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Topic: Easy: Two Coin flips (Read 12586 times) |
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Re: Easy: Two Coin flips
« Reply #25 on: Aug 8th, 2002, 12:15pm » |
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Possible results:
HpHd, TpHd, HpTd, TpTd
"one of the coins came up heads" == "either P or D or both came up heads", which means TpTd is not possible. The other three are equally possible, and in only one case are both coins heads. 1/3.
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JW
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To all of those people saying that 'one of the coins came up heads' equates to the other coming up tails I think that is a misinterpertation. If both coins came up heads then the statement 'one of the coins came up heads' would still be true since one of the coins did indeed come up heads, the fact that two of the coins came up heads does not change this.
Now as for all of the people questioning the motive of the flipper I say that just makes your conclusion less accurate. Since we do not know the motives of the flipper (if he/she is looking only at the dime while making the statement) then to venture a guess as to the motives is foolish. The best answer is 1/3 because the only information given to you is that ONE of the coins came up heads. Given this and the fact that there are three possible scenarios involving ONE of the coins coming up heads (HT, TH, HH) the chance of both coins coming up heads is 1/3.
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Wanjoon
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Re: Easy: Two Coin flips
« Reply #27 on: Aug 18th, 2002, 1:06pm » |
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Oops, didn't see this thread.
But seriously guys, the answer is really 33 1/3 %. This is an elementary probability problem used in statistics and discrete mathematics courses.
TH and HT are NOT the same, precisely because there are two different ways to get a heads and tails combination, whereas there is only one way to get HH.
If you had a six-sided die that replaced the 5 side with another 6 side, would you still say that the probability of getting a 6 was only 1/6? Of course not.
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crawfish
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Re: Easy: Two Coin flips
« Reply #28 on: Aug 23rd, 2002, 11:00am » |
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The answer is 50%.
If I asked the question as follows: "If I flipped two coins and one came up heads, what is the probability the other also came up heads", the answer would definitely be 33%. But in this case the coins have already been flipped, and we are given the state of one of the coins. Thus, we can remove one coin from consideration and evaluate the second coin alone as a random occurence of one coin flip.
At least, I think so. 
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S. Owen
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Re: Easy: Two Coin flips
« Reply #29 on: Aug 23rd, 2002, 11:32am » |
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The question is whether "one came up heads" refers only to the state of one of the two coins (in which case the answer is 50%), or to both - that is, "at least one of the two came up heads." (in which case it's 33%).
It makes no difference whether the coins were already flipped, or whether you are talking about some hypothetical situation. That is, the answer is 33% even if the flipper has already flipped the coins, provided the flipper is reporting on the state of both coins ("at least one...").
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Dustin D.
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Re: Easy: Two Coin flips
« Reply #30 on: Aug 23rd, 2002, 9:34pm » |
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The question isn't clear. Depending on how you interpret it, the answer is either 25% or 50%. The outcome of the first coin toss does not effect the outcome of the second. So if you're strictly talking about what the chance is of the second coin being heads, its outcome is independent of the first coin, which means 50%. However, as I read the question, I see a lot of emphasis on the word "also," thus implying you're looking at the probability of both coins coming up heads. This means we're looking at the 4 possible cases, HH, TT, HT, and TH. Obviously HH has a 25% chance.
The 33% argument is invalid, because you can't discount a possible outcome. You don't erase the odds of getting heads or tails just by finding out information about the outcome after it happened. Probability deals with what might happen before it happens, so you can't judge probability after the fact.
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S. Owen
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Re: Easy: Two Coin flips
« Reply #31 on: Aug 24th, 2002, 7:00am » |
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So you are saying that there is a nonzero probability that there were two tails, given that "one came up heads"? Whoa!
You can reason about the probability of an outcome, even if it has already happened. Maybe you would find the question more acceptable if it said "if I were to flip two coins, and one were to come up heads, what would be the probability that the other is heads?" But this is the same thing.
If you are still not convinced, write a computer program that simulates this situation (your program does have to discard TT flips, yes), and you will see that it is 33%.
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flapp
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Re: Easy: Two Coin flips
« Reply #32 on: Aug 25th, 2002, 11:37pm » |
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The key is not the statement "One of the coins came up heads", but rather "what is the chance that the other coin <B>also</B> came up heads?"
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S. Owen
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Re: Easy: Two Coin flips
« Reply #33 on: Aug 26th, 2002, 6:59am » |
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Can you elaborate?
It was argued above that "also came up heads" means that we are talking about the probability that both came up heads (and then, that the answer is 25%, which is wrong).
There is a potential ambiguity in "one of the coins came up heads" but it is not resolved by "also"... in fact, remove the word and the riddle is the same.
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ggkrause
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Re: Easy: Two Coin flips
« Reply #34 on: Aug 26th, 2002, 10:50am » |
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This riddle is an example of conditional probability. What you are trying to determine here is the probability that (A) BOTH coins have landed heads-up GIVEN that (B) AT LEAST one coin has landed heads-up. The formula is:
P(A|B) = P(A <intersect> B)/P(B) [sorry, can't find the upside-down 'U']
The intersection of A and B is the case HH. Its probability is 1/4.
The probability of B is 3/4 (all cases except TT).
(1/4)/(3/4) = 1/3
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Emprod
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Re: Easy: Two Coin flips
« Reply #35 on: Aug 31st, 2002, 4:46pm » |
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Don't overthink it.
Plug the exact facts given by this problem into a machine, and it will tell you 0%.
0%.
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Jonathan_the_Red
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Re: Easy: Two Coin flips
« Reply #36 on: Sep 1st, 2002, 4:30pm » |
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Kenny has it right here: the solution cannot be determined without knowing the algorithm used by the flipper to decide what to say after flipping. In this, it is kind of similar to the Monty Hall problem, which cannot be solved unless you are assured that Monty will always open a door to reveal a goat before giving you the option of switching.
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Alex
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Re: Easy: Two Coin flips
« Reply #37 on: Sep 12th, 2002, 8:11pm » |
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we are analyzing the question, not what it could say... so i'm thinking the question is "what is the chance that the other coin also came up heads?"
not what is the chance that they BOTH came up heads.
so the chance that the <b>other</b> also came up heads in 1/2 not 1/4 or 1/3!
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David Balmain
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Re: Easy: Two Coin flips
« Reply #38 on: Sep 19th, 2002, 9:51pm » |
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Lets take the motivation of the flipper out of the equation.
Consider this. The flipper flips two identical coins and by mistake reveals that one of the coins is a head.
If anyone thinks that the probability of the other coin being a head is not 50%, please tell me why? If not, please explain how this is different to the original problem?
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S. Owen
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Re: Easy: Two Coin flips
« Reply #39 on: Sep 20th, 2002, 5:45am » |
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If you mean that, for example, the flipper accidentally gave you a peek at one of the coins, and it was heads, and asks what the chance is that the other is heads, then yes it's definitely 1/2. This is equivalent to one interpretation of the original problem.
However if the flipper somehow lets it slip that "at least one of the two coins came up heads" and then asks what the chance is that the other is heads, the answer is 1/3. This is equivalent to the other interpretation, probably the intended interpretation, of the original problem.
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Zaius
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Re: Easy: Two Coin flips
« Reply #40 on: Nov 16th, 2002, 1:11am » |
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I think it's 50%. The problem with the HT, TH, HH theory is that in this particular case HT and TH are identical, since we do not care which coin was tails and which was heads. In other words, the only 2 possibilities are a combination of tails and heads (in whatever order) and a combination of heads and heads. Only 2 possibilities: the other coin is tails, or the other coin is heads. And either possibilities are equal in value (2 sides to a coin).
This is, of course, assuming that the riddler does not mean "one and only one of the coins came up heads" when saying "one of the coins came up heads". In that case, the other coin would be tails, and the answer would be 0.
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S. Owen
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Re: Easy: Two Coin flips
« Reply #41 on: Nov 16th, 2002, 10:04am » |
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While HT and TH are in some sense indistinguishable to you, it's not valid to count them as one outcome. One heads and one tails happens twice as often as two heads. If HT and TH were the same thing, then if you repeatedly flipped two coins, you would see a heads and tails 1/3 of the time. You can confirm however that you actually get it 1/2 the time.
For this problem, try repeatedly flipping two coins and discarding two-tails outcomes - you will find that two heads come up once for every two times a head and tail come up, on average - that is, 1/3 of the time.
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babe72724
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Re: Easy: Two Coin flips
« Reply #42 on: Nov 22nd, 2002, 4:58pm » |
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if there are two coins and one was heads then the other one would have 33% chance of bein heads wouldn't it? because the heads side is heavier than the tails side therefore making the tails side show more. then the heads side would only show once every three tosses. isn't the weight of one a factor of the other?
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KAuss
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Re: Easy: Two Coin flips
« Reply #43 on: Nov 25th, 2002, 11:53pm » |
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Here is my crack at this given ONE assumption...
1) The flip of the coin in question is a fair coin and gaining a head or a tail from a result of the flip is a fair 50%
there is actually 2 answers to this question.
BEFORE the flip, your chnaces are 50% Reason being is because no matter how many million times you flipped a coin before you flipped this one, no matter how many times you hit heads, or how many times you hit heads in a row, your NEXT flip IS, HAS been, and forever WILL be 50% because when you flip a fair coin, you'll only have either a head or a tails.
So before you flip, your chances are 50%
AFTER you flip with results in hand, AND if you ask the question of the coin's possibility to land heads, then your chances are STILL 50% due to physical abilities of the coin. BUT, if you are asking what are the chances you get 2 heads in a ROW, then your chances are 1/4 (if the first coin is not revealed). The reason being is that the very same reason the coin is just as easy to fall as tails on every flip, to consecatively land on heads in a row will be increasingly harder as you flip more heads. As you get more biased and biased, you're defying the 50% probable which we all understandingly know.
And to put this in betting terms and to stretch the situation to the extreme...
If you were to bet someone after flipping 1 million heads in a row if their NEXT FLIP will be a head, your chances are 50% < you cannot argue this if you're judging only a single flip no matter what they flipped before hand.
If you were to bet someone to FLIP 1 MILLION heads in a row, your chances are 50%^n or 1/2^n where n = the number of times in a row in this case 1 Million Your chances to flip 2 heads in a row is actually 1/4 but since you already got 1 head, you do 2 - 1/4 which gives you 1/2 since you only got 2 flip probables left.
But if you are to bet someone 1 MILLION in a row, but some how know they just flipped 999,999 heads in a row, then your chances are back to 50% because the flipper no longer has to worry about achieving 999,999 more heads after the next one. It'll all boil down to ONE last flip in which a fair coin will give you 50%... (The reason for this is because you actually do a count down effect. Where after the first head, you only need 999,999 more which means your chances are 1/2^999,999 and you keep going down till you're 1/2^1 = 1/2, initially it'll be a good bet to bet against someone but as they land more and more, their chances actually get easier and easier till it ultimately boils down back to 1/2)
^ If that didn't do it, I must suck in expressing the english language... (Well my test scores already show me I do but anywho  )
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| « Last Edit: Nov 26th, 2002, 12:01am by KAuss » |
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Icarus
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Re: Easy: Two Coin flips
« Reply #44 on: Nov 26th, 2002, 4:31pm » |
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KAuss, to apply your analysis to this question, you would have to change the question to "Flip a coin twice. The first time it is heads. What is the probability it will be heads a second time?" Everyone agrees that the answer to this question is 50%. But that question is not equivalent to the one asked here. Try this: Flip two coins several times. Count the number of occurences where at least one of the coins was heads. Count the number of occurences where both coins are heads. You will find the ratio of (both heads) to (at least one head) is approximately ~1/3. (The more flips, the closer you will get.)
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KAuss
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Re: Easy: Two Coin flips
« Reply #45 on: Nov 27th, 2002, 7:24pm » |
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But it's like many of the people here have stated, there is a logical mathmatical way to figure out probability....
If you want to find out what the probability is to flip 2 heads in a row, it's 1 / 2^n where n is 2 in this case for 2 heads...
so the answer is 1/4 of the time you should get heads since there are 4 probables..
c1 c2
H T
H H
T T
T H
as you can clearly see, there are four outcomes to flipping a fair coin 2 times... and H, H is one out of the 4 options... so it only makes sense to say 1 / 4...
My point was, that each time you hit heads, you subtract the n... which means that if you bet someone who wants to flip 3 heads in a row, which is 1 / 8 chance of happening, after the first head, it'll be a 1 / 4 chance...
and reguardless of the # of flips in a row they need, a billion or a trillion, if they got those many heads previous to the last flip, the last flip will ALWAYS be 50%...
I think you're saying what I'm saying, if so then forget this post =P
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Icarus
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Re: Easy: Two Coin flips
« Reply #46 on: Nov 29th, 2002, 11:08am » |
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Unfortunately, we are not saying the same thing. My point is, you are misunderstanding the situation. The experiment was to bring to your attention where the misunderstanding was. It is true you can figure out the probabilities here by logic. But only if you start from the correct point. Pythagorus very logically deduced extremely stupid concepts about how the world worked, because he was believed it beneath him to ever check himself with experiments. The answer to the puzzle is 1/3. The reasons for this have been well-explained already in the thread, and the experiment will confirm the result. The problem with your logic is that you are examining a situation which is slightly different than the one in the puzzle. In particular, in relation to what you said in your last post, note that the TT case has been precluded because you are told that did not happen. But that is all you have been told. The other 3 cases are still equally possible, so 1/3 is probability of 2 heads.
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redPEPPER
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Re: Easy: Two Coin flips
« Reply #47 on: Dec 16th, 2002, 10:21am » |
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The answers 1/2 and 1/3 are arguably both valid. It boils down to why (how) the person said that one of the coins was a head.
We flip two coins, there are 4 possible occurences: HH, HT, TH, TT.
We receive an additional information. Either this information means that one of the coins, arbitrary chosen (say, the first one), happens to be a head, then that removes TH and TT, and leave us with a 1/2 probability. Or the information means that both coins were checked to see if one (any one) of them is a head. It happens that one is. That only removes TT, and the probability is then 1/3.
After reading this thread I'm guessing that the original intent was to go for a 1/3 answer, but the riddle would certainly benefit from a more accurate wording. I liked Mark Schnitzius' idea of a question: "- Is at least one of them a head? -Yes."
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Zaius
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Re: Easy: Two Coin flips
« Reply #48 on: Jan 31st, 2003, 11:45pm » |
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First we need to understand that the flipper kept flipping his coins until at least one coin came up heads, since having both coins come up tails doesn't work with this question. Thus, he can only come up with 3 possibilities, all equally likely:
TH
HT
HH
Second, we assume that the flipper really says: "At least one of the two coins came up heads, what is the chance that they are both heads?" If the flipper instead meant "One and only one of the two coins came up heads," then we would KNOW the other would be tails and the experiment would be invalid. In other words, I don't see how the flipper could be interpreted ANY other way that with my interpretation. There is no reason to think that the flipper was only looking at one coin before announcing that he saw a head.
In his list of possibilities, again all equally likely: TH, HT, HH, only one of them has two heads, and it only happens 1/3 of the time.
Therefore, yes it's 1/3.
*Humility Disclaimer*: I may be wrong
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Chronos
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Re: Easy: Two Coin flips
« Reply #49 on: Feb 3rd, 2003, 4:04pm » |
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Quote:| First we need to understand that the flipper kept flipping his coins until at least one coin came up heads, since having both coins come up tails doesn't work with this question. |
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But there are other ways to handle this. We don't know what the flipper is doing. Maybe he flips two coins, picks one at random to look at, and then says either "At least one is heads" or "At least one is tails". In this case, the 50-50 answer is correct. Maybe he keeps on flipping until at least one is heads, reflipping all cases of TT, and then says "At least one is heads". In this case, there's a 1 in 3 chance that both are heads. Heck, maybe he flips them only once, looks at both coins, and says one of three things: Either "They both came up tails" for TT, "They both came up heads" for HH, and "At least one came up heads" iff it's HT or TH. In this case, givent hat he said that, there's no chance of HH.
Without knowing what he's doing, no meaningful answer is possible.
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