Logical Test : Coinflip

Shoop wrote:

When you know at least one of the coins is heads, you can deduce its not TT, so you are left with :
HH
HT
TH

Wrong it can only be
HH
HT

How can it possibly be TH if the first coin we already know was H?

Tell me he's trolling

VsKatshuma wrote:

Tell me he's trolling

Nvm your right i re-read the OP and it is 1/3 so gg to me. But technically its still 1/2 statistically its 1/3

BEeeH wrote:

turdens -.-

Quote :

There is 4 possible outcomes when you flip the coins.

(h,h) (h,t) (t,h) (t,t)

Given atleast one is heads we can remove (t,t).

That leaves us with 3 alternatives:

Alt 1
Coin A: heads
Coin B: heads

Alt 2
Coin A: heads
Coin B: tails

Alt 3
Coin A: tails
Coin B: heads

Now to the trixers, Alt 2 and Alt 3 arent always possibilities.

We cannot know which coin that's heads, but lets assume that Coin A is. That leaves us with only 2 alternatives because Alternative 3's Coin A = tails. So, IF Coin A was heads, there is 1/2 chance for coin B to be heads aswell.

If we assume that Coin B was heads, it would take out Alt 2, and leave us with only 2 alternatives. The chance would also be 1/2.

50% chance bros

This might make sense to some when you first look at it since it creates an illusion. What it says is basically that if we know coin A is heads then it's 50 % both coins are heads and that if coin B is heads it's also a 50 % chance both coins are heads. Since we know that either A or B is heads it should be a 50% chance of both coins being heads every time since either scenario A or scenario B will happen every time.

The flaw with this theory is that you assume it's 2 separate events. It becomes easy to see if you break it down a little.

Scenario 1 (Coin A is heads) has 2 possible outcomes.

1.(H,T)
2.(H,H)

Scenario 2 (Coin B is heads) has 2 possible outcomes.

1.(T,H)
2.(H,H)


Here's the flaw, the both bold outcomes is the EXACT same outcome. Remember the only requirement to be in Scenario 1 was that coin A should be H. Outcome 2 in Scenario 2 fill this requirement and should therefore be viewed as an outcome in Scenario 1. The flaw is created when you start to view scenario 1 and scenario 2 as 2 separate events.

XXXandBEER wrote:

Shoop wrote:

When you know at least one of the coins is heads, you can deduce its not TT, so you are left with :
HH
HT
TH

Wrong it can only be
HH
HT

How can it possibly be TH if the first coin we already know was H?

I'll answer, ignoring the possibility of a troll.

Quote :

I flip 2 coins and you ask me if at least one of them is heads? I answer yes.

What is the chance that the other coin is heads aswell? (Given that you can trust my answer).

I realized I had it wrong after I re-read the question (didn't bother looking at other posts, la). But if I changed it to 1/3rd I'd look like a newb.

LIST OF SHAME AKA RETARD-MONGOS THAT GOT IT WRONG


1. Chakra
2. Beeeh
3. Beer
4. Lakes
5. Night
6. Trollpro



BOOOOOOOo boOOOOOO BOoooo SHAME ON THEse GEEKS OF HELL!

HALL OF FAME

Sidey Kats Stekoo

YEEEEEEEEEEEEEEEe

In this research paper done by a group at Stanford: http://www-stat.stanford.edu/~susan/papers/headswithJ.pdf

There is a statistical bias in a coin flip such that the side the coin is launched from has a ~51% chance of landing on the launched side. Therefore, assuming the person flipping is not trained to introduce his own bias, and the quarter is a standard U.S. coin with little imperfection, my answer is 51% or 49% depending on what side it started on.

The 1% comes from the fact that the coin will spend either more time in the launched state than the other state, or an even amount of time in the launched state and the other state, but NEVER more time in the other state than the launched state.

Illustration:
HTHTHTHTH
HTHTHTHT

In the first, the coin goes through the series of states and ends up with more time in heads than tails. (5 heads, 4 tails)
In the second, the coin goes through equal time in heads and tails. (4 heads, 4 tails)
If you start with heads, there's no possible sequence where tails will have more time than heads.

Shoop wrote:

Aight kids its time for answer. And yeah many of you were right. It is ofcourse 1/3 chance.

Its pretty simple. As many of you explained, there are originally 4 different possibilities.

HH
HT
TH
TT

When you know at least one of the coins is heads, you can deduce its not TT, so you are left with :

HH
HT
TH

And in just 1 of these 3 cases is the other coin also heads. So 1/3. Well played, Sidey was first with answer and Stek was first with showing he understood the reasons behind the answer.

As EnStekPaNnaWOLF said, HT and TH are the same event. Therefore it reduces down to HH or (HT = TH). This follows from the way you phrased the question: what is the chance the other coin is heads as well. Therefore the ordering of the flips doesn't matter.