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I picked one of two coins at ran-
dom, then tossed it three times. It
landed heads up each time. If one
coin was fair and the other two-
headed, what is the probability
that I picked the two-headed coin?

Anyone know how to do this?

By LeavingSoon on Thursday, January 23, 2003 - 09:57 pm: Edit

I'm an idiot when it comes to math, but--and this probably isn't the right answer--why wouldn't it be 1/2? two coins, chance of picking one of them = 1/2. Sorry, but I hate math! (Well, don't hate it, but get frustrated with it!)

By hi on Thursday, January 23, 2003 - 10:07 pm: Edit

my answer: 8/9

By LeavingSoon on Thursday, January 23, 2003 - 10:09 pm: Edit

go with that one.

By AzN_Cowboy on Thursday, January 23, 2003 - 10:09 pm: Edit

Well, the fact that it turned heads three times changes the probability of the coin selection. Suppose we had not tossed the coins. Then the probability would be 1/2.
I reach a dead end when I do this problem.
First of all, there is a fifty-fifty probability initially.
Then, there is a 1/8 probability of receiving three heads if I chose the fair coin.
Furthermore, there is a 100 percent probability of receiving three heads if I chose the two-headed coin.
All that remains is to put these numbers together. It is this part of the problem which gives me trouble.
Any help?

By hi on Thursday, January 23, 2003 - 10:09 pm: Edit

with the one-headed coin there is one way, out of 8 possible flip combinations, to get 3 heads in a row. for the 2-headed coin, it is 8/8. so among the 2 coins there are 9 ways to get 3 heads in a row. 8 of those 9 ways are with the 2-headed coin.

By hi on Thursday, January 23, 2003 - 10:10 pm: Edit

i am right. i have never been wrong.

By AzN_Cowboy on Thursday, January 23, 2003 - 10:10 pm: Edit

Jeese...
Hi, how did you figure it out?
Do you have the Math League Contest Book or something?
Or did you really figure it out? If you did, show me your work. (AND DON'T COPY OUT OF THE BOOK!!!)

By hi on Thursday, January 23, 2003 - 10:12 pm: Edit

um, i thought logically... i dont have the contest book or whatever.

i gave you my steps above
with the 1-headed coin, chances to get 3 heads in a row is 1/8.
for the 2-headed coin, it is 8/8.

so, in total there are 9 ways. 8 of the 9 ways are with the 2-headed coin. thus, the solution is 8/9.

By AzN_Cowboy on Thursday, January 23, 2003 - 10:15 pm: Edit

Jeese...You must've done really well in the Math League competitions.
What year in high school are you?

By hi on Thursday, January 23, 2003 - 10:21 pm: Edit

I'm a senior. I haven't done any Math League or whatever. I've taken a few AMC tests though. As a freshmen i scored 100 on the AMC 10. As a junior i got a 109 on the AMC 12 and a 2 on the AIME. Pretty decent, not great.

By AzN_Cowboy on Thursday, January 23, 2003 - 10:28 pm: Edit

Cool. Any advice for a Freshman regarding the AMC tests? I'm planning on taking the tests, but I have to take them at different schools because my school cares little for Math Competitions.

By freak on Thursday, January 23, 2003 - 10:39 pm: Edit

If I am not mistaken, the answer is probably 1/2. If you do a tree diagram of all the possibilities...

Prob. of choosing two-headed coin: 0.5
Prob. of choosing fair coin: 0.5
Prob. of flipping 3 heads on two-headed coin: 1.0
Sum of probs. of all possible flips with fair coin: 0.0625*8 = 0.5
Sum of all possible probabilities for choosing a coin and flipping it three times: 0.5*1.0 + 0.5 = 1.0

(Probability of choosing a two-headed coin)/(Sum of all possibile probabilities)= 0.5/1.0 = 0.5

Conclusion: The probability of choosing a two-headed coin is 0.5 or 1/2.

By freak on Thursday, January 23, 2003 - 10:45 pm: Edit

My advice for you on the AMC tests is to be really careful about your answer choices. In any case, learning to program your calculator may be of some assistance when you want to test answers to questions such as: what is the sum of every 3rd number from 1 to 2003. Watch for key words such as "prime" or "greater than 0," etc. Remember that 2 is a prime number and 1 isn't, and remember other miscellaneous facts like that.

I was AMC school winner last year with score of 122.5, never had much luck on AIME. Currently a senior this year. Good luck.

By LeavingSoon on Thursday, January 23, 2003 - 10:47 pm: Edit

yay!

By Jason Toff on Thursday, January 23, 2003 - 10:50 pm: Edit

i'd think it was 2/3rd. if you pick 1 of 3 coins at random, it doesnt matter the turn out. your chances of getting a 2 headed coin is # of success / # of possible. There are 2 successes since there are 2 2-headed, and 3 total (adding the 1 fair coin)...

why isn't this right? where's the flaw

By freak on Thursday, January 23, 2003 - 10:52 pm: Edit

Jason Toff--
"I picked one of two coins at random"

There are two coins, that's the flaw.

By hi on Thursday, January 23, 2003 - 10:54 pm: Edit

freak, you are wrong. and im surprised since you got quite a good AMC score.

IGNORE THE FACT THAT A COIN WAS IMMEDIATELY CHOSEN. that is what is confusing you.

now consider all the ways which 3 heads could turn up in a row. 9 ways, 8 of which are by unfair coin. There are 16 ways of flipping (not necessarily getting 3 heads in a row) in total, all having with the same chance (1/16). 8/9 times you perform this specific situation (getting 3 heads in a row) it will be with the unfair coin. thus the answer is 8/9.

it's not 1/2.

By freak on Thursday, January 23, 2003 - 10:59 pm: Edit

I believe that's where you are mistaken. There are only two ways of getting 3 heads to turn up in a row. There are only 9 ways of flipping since there is a probability of zero that you will get any tails with the two-headed coin. You cannot ignore the fact that a coin is immediately chosen because we are trying to determine the probability of actually choosing a coin.

By hi on Thursday, January 23, 2003 - 11:07 pm: Edit

fine. don't ignore the fact that it's immediately chosen. it's just not needed in order to find the answer. You are wrong when you are considering that there's only 2 ways to get 3 heads to turn up in a row.

a quick proof that your answer is incorrect:
The odds are 1/8 that 3 heads will occur in a row with the fair coin.
The odds are 100% that 3 heads will occur in a row with the unfair coin.
Thus if 3 heads appear in a row, it is far more likely that it is the unfair coin that is being flipped.

By freak on Thursday, January 23, 2003 - 11:19 pm: Edit

It is necessary to immediately choose a coin because the possibility of choosing the coin must be multiplied by the possibilities of flipping the coins.

There are indeed only 2 ways to get 3 heads to turn up in a row:
3 heads in a row with the two-headed coin
3 heads in a row with the fair coin

Remember that we are discussing the _probability_ that we choose the two-headed coin...hold on a second, I believe that this may be a trick question...azn cowboy?

By hi on Thursday, January 23, 2003 - 11:26 pm: Edit

make a huge tree branch, starting with choosing the fair and unfair, and then going on to the coin flippings. You will see at the end that 9/16 routes lead to 3 heads in a row. 8 of those 9 are by the unfair coin.

can someone tell freak that i am right?

By hi on Thursday, January 23, 2003 - 11:28 pm: Edit

"There are indeed only 2 ways to get 3 heads to turn up in a row:
3 heads in a row with the two-headed coin
3 heads in a row with the fair coin"

If you are analyzing the problem correctly, you should be dealing with 9 possible ways to get 3 heads in a row. It can also be said that there is only 1 way to get 3 heads in a row:

3 heads in a row with either of the coins.

you are looking at the problem from a skewed perspective.

By freak on Thursday, January 23, 2003 - 11:35 pm: Edit

Only 2 routes...

If you are going to draw a tree branch for the two-headed coin, you will end up with 8 branches, but you have to take away 7 of those because those are zero probability since it is impossible to get a tail with the two-headed coin. Thus you only have 2 possible ways, 1 for the two-headed coin, 1 for the fair coin.

By hi on Thursday, January 23, 2003 - 11:43 pm: Edit

no!!! god!!!

IM GOING TO DRAW YOU THE DAMN BRANCH OF THE UNFAIR COIN SO YOU REALIZE THERE ARE 8 (YES 8) POSSIBLE ROUTES!!!!

----h
--h-
----h
h-
----h
--h-
----h

----h
--h-
----h
h-
----h
--h-
----h


Ok, now count the possiblities for 3 flips of the unfair coin. Yes, 8! All of them which yield the combination of 3 heads in a row!

(sorry about all the dashes, the forum wouldn't let me space it out the way i wanted.)

By freak on Thursday, January 23, 2003 - 11:57 pm: Edit

Okay, your tree branch drawing makes absolutely zero sense to me. But I see where you're getting at now...

By Pat57575 (Pat57575) on Friday, January 24, 2003 - 12:10 am: Edit

I think hi is right. Consider an alternative example:

Say the randomly chosen coin landed on tails three times. Even though your initial chances of choosing the fair coin were 1/2, you know the probability you chose it now is 100% or 1/1 because there is one way to get 3 tails with the fair coin and there are no ways with the double headed one.

In the same way you can figure there is an 8/9 chance you picked the two headed coin. (after looking at the results).

notice the question reads "what is the probability" and not "what was the probability"

By Hi is right on Friday, January 24, 2003 - 03:45 pm: Edit

I am absolutely sure it's 8/9.

I am not going to show you the arguments for 8/9, because hi has shown as clear as anyone can get that the probability is 8/9.

What I am going to show is the fallacy of freak's argument:

Question:

What is the probability of picking a 2-headed coin?

Freak's Argument:

Premise: The probability of choosing the 2-headed coin is 1/2, and the probability of choosing the regular coin is 1/2.

The sum of the probabilities is because of... (we all agree that the sum of the probabilities here is 1, so I'll skip it).

Conclusion: The probability of the 2-headed coin thus is
0.5 / 1 = 0.5

The premise and the conclusion are the same--this is a CIRCULAR ARGUMENT!

Here's an alternate question that further demonstrates why freak's argument was wrong:

There are two coins. One is two-headed and the other two-tailed. One flips the coin 3 times, and you get 3 heads. What is the probability that the coin chosen was 2-headed?

Well, I'll assume that everyone will go, duh, the coin chosen must be 2-headed, because you never get heads with 2-tailed coin, so the probability is 1.

Now I'll use the same argument Freak used before:

The probability of choosing 2-heads coin: 1/2
The probability of choosing 2-tails coin: 1/2

The sum of the probabilities is 1.

So the probability of choosing 2-heads coin is
0.5 / 1 = 0.5!

I hope this settles the argument once and for all. (I have nothing personal againt you, "Freak," I am just showing the problem with your argument, that's all).

By math on Friday, January 24, 2003 - 07:11 pm: Edit

I got an 800 on my math SAT 1 and 2C, and I just calculated that it's 8/9. Go with that answer.

By AzN_Cowboy on Friday, January 24, 2003 - 10:43 pm: Edit

math: That proves nothing. Math competitions actually make you think outside of the box.
Hi is right. I have the answer key.

Thanks for the help guys.

By 1111403 on Monday, January 27, 2003 - 05:34 pm: Edit

hi is right.

Freak, the question is not "what is the probability of choosing the unfair coin?" (the answer to THAT question IS 1/2) but rather the question is: "if i get 3 heads in a row, what is the probability that the coin i used is the unfair one?" (the answer is 8/9)

By Bougie (Bougie) on Thursday, April 03, 2003 - 07:55 pm: Edit

Hi is right, i just scored 109.5 on the AMC and 3 on the AIME, 8/9 is correct. Simply, out of the nine ways you can get 3 heads in a row, only one is with the fair coin. Why don't you try it with two pennies? mark one to indicate it's unfairness, then see who's right.