Two Coins: Probability

[DeletedUser16008]

The odds dont change its still 50% not 33.333333 etc

You cant have 3 other possibilities with one coin seeing as the other one has already been decided ie heads...

Read the question as it is without making up anything else and its a simple 2 option heads or tails... thats 50/50 .. not rocket science or 1/3

 

[DeletedUser9470]

The odds dont change its still 50% not 33.333333 etc

You cant have 3 other possibilities with one coin seeing as the other one has already been decided ie heads...

Read the question as it is without making up anything else and its a simple 2 option heads or tails... thats 50/50 .. not rocket science or 1/3

ok then. try it yourself
i found this experiment done already:

here is a nice sample space, of tosses for a pair of coins, with 78 entrants

TH HT TH HH HT HH
HH TT TH HT TT TT
TT TH TH HH TH TT
HH TH TT TH TH HH
HT HH HH TT HH HT
HT TT HT TH TH HH
TT TT TT HH TT HH
TT TH TH HH TH HT
HT HH TH HT HT TH
TT TT TH HH TT HT
TH HH HH TH TH TT
TT HT HH TH HT TT
HT TH TH HH HT HH

19 of these are TT so they cannot be used. That leaves us with 59
Of those 59, 21 of them are double heads. 21/59 is almost 36% very close to 1/3 not so close to 1/2

Also you can now see that HT and TH are 2 completely different outcomes that are separately counted and not ruled out.

If you don't trust me try it yourself.
the problem starts by flipping 2 coins, not by knowing one coin and flipping the other...

 

[nashy19]

I knew that would happen if TT was ignored. That's why I said at the start "however I'm not sure whether TT should just be ignored, because there is a 50% chance one coin is tails, in which case the next coin must be heads."

One of them has to be heads, so should TT be automatically TH/HT?

21/78 = 0.26, close to 25%.
My intuition has been saying a quarter forever but I can't work out exactly why.

 

[DeletedUser9470]

I knew that would happen if TT was ignored. That's why I said at the start "however I'm not sure whether TT should just be ignored, because there is a 50% chance one coin is tails, in which case the next coin must be heads."

One of them has to be heads, so should TT be automatically TH/HT?

21/78 = 0.26, close to 25%.
My intuition has been saying a quarter forever but I can't work out exactly why.

"at least one is heads, whats the poss both are?" = out of all the "at least one is heads" possibilities, what percentage would be HH?
= 1/3


Awesome brain teaser man. most people cannot/wont understand it.
this question is one of the questions my maths teacher posed us right at the start of the probability maths course he made us learn, it takes a lot of thinking to get your head round probability and personally I failed. This teaser though is pretty basic compared to what is thrown at you afterwards.

thing with probability laws is that they actually verify irl.

 

[DeletedUser]

-Neo-, I think you should stop debating it; you're starting to make sense to me, and I'd hate to have to admit that I might be wrong!

 

[nashy19]

"at least one is heads, whats the poss both are?" = out of all the "at least one is heads" possibilities, what percentage would be HH?
= 1/3

If it's taken to mean one is heads no matter what, as a condition on top of the usual probability. The condition would not affect the likelihood of double heads because there's already a sufficient amount of heads, but it would affects TH/HT putting them at 37.5% each and 25% for HH (which is confusing because that's the same as a normal two coin flip).

If you take it as "one of the coins is heads" (which it actually says), as in one of the flipped coins landed on heads, then it would be 50%.

Does that cover everything?

The language might still be confusing

 

[DeletedUser]

If quantum mechanics were to apply, the coin could be in additional state, namely, the superposition of heads and tails. Therefore, it's a 25% chance...you're right nashy.

 

[nashy19]

If quantum mechanics were to apply, the coin could be in additional state, namely, the superposition of heads and tails. Therefore, it's a 25% chance...you're right nashy.

I'll leave it up to you to work out the chance of quantum physics applying to the coin tosses.

 

[DeletedUser]

I guess I'm the only brave enough to.

 

[DeletedUser15057]

Supposed that a coin tossing machine was set up to toss both coins up with exactly the same force applied on the same area of the of the coins, for them to spin exactly the same number of turns before both landing on exactly the same spot every time in order to get a heads result for both coins.

How would that then affect the probability??

 

[DeletedUser]

Okay, here's where your mistake is Neo:

The chance is 25% that HH would come up.

You are told one of them is heads. It is then asked, "what is the chance both are heads?"

Well then, if you consider you know one of them is heads, then the remaining coin is what you're determining, which puts it at 50% heads or tails.

If you ignore the information they provided, that one of them is heads, then the chance both of them is heads is 25%.

If you know one of them is heads, but you are ONLY determining the chance both would be heads, it is STILL 25%.

So, as you can see, it is NEVER 33% (or 36% as your bad math resulted).

 

[DeletedUser]

I agree with Neo on this one. As already stated there are four possible outcomes.

HH HT TH TT

We can assume that it's an evenly weighted coin with a 50% chance of H and 50% chance of T (and it will never land on its edge).

This tells us that the each option has an equal chance.

So the chances of HH with no conditions set is 1/4 = 25%
If we know the first coin is H then we have HH or HT and the chance is 1/2 = 50%
If we know that at least one coin is H then we have HH, HT, or TH and the chance is 1/3 = 33%

In this particular case we are told "At least one is heads". Therefore the chance is 33%.

If the question was worded differently the answer could be different. But it's pretty clear that we know a coin is heads but don't know which one.

 

[DeletedUser]

Ah, but you made the mistake. It asks what is the probability that both are heads. That probability is 25%, regardless of whether you know one of them is heads. The probability doesn't change.

 

[nashy19]

Supposed that a coin tossing machine was set up to toss both coins up with exactly the same force applied on the same area of the of the coins, for them to spin exactly the same number of turns before both landing on exactly the same spot every time in order to get a heads result for both coins.

How would that then affect the probability??

Heads most of the time, depending on of the accuracy machine, the chance of an earthquake or the universe imploding ect.

 

[DeletedUser16008]

:laugh::laugh: Amusing, you cannot have a 1/3 chance of something that has only two outcomes ie 50% ... you know one coin is heads therefore the second has only two options heads or tails ... dosnt matter which coin comes down heads you know that will happen so it leaves just one so its 50%.

There were NOT 4 possible outcomes in the beginning as two have been taken away by stating "at least one is heads" it remains only to calculate the other coin odds it dosnt matter which one ... ill say again you cannot have a coin that is only heads or tails, thats TWO options not 3 ..having a 1/3 chance lol it dosnt work putting 3 into 2.

If it had been about 2 coins random at the beginning then it would have been 25% but its not and once you take away one coin it leaves a simple evens odd of 50%.

Hells is spot on & its really amusing to see the figure of 1/3 being used, it can NEVER be that whatever you do with the coins unless you include the edge as an option. its either 25% with two or 50% with one thats it.

Saying all that the wording does not even say the coins are usual heads and tails ... one could be a blank the other double headed so.... lol its all in the wording

 

[DeletedUser9470]

like i said
try it out for yourself.
your bad maths doesnt excuse not verifying what you say.

 

[DeletedUser]

Supposed that a coin tossing machine was set up to toss both coins up with exactly the same force applied on the same area of the of the coins, for them to spin exactly the same number of turns before both landing on exactly the same spot every time in order to get a heads result for both coins.

How would that then affect the probability??

This would have to be performed in a vacuum and the atomic structure of the coins would have to be identical. Then the probability would be 100%......oh, wait, there's something called the uncertainty principle.......

 

[DeletedUser]

I gave the answer in post #9.

I'm sorry that Victor & Hellstromm cannot figure out what is basically one of the most simple probabilistic problems possible. C'mon guys, even NEO gets it!
It does explain why the discussions threads never go anywhere though. Even when there's a simple provable answer and it's been explained, some people will fail to either grasp or accept it. And it shows that people can be very opinionated and dogmatic even on subjects of which they have no understanding. It also explains why casinos make such a good living.

I'm sorry if that sounds patronising, but I have studied maths to quite a high level as part of a physics degree course and I do know what I'm talkiing about.

If you're not convinced, keep reading my earlier post until you get it, or ask a mathematician to explain it to you!

Some people seem to have got confused because Head/Tail is a binary combination and that there are 2 coin tosses. If it helps, imagine that 3 coins are tossed and you are told that at least one of them is heads. What is the probablity that all 3 are heads?
The answer is 1 in 7.

 

[nashy19]

1. Flipping two coins, flipping both coins again if the result is double tails because there must be at least one heads.
2. Flipping two coins where at least one of the coins is guaranteed to be heads because there must be at least one heads.

There is a difference.

 

[DeletedUser]

I'm sorry if that sounds patronising, but I have studied maths to quite a high level as part of a physics degree course and I do know what I'm talkiing about.

If you're not convinced, keep reading my earlier post until you get it, or ask a mathematician to explain it to you!

Lol, trying to pull out credentials? Should I as well? *smirk*

Anyway, on page 1 I already pointed out the link posted by Nashy, but people wanted to continue to debate it, so I returned for more fun. There is no correct answer, because the question is sufficiently ambiguous. --- http://en.wikipedia.org/wiki/Boy_or_Girl_paradox