## Flipping coins

Posted by Chris on November 18, 2010 – 6:12 pm

It’s ages since I’ve seen this one.

After flipping two coins, at least one of them shows heads. What is the probability that the other one also shows heads?

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Posted by Chris on November 18, 2010 – 6:12 pm

It’s ages since I’ve seen this one.

After flipping two coins, at least one of them shows heads. What is the probability that the other one also shows heads?

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November 18th, 2010 at 8:01 pm

With 2 coins u can have combinations of TT,TH,HT,HH of tails n heads. Since one of them is already a head we cannot have TT. That leaves us with 3 combinations with just one having both heads. So i will go with one over three.

November 19th, 2010 at 12:20 am

33.333%

November 19th, 2010 at 2:46 am

1/4

November 19th, 2010 at 4:34 am

33.33%

November 19th, 2010 at 6:37 am

Three right answers.

Rohan. I’ll nit-pick your answer (but only because you’ve given me some material to work with and I’m sure you’re smart enough to deal with it). Whilst you have the right answer (and I’m quite sure you really do understand why), you have omitted a crucial observation/statement.

I could say, if I throw a die, it could show a 6 or not, that’s 2 possibilities, so that’s 1 in 2 of throwing a 6.

So what did you miss?

November 19th, 2010 at 6:48 am

The odds of each coin being heads is 1/2. Ergo, It is irrelevant that one of the coins is heads. The remaining coin still has a 50/50 chance of being heads.

November 19th, 2010 at 8:57 am

@ Chris: I think i got what u r looking for. The difference between possibility n probability.

When we flip 2 coins, there is an equal probability (25%) of getting TT,TH,HT,HH (as opposed to 1/6 chance of getting a 6 when we throw a die & 5/6 chance of not getting a 6)

When one of them is a head, probability of TT reduces to 0, giving us a 33.33 % chance each of getting TH,HT or HH

November 19th, 2010 at 9:19 am

Hi Nimble11,

Unfortunately you’re wrong, I struggled with this myself when I saw it before. The trouble is that we all know that the probability of a coin landing heads up / tails up is even, so it’s natural to apply that knowledge to this question without thinking any more deeply than that, but the question is worded to skew the odds.

Rohan explains it perfectly well (I can’t see the ommission that Chris is indicating) but maybe think of it this way: If I met you in a pub and offered you the following wager would you take it???….

You put £10 on the table, I’ll put £15 on the table.

We each then simultaneously flip a separate coin.

If one or both of the coins is a head I’ll take away a coin that shows heads, then, if the remaining coin is a head you take the pot, if it’s a tail I take the pot.

If your answer of 50/50 was correct then you’d be making a profit because youd be winning 1/2 the time, however the odds are skewed in my favour because I get to choose which coin I take according to whichever shows heads. It could be your coin, or it could be my coin, but the choice is determined after the flip, not before, and that’s where my advantage is.

On average one in every four flips will be tail / tail and neither of us would take the pot.

Of the other three flips only once will the result be head / head which will result in you winning, so you win once every four flips giving you £15 profit, and I win twice in every four flips giving me £20 profit.

Try it on your drinking buddies!

November 19th, 2010 at 10:13 am

Hi Rohan. That’s close enough. The point being that TT, HT, TH and HH are

equally likely. Taking away the TT doesn’t stop HT, TH and HH from being equally likely. As 1 of the 3 possibilities must come up, they must each have a probability of 1/3.Thank you.

November 20th, 2010 at 1:55 am

Just to weigh in here….

I was never any good with probabilities, and I can see the way the question is worded to make you think it is 50/50….

But my mind-set tells me it has to be 50/50 because one coin is independent of the result of the other coin. So even though I am wrong, I am still going to say 50/50 – 1 in 2 chance of the other coin being a head.

Dual Aspect has given a really good demonstration of why I am wrong, and I shall find out by tomorrow morning if this pub trick works to fleece friends and fools alike! Or if I am considerably poorer! I am sure Dual Aspect is good for my losses!

November 20th, 2010 at 4:55 am

Karl. Bet even money on the other coin being tails (it’s twice as likely to be tails as heads). You’ll be loaded (both ways) before long.

November 20th, 2010 at 6:35 am

well thats easy its even chance because it can be one or the other and there is only 1 head and one tails

November 20th, 2010 at 1:44 pm

Hi Tom. Nope! Each possibility has a different probability. The probability of the other coin being a heads is 1/3 and of being a tails is 2/3.

November 20th, 2010 at 5:15 pm

ITS 50/50 AS THERE IS ONLY 2 FACES GIVING EACH SIDE THE SAME CHANCE TO APPEAR

November 20th, 2010 at 5:24 pm

Dan, there is no need to SHOUT the wrong answer. I’ve pointed out several times, not all possibilities are equally likely. See Rohan’s posts (1 and 7) for a good explanation of why it’s 1/3. Dual Aspect gives another useful take on it (especially in the last half of his post).

November 21st, 2010 at 8:38 pm

its a 50% posibility duhh! it doeant change one coin 2 possibilities! 50/50

November 22nd, 2010 at 6:08 am

rina. duhh! it’s 1/3 as shown by Rohan using proper logical/probabilistic reasoning. Dual Apect has also shed light on what’s going on.

You have used the same kind of faulty reasoning that I illustrated in the third paragraph of my post #5. Yes there are two possibilities, but they are not equally likely. The probability of heads is 1/3, probability of tails is 2/3.

November 22nd, 2010 at 9:34 am

Another way to see it. If you did 100 flips, then on average 25 would be HH, 25 would be TT and the remaining 50 would be HT/TH. We now discard the 25 TT (as at least one isn’t a heads). That leaves 75 flips of which at least one is heads. That makes the probability of flipping two heads given that at least one is heads be 25/75 = 1/3.

November 22nd, 2010 at 10:42 am

Actually Chris im sorry to inform you that Rohan’s method is not quite accurate ==> false !!

what if we wanted the probability of the second coin being a Tail ?

Rohan’s predicted answer will be 2/3 because TH + HT = 1/3 + 1/3 chance of occurance.

But what if the coin was loaded in a way so P(T)= 1/5 & P(H)=4/5 ?

Then his answer is wrong because he is working as if the coins have equal probabilities whereas they don’t having one of them loaded !!

The proper way to think of it is :

While we already know that one of the coins is a head !

then we are 100% sure that this coin is a head !!!

meaning ==> you must solve for the second coin given you know the outcome of the first coin !!

You must compute :

P( other coin being a head / 1 coin is a head )

The way to read it : probability of having a head GIVEN that we already have a head !!!

for simplification purposes:

let us assume that the 1st coin turns out head

the 2nd is unknown

So P( 1st coin is a head ) = 1 because ==> we are 100% sure

Our only available options are :

HT ===> P( 2nd coin T / 1st coin H ) = 1 * 2/3 = 2/3

HH ===> P( 2nd coin H / 1st coin T ) = 1 * 1/3 = 1/3

Same thing for Chris.

Your assumption is also false !!

try it when you have P(T)=1/5 & P(H)=4/5 you surely will get a wrong answer.

When we deal with a loaded coin or die meaning that probabilies are not equal for every outcome, we cannot generalize we have to go deep into the calculations.

Thank You !

November 22nd, 2010 at 12:03 pm

Hi Ribs. Under the remark from post #7, Rohan’s method is pretty correct. If you don’t assume that you work with a fair coin, you don’t have enough information to answer the problem! Now, if you want to generalize it to an arbitrary coin with probability p for H and 1-p for T, then we have 3 possible outcomes (we don’t care exactly which coin shows what) TT – probability (1-p)^2, HH – probability p^2 and HT – probability 2p(1-p). Now according to the Bayes formula for conditional probability, that you state, the probability that the other coin is also H is p^2/(p^2+2p(1-p))=p/2-p.

When p=1/2 we get the above (1/2)/(3/2)=1/3.

November 22nd, 2010 at 1:11 pm

Hello Slavy regarding Bayes formula i am not sure that you are using it correctly , so i can’t comment on that !

There are different methods to find the answer and i am sure that my answer is correct !

The answers of Rohan & Chris are incorrect !

Chris did specify that the other coin has a probability twice of having heads than tails ( post # 11 ).

So we do have enough information to compute this one.

If you try my example P(H)= 4/5 & P(T)= 1/5 you will find that their methods are incorrect.

Enjoy amigos !

November 22nd, 2010 at 1:12 pm

Need to correct something :

Twice of having tails than heads !!

Sorry for that

November 22nd, 2010 at 2:08 pm

Hi, Ribs. Sure, on your example all the solutions are incorrect, since you consider a biased coin. If you follow my general formula, you will see that in this case the answer is (4/5)/(2-1/5)=4/9, not 1/3.

November 22nd, 2010 at 2:26 pm

Hi Ribs. Of course I was assuming the coins were fair – if not the question can only sensiibly be answered algebraically as there isn’t enough information to do otherwise. In fact I had already partially dealt with your point way back in post 5. When answering problems like these, it is perfectly reasonable to assume that the coins are equally likely to land heads or tails. It would have been reasonably easy to stay with Rohan’s method and calculate the probailities for each of the states. (Addendum, Feb 17, 2011: I have now posted “Firls, firls, firls”, and that shows how to use Rohan’s method when e.g. head/tails are not equally likely).

Let the coins be A and B. Let P(A=H) = p and P(B=H) = q. Then P(HH) = pq and

P(TT) = (1-p)(1-q). For the original problem, P(HH given Hx or xH) = P(HH)/(1-P(TT)) = pq/(1-(1-p)(1-q)).

If the question had asked what is the probability that the other coin is a tails, then the answer is 2/3 (assuming fair coins).

Using your p = q = 4/5 we get P(HH given Hx or xH) = 2/3.

If the coins had been picked at random from real coins in circulation, then the probability of them being biased by more than 0.1% (say) (yet alone to such an enormous degree as you have chosen) is so close to 0, that it is of no practical consequence and Rohan and my arguments and assumptions stand.

Our methods are not incorrect. Rohan simply didn’t state (in his first post) that he was assuming fair coins. It was clear from his second post and in most of my posts that we were assuming fair coins.

Of course I could also torture English by saying that your method is wrong because if the coins were fair after all, you’d have calculated the wrong answer. But I won’t say that as there is nothing wrong with the methods; it’s only the assumed data that is being disputed.

November 22nd, 2010 at 2:30 pm

Hi slavy. I concur with Ribs’ that your equations are in error. I think you need some coffee

November 22nd, 2010 at 2:44 pm

Hi Chris. If you set p=q in post #24 and since 1-(1-p)^2=p(2-p) you get exactly mine formula (i.e., in this partial case pq/(1-(1-p)(1-q))=p/(2-p)). So, taking p=4/5 we derive 4/9, not 2/3. I never drink coffee

November 22nd, 2010 at 2:47 pm

OK – I told you I am bad in arithmetic 2-4/5=6/5, not 9/5, so yes you are right! But my formula is correct (just my computations are terrible).

November 22nd, 2010 at 3:10 pm

Hi slavy. I wasn’t paying enough attention either. I saw 2-p and 4/9 so jumped to conclusions. I agree your maths is right (but your original arithmetic was wrong – twice ).

Now I know why you get arithmetic wrong – no coffee.

November 22nd, 2010 at 4:20 pm

Hi Ribs. I’m not sure if you’ve quoted me correctly. I didn’t

specifythat the other coin had a probability of 2/3 of being a tails.I only said that the probability of the other coin being a tails is 2/3

ifat least one coin is a heads. Before discarding the TT case, each coin will independently be a H or a T with probabilty 1/2.November 22nd, 2010 at 5:57 pm

50%

December 28th, 2010 at 9:55 pm

I was rummaging around in the past posts, and noticed that this one had lots of responses, most of them pretty nonsensical. The probability of tossing a coin and getting a head is 50% regardless of how many coins you’ve tossed before (one or a thousand), and regardless of whether the results of the previous tosses were heads or tails. Each coin toss is an independent occurrence. If the question is: what is the probability of tossing two coins and getting two heads? The answer is .5 x .5 = .25 or 25%.

January 1st, 2011 at 4:11 pm

Hi Al. You haven’t answered the question that I posted.

January 6th, 2011 at 2:17 am

the tricky part of the question lies in the “at least one of them shows heads”. if you flip coins a and b, then look at them both and sees that a head is on either or both coins, this fulfills the requirement. that’s why it is 1/3. ht, th, and hh are the only possibilities (3) and only hh satisfy the criterion. but, if the question were to ask, “look at first coin w/o looking at the second, and it is a head”, then the probability of coin b being head also is 1/2, as the outcome of coin a does not affect the outcome of coin b.

January 6th, 2011 at 6:10 am

The paradox is, to my mind, that it is twice as likely that the other coin is showing a tail (rather than a head) even though you are discarding cases where you get tails.

February 16th, 2011 at 4:19 pm

Do this :

Take 8 coins and set them up on a table in sets of two in the four possible outcomes of

HH HT TH TT

remove the TT set and you are left with

HT TH HH

Cover up a H for each set and count the remaining coins as to whether they are H or T

There will be 1 H and 2 T s every time

Jon

February 16th, 2011 at 5:16 pm

Hi Jon. That’s quite nicely put.

My only (virtually standard) criticism is that you didn’t mention that the four possible outcomes are equally likely. Failure to recognise that crucial feature is, probably, the primary reason that many posters get probability questions wrong; and that’s the real reason that I have mentioned it. Just check through the posts (of just about all the probability questions that have been posted) to see plenty of examples.