because it's so simple that a lot of people with a limited background in statistics assume they know the correct answer. The question the OP actually posted is an INEFFECTIVE TROLL because all of the idiots start arguments never even show up.

The only people who stick around to figure out the answer to a complex problem are the ones who actually have the mathematics and statistical background required to get the correct answer.

tl;dr - This thread says more about trolls than it says about probability.

>>2314023
There's no need to reflip because I make a table for all possible results and cut the lines without the required number
for example if two coins the table is
11
10
01
00

supposing I cut for sum < 1 i get
11
10
01
after the cut

>>2314007
It's similar to the approximation I got by hand in
>>2313947
>>2313997
I guess this is the correct answer.

>>2314007
A bit embarrassed I didn\'t notice that the randomization of the choice of coins doesn\'t affect the result. Simpler, and I hope it doesn\'t wrap this time...

from fractions import Fraction

def outcomes(n):
if n == 0:
yield []
else:
for x in outcomes(n-1):
yield [False] + x
yield [True] + x

num = 0
den = 0
for coins1 in outcomes(10):
if sum(coins1) >= 3: # at least 3 heads
coins2 = coins1[0:5]
if sum(coins2) >= 2: # at least 2 were heads
coins3 = coins2[0:3]
if sum(coins3) >= 1: # at least one is heads
den += 1
if sum(coins2) - sum(coins3) == 2:
num += 1

print(Fraction(num, den))

It\'s the same answer, but hopefully easier to debug so we can resolve the conflict with >>2314006

>>2314007
A bit embarrassed I didn\'t notice that the randomization of the choice of coins doesn
\'t affect the result.  Simpler, and I hope it doesn\'t wrap this time...

from fractions import Fraction

def outcomes(n):
    if n == 0:
        yield []
    else:
        for x in outcomes(n-1):
            yield [False] + x
            yield [True] + x

num = 0
den = 0
for coins1 in outcomes(10):
    if sum(coins1) >= 3: # at least 3 heads
        coins2 = coins1[0:5]
        if sum(coins2) >= 2: # at least 2 were heads
            coins3 = coins2[0:3]
            if sum(coins3) >= 1: # at least one is heads
                den += 1
                if sum(coins2) - sum(coins3) == 2:
                    num += 1

print(Fraction(num, den))

It\'s the same answer, but hopefully easier to debug so we can resolve the conflic
t with >>2314006

>>2313947
this is not taking into account that 10 coins were flipped and at least 3 were heads
the events hhhhh, hhhhf, hhhfh, etc, should not have equal probability

>>2313743
Poor squidgirl. It'll be okay. We love you even when you're not flipping coins.

Okay, I found a problem in >>2314006
It\'s computing the probability all three of your coins are heads. The problem asks for the probability all two of your friend\'s remaining coins are heads.

>>2314105
oops, my bad
I didn't read carefully
ignore that program then, I suck balls

>>2314073
I know, but if >>2314007 is right, it's a good approximation.
Also, I'm not quite sure, but the chances of my friend picking H instead of F when he pick 5 coins should be a little higher, aren't they?

>>2314105
fixed
28.32% it is

>>2314144
Ha. Surprisingly close to the answer to the original troll question.

BUT WHAT IF IT\'S NOT ANY THREE COINS? WHAT IF IT\'S THE FIRST THREE? THIS PROBLEM IS OPEN TO INTERPRETATION, I READ IT ON WIKIPEDIA!

>you flip ten coins, at least three of which come up heads
Expected number of heads is 3 + 7/2 = 6.5

>friend randomly selects five of those coins, at least two of which are heads
OK, he gets 2 heads, leaving 8 coins, so the expected number of heads remaining is 4.5. He takes three of these so he has an expected 2 + 3*4.5/8 = 3.6875 heads.

>you randomly take three of your friend's coins and find at least one of those coins is heads
All right, you get one of his heads, so now he has 2.6875 heads remaining, out of 4 coins. Taking two random ones halves this, so now he has 2 coins with an expected value of 1.34375 heads. Each coin has a 0.671875 chance to be heads, so this squared is 0.451416015625.

~45%.

>>2314405
>Using the noninteger expected values for discrete distributions