If 1 billion people, all each flip a coin 100 times...

If 1 billion people, all each flip a coin 100 times, what are the chances that at least 1 of them will get heads on every single flip? Also say how you got your answer.

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en.m.wikipedia.org/wiki/Gambler's_fallacy
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email your professor that you just inquired for the answers on a mongolian gong sharing forum and that you would like a 0 on the assignment

50/50% Either they didn't do nuffin or they don't think it be like it is, but it do

Very small.

Start by 0.5^100

And then this very small nimber x 100 billion I assume

Roughly 7.9x10^-22

If one person flips a fair coin once the outcome of getting heads is [math] 1/2 [/math], if he flips it again the probability is the same so the probability of getting 2 heads in a row would be [math] 1/2 \times 1/2 = \left ( 1/2 \right )^{2} [/math] doing this 100 times then is clearly [math] \left ( 1/2 \right )^{100} [/math] now just multiply by 10^9 [eqn] \frac { 10^{9} } { 2^{100} } = \frac { 2^9 \times 5^9 } { 2^{100} } = \frac { 5^9 } { 2^{91} } [/eqn] Which is much less than 1.

1,000,000,000(.5^100)

1 - ((1-chance of getting 100/100 heads)*1 billion)

no

yes

en.m.wikipedia.org/wiki/Gambler's_fallacy

Leave it to Veeky Forums to make getting head sound unexciting.

Where to learn probability and stats?

In middle school.

Isn't it

[eqn]\sum_{k=1}^{1000000000} {{1000000000} \choose {k}} * (0.5^{100})^{k} * (1-(0.5^{100}))^{1000000000-k}[/eqn]

Probably bungled the commands but I think this should be it.
There is probably a much easier way to do this.

lmao dude

I like these odds!

What the hell

If I flip a coin 10 times, what are the odds that heads will land EXACTLY 5 times?
If I flip a coin 10,000 times, what are the odds heads will land exactly 5,000 times?

The chance of getting heads 100 times is 0.5^100

You have a billion chances of doing so so you multiply by one billion

0.5^100*1000000000

Wolfram alpha gives:

7.888609052210118054117285652827862296732064351090230 × 10^-22

Pretty low my negro.

this is correct

It is rather blatantly not.

It needs to be to the power of 1 billion.

(1-(0.5^100))^1000000000 is the chance for a perfect run of no heads
1 - that is the chance of getting at least one.

That would be if we wanted to see the odds of no one getting a heads in 1 flip each.

It's still a non zero chance that everyone would come up heads

The correct answer is
1 - (1 - (1/2)^100)^(10^9)
It's the probability that 10^9 people all don't get 100 heads in a row negated.
¬∀¬P = ∃P
Multiplying by a billion is obviously wrong. It would mean if you got enough people the probability would go over 1 which is ridiculous.

>1 - (1 - (1/2)^100)^(10^9)

I took that as you wrote it and just pasted it on wolfram alpha

The result is
7.888609052210118054114174145192043233517423708099153... × 10^-22

Literally the same as mine.

Congratulations, you are a smart idiot. You were not born an idiot but you made yourself into an idiot by thinking you were smarter than me.

It's a fluke, your number is wrong.

0.50 ^^ 100. 7.8% chance.

>If 1 billion people, all each flip a coin 100 times, what are the chances that at least 1 of them will get heads on every single flip? Also say how you got your answer.
If quantum mechanics are taken into consideration, the probrability is 100%.

If not, very low.

One coin can be head or tails, so 2 options
Possible results: 2^100
Chance for a specific result:1/2^100
Chance with 1 billion people: 1/(2^100:1000000000)
>basically zero

FPBP

This is rather an IQ question of sorts, a mind-bending question that does not require advanced technical skills in maths, per-se.

The trick is to realise that the number of times the coin is flipped is irrelevant. The odds are 50/50 no matter how many times you do it.

So the chance of getting head on 100 flips is the same as the chance of getting heads on a single flip, ergo 50/50.

Thus if a billion people flip the coin 1 time, the chance they will get heads on every flip is 50/50.

Now since a billion people are flipping the coin 100 times, ergo 1 person, 100 times, you convert the 50/50 ratio into math. 50/50 = 1.

Thus the chance of one person getting heads on every flip out of a billion people flipping the coin 100 times is 1.

Tricky good question

Cheers

If you have 2 persons

What are the chances either of them gets heads in their first attempt?

10e9*.5^-10e2

I got the same expression for the probability as
but the numbers are hanging in mathematica on my computer.
appears to be close to the number I'm coming up with. In fact, as I increase the powers of 10 from 5 to 6 to 7, the probabilities go up from 7.886 * 10^-26 to 7.886 * 10^-25 to 7.886 * 10^-24. Seems reasonable that the actual answer will be 7.886 * 10^-22

This board is such shit

>if you got enough people the probability would go over 1 which is ridiculous.
What? How's that ridiculous?

Oh boy. I hope you're trolling.

...

This user gets it.

The answer is always 50/50. Don't fall for the trolls in the thread OP.

It really doesn't matter how many people are flipping coins, its still going to be really tiny.

I duno

bump

if a trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion trillion people do it, it would be big

A probability just tells you [good scenarios / possible scenarios]

For anyone interested

The chances are 50/50 when 10^30 people do it

>chance

Shit, dude, there's a 50/50 chance that all of them get heads every single time.

This question is pretty much too tedious to answer since you said "at least 1". That means everything up to all 1 billion people can get it. Thus, you would have to do a sum of a billion nCr calculations. I can probably solve it with a program but I don't want to devote that much processing time

Are you retarded?
Let's say the number of flips is 2 instead of a billion. What is the probability of getting 2 heads?
Hint: it's not 50/50

It is obviously 50% if you actually think about it. But I guess that is too much to ask for on summer Veeky Forums nowadays

ITT: brainlets think there is a 50% chance of getting 100 heads in a row as long as there is another person present

see

first post best post

second post second-best post