Puzzle/logic question

[TallBill]

Originally posted by: Chiropteran

Originally posted by: TallBill

But we're not picking the trick coin. We're checking the odds given the fact that 4 flips have already come up heads, which drives up the odds of the trick coin already being selected. Hence, Bayes theorem.

The odds of any coin giving you heads 4 times in a row is 100%, because it is stated in the puzzle. There is zero chance of getting tails.

Sigh, I don't want to call you a fucking moron because you just are not grasping the concept. It's not an easy one, I just happened to learn it just last week.

You are definitely borderline though.

 

[GodlessAstronomer]

Originally posted by: Chiropteran

Originally posted by: TallBill

But we're not picking the trick coin. We're checking the odds given the fact that 4 flips have already come up heads, which drives up the odds of the trick coin already being selected. Hence, Bayes theorem.

The odds of any coin giving you heads 4 times in a row is 100%, because it is stated in the puzzle. There is zero chance of getting tails.

This just means that any experiments in which 4 heads are not flipped are disregarded.

Edit - to be clear, the problem doesn't say "you have flipped 4 heads, what are the odds of flipping another". The problem says, "if you flipped 4 heads, what are the odds of flipping another". It's subtle.

 

[Mo0o]

Originally posted by: GodlessAstronomer

Originally posted by: Chiropteran

Originally posted by: TallBill

But we're not picking the trick coin. We're checking the odds given the fact that 4 flips have already come up heads, which drives up the odds of the trick coin already being selected. Hence, Bayes theorem.

The odds of any coin giving you heads 4 times in a row is 100%, because it is stated in the puzzle. There is zero chance of getting tails.

This just means that any experiments in which 4 heads are not flipped are disregarded.

Edit - to be clear, the problem doesn't say "you have flipped 4 heads, what are the odds of flipping another". The problem says, "if you flipped 4 heads, what are the odds of flipping another". It's subtle.

Bayes theorum still applies in either case I believe

 

[Chiropteran]

Originally posted by: Mo0o

Originally posted by: Chiropteran

Originally posted by: TallBill

But we're not picking the trick coin. We're checking the odds given the fact that 4 flips have already come up heads, which drives up the odds of the trick coin already being selected. Hence, Bayes theorem.

The odds of any coin giving you heads 4 times in a row is 100%, because it is stated in the puzzle. There is zero chance of getting tails.

please read my comment

I did, the problem with your comment is that it makes no logical sense.

If I pick a coin at random, occasionally picked a coin that flips heads & tails, and occasionally that coin flips heads 1000000/1000000, yes I would assume the one flipping heads all the time was the trick coin.

But if I picked a coin at random multiple times, and EVERY TIME it flipped heads 1000000/1000000, and I knew only one coin was a trick coin? Well I wouldn't know what to think, because the results would be totally illogical. If I am picking a coin at random I should only get the trick coin 5% of the time.



Let me try an analogy.

You have a 6 sided die.

You roll the die 100 times, recording each number. If you roll a 6 at any time during these 100 rolls, you start over and throw away your previous results, until you successfully roll the die 100 times without ever rolling a 6.

What is the chance of recording only numbers one through five with no sixes at all?


Now do the same thing, with a trick die, that has the numbers 1-5 and the 6th face just says "roll again".

Whats the chance of recording only the numbers one through five using the trick die?

Now lets repeat the process, except you pick one of the dice at random, either the real one or the trick one. Same as before, if you roll a 6 you start over.

Now, looking at your list of results, which in either case is going to contain 0 sixes, can you tell which die was used?

 

[GodlessAstronomer]

Originally posted by: Chiropteran

Originally posted by: Mo0o

Originally posted by: Chiropteran

Originally posted by: TallBill

But we're not picking the trick coin. We're checking the odds given the fact that 4 flips have already come up heads, which drives up the odds of the trick coin already being selected. Hence, Bayes theorem.

The odds of any coin giving you heads 4 times in a row is 100%, because it is stated in the puzzle. There is zero chance of getting tails.

please read my comment

I did, the problem with your comment is that it makes no logical sense.

If I pick a coin at random, occasionally picked a coin that flips heads & tails, and occasionally that coin flips heads 1000000/1000000, yes I would assume the one flipping heads all the time was the trick coin.

But if I picked a coin at random multiple times, and EVERY TIME it flipped heads 1000000/1000000, and I knew only one coin was a trick coin? Well I wouldn't know what to think, because the results would be totally illogical. If I am picking a coin at random I should only get the trick coin 5% of the time.



Let me try an analogy.

You have a 6 sided die.

You roll the die 100 times, recording each number. If you roll a 6 at any time during these 100 rolls, you start over and throw away your previous results, until you successfully roll the die 100 times without ever rolling a 6.

What is the chance of recording only numbers one through five with no sixes at all?


Now do the same thing, with a trick die, that has the numbers 1-5 and the 6th face just says "roll again".

Whats the chance of recording only the numbers one through five using the trick die?

Now lets repeat the process, except you pick one of the dice at random, either the real one or the trick one. Same as before, if you roll a 6 you start over.

Now, looking at your list of results, which in either case is going to contain 0 sixes, can you tell which die was used?

Sigh. You're asking a totally different question! The question was never "did you pick a trick coin". It's merely asking this:
Given that when you chose a coin at random you had a 1/20 chance of pulling the trick coin. Now if your randomly chosen coin happens to come up heads 4 times in a row, what are the chances of the next flip giving heads?

Yes, the results are counter intuitive. But unless you can find a logical flaw in either KoolAidKid's maths or my software then I'm afraid the mathematics is squarely on our side.

I could never understand why people can refuse to believe mathematical statements given concrete proof. In this case we have both mathematical proof AND empirical proof, yet you let your intuition guide you instead of accepting that you have a flawed human brain which sometimes doesn't intuit things quite correctly.

 

[Mo0o]

Originally posted by: Chiropteran

Originally posted by: Mo0o

Originally posted by: Chiropteran

Originally posted by: TallBill

But we're not picking the trick coin. We're checking the odds given the fact that 4 flips have already come up heads, which drives up the odds of the trick coin already being selected. Hence, Bayes theorem.

The odds of any coin giving you heads 4 times in a row is 100%, because it is stated in the puzzle. There is zero chance of getting tails.

please read my comment

I did, the problem with your comment is that it makes no logical sense.

If I pick a coin at random, occasionally picked a coin that flips heads & tails, and occasionally that coin flips heads 1000000/1000000, yes I would assume the one flipping heads all the time was the trick coin.

But if I picked a coin at random multiple times, and EVERY TIME it flipped heads 1000000/1000000, and I knew only one coin was a trick coin? Well I wouldn't know what to think, because the results would be totally illogical. If I am picking a coin at random I should only get the trick coin 5% of the time.



Let me try an analogy.

You have a 6 sided die.

You roll the die 100 times, recording each number. If you roll a 6 at any time during these 100 rolls, you start over and throw away your previous results, until you successfully roll the die 100 times without ever rolling a 6.

What is the chance of recording only numbers one through five with no sixes at all?


Now do the same thing, with a trick die, that has the numbers 1-5 and the 6th face just says "roll again".

Whats the chance of recording only the numbers one through five using the trick die?

Now lets repeat the process, except you pick one of the dice at random, either the real one or the trick one. Same as before, if you roll a 6 you start over.

Now, looking at your list of results, which in either case is going to contain 0 sixes, can you tell which die was used?

Ugh christ I dont even know what to say to you now. You're inventing random rules.

The problem says you pick ONE coin from the pile and flipped it 4 times which came up heads everytime. And now you're about to flip it a 5th time. THAT'S IT.

The results of your last 4 flips matter!! The more times the problem told you you flipped heads, the more likely you have inadvertently picked the trick coin and not a normal coin.

If the problem said you pick a random coin and flipped it a million times and it came up heads everytime, even you would intuitively see the chance of u getting a head on the next flip is higher than 52.5% because the probability of you having selected teh trick coin is much higher than 1/20. So it stands to reason that the # of heads you accomplished (as stated by the problem) matters in the probability result.


Can the OP just post the answer so we can end this nonsense.

 

[CoinOperatedBoy]

Originally posted by: Mo0o
Can the OP just post the answer so we can end this nonsense.

OP doesn't know.

 

[Chiropteran]

Originally posted by: GodlessAstronomer


Sigh. You're asking a totally different question! The question was never "did you pick a trick coin". It's merely asking this:
Given that when you chose a coin at random you had a 1/20 chance of pulling the trick coin. Now if your randomly chosen coin happens to come up heads 4 times in a row, what are the chances of the next flip giving heads?

Yes, the results are counter intuitive. But unless you can find a logical flaw in either KoolAidKid's maths or my software then I'm afraid the mathematics is squarely on our side.

I could never understand why people can refuse to believe mathematical statements given concrete proof. In this case we have both mathematical proof AND empirical proof, yet you let your intuition guide you instead of accepting that you have a flawed human brain which sometimes doesn't intuit things quite correctly.

I can understand how you came up with your answer, and I agree it makes sense, but I still don't like the wording of the OP.


Revised:
A jar has 20 coins, 19 are normal and one has heads on both sides. You pick a random coin from the jar and flip it 4 times. If it came up as heads four times in a row, what are the chances that the next flip will come up heads?

If that is how the OP was worded, I would agree 100% with your answer.

However, it wasn't worded that way.

Alternate revision:
A jar has 1000 coins, 950 are normal and 50 have heads on both sides. You pick a random coin from the jar and flip it 4 times, recording the results on a piece of paper next to the coin. Repeat 999 times, until every coin has been flipped 4 times. Now go back to one of the coins that flipped heads 4 times in a row, what are the chances that the next flip will come up heads?

Do you see the difference been my revised wording and the OP?


Edit: here is the flaw. Attempt to work through the puzzle in real life-

1 pick a coin.

2 flip coin

3 see coin flip as tails

puzzle fails, can't actually be done in real life.

 

[Mo0o]

Originally posted by: Chiropteran

Originally posted by: GodlessAstronomer


Sigh. You're asking a totally different question! The question was never "did you pick a trick coin". It's merely asking this:
Given that when you chose a coin at random you had a 1/20 chance of pulling the trick coin. Now if your randomly chosen coin happens to come up heads 4 times in a row, what are the chances of the next flip giving heads?

Yes, the results are counter intuitive. But unless you can find a logical flaw in either KoolAidKid's maths or my software then I'm afraid the mathematics is squarely on our side.

I could never understand why people can refuse to believe mathematical statements given concrete proof. In this case we have both mathematical proof AND empirical proof, yet you let your intuition guide you instead of accepting that you have a flawed human brain which sometimes doesn't intuit things quite correctly.

I can understand how you came up with your answer, and I agree it makes sense, but I still don't like the wording of the OP.


Revised:
A jar has 20 coins, 19 are normal and one has heads on both sides. You pick a random coin from the jar and flip it 4 times. If it came up as heads four times in a row, what are the chances that the next flip will come up heads?

If that is how the OP was worded, I would agree 100% with your answer.

However, it wasn't worded that way.

Alternate revision:
A jar has 1000 coins, 950 are normal and 50 have heads on both sides. You pick a random coin from the jar and flip it 4 times, recording the results on a piece of paper next to the coin. Repeat 999 times, until every coin has been flipped 4 times. Now go back to one of the coins that flipped heads 4 times in a row, what are the chances that the next flip will come up heads?

Do you see the difference been my revised wording and the OP?

Thats not what the OP's question stated at all.

 

[GodlessAstronomer]

Originally posted by: Chiropteran
Do you see the difference been my revised wording and the OP?

OP:

jar with 20 coins: One of the coins is a trick coin that has both sides heads. You pick a random coin from the jar and flip it 4 times, and each time it comes up heads. What are the chances that the next flip will come up heads?

Your ideal:

A jar has 20 coins, 19 are normal and one has heads on both sides. You pick a random coin from the jar and flip it 4 times. If it came up as heads four times in a row, what are the chances that the next flip will come up heads?

They're nearly identical, what is it you didn't like about the OP wording?

 

[GodlessAstronomer]

Originally posted by: Mo0o

Thats not what the OP's question stated at all.

I think he might just be trolling now. You can't just completely rewrite the question

 

[Mo0o]

Originally posted by: GodlessAstronomer

Originally posted by: Mo0o

Thats not what the OP's question stated at all.

I think he might just be trolling now. You can't just completely rewrite the question

This feels like all the endless "science/math" debates on teh internet. The losers eventually just start nitpicking the OP's premise in order to back up their absurd position

 

[Rebasxer]

I'm not gonna attempt to deal with the numbers, but can someone atleast tell me if conceptually I understand this?

Because there is one trick coin that was possible to select, and you cannot disprove that you did not draw the trick coin by the first 4 heads outcome, you have to factor in the possibility that you have to find out the probability of having the trick coin and then factor that into the probability of flipping heads?

 

[darkxshade]

50%

Because logically after the first flip, one should be able to tell if he's flipping the trick coin or not and it would be pointless to keep flipping it if it is since it's not gonna ever turn tails. Therefore, it must be a regular coin and so the chances of that 5th flip is 50% because it is also independent of the first 4 flips.

 

[Mo0o]

Originally posted by: Rebasxer
I'm not gonna attempt to deal with the numbers, but can someone atleast tell me if conceptually I understand this?

Because there is one trick coin that was possible to select, and you cannot disprove that you did not draw the trick coin by the first 4 heads outcome, you have to factor in the possibility that you have to find out the probability of having the trick coin and then factor that into the probability of flipping heads?

ya

 

[GodlessAstronomer]

Originally posted by: Mo0o

Originally posted by: GodlessAstronomer

Originally posted by: Mo0o

Thats not what the OP's question stated at all.

I think he might just be trolling now. You can't just completely rewrite the question

This feels like all the endless "science/math" debates on teh internet. The losers eventually just start nitpicking the OP's premise in order to back up their absurd position

What's really strange is that people just refuse to believe that they were wrong to begin with. I hope people like that don't actually study mathematics! If you look at the first page of this thread, my first comment was completely wrong, but I have the backbone to admit that my thinking was flawed.

 

[CoinOperatedBoy]

I think a more detailed explanation would be helpful for the people who don't get it. It's not one of those trick questions where the number of flips is meaningless. And saying this is impossible real world is false. It's just conditional probability. Here's how the other posters applied Bayes' Theorem and came up with the right answer.

Event A = You choose the trick coin.
Event B = You flip the coin 4 times and get heads each time.

P(A) is the probability of choosing the trick coin. This is given as 1/20.
P(A) = 0.05

P(A') is the probability of choosing a normal coin. This is given as 19/20.
P(A') = 0.95

P(B|A) is the probability of getting 4 heads in a row given that you chose the trick coin. Since the trick coin can only produce heads:
P(B|A) = 1

P(B|A') is the probability of getting 4 heads in a row given that you chose a normal coin.
P(B|A') = (1/2) (1/2) (1/2) (1/2) = 1/16 = 0.0625

P(B) is therefore the probability of getting 4 heads in a row with either type of coin.
P(B) = P(B|A) P(A) + P(B|A') P(A') = (1) (0.05) + (0.0625) (0.95) = 0.109375

P(A|B) is the probability that you chose the trick coin, given the first four flips were heads. Applying Bayes' Theorem:
P(A|B) = (P(B|A) P(A)) / P(B) = ((1) (0.05)) / (0.109375) = 0.457142


The probability that the fifth flip will be heads can then be calculated by:
P(A|B) P(flipping the trick coin will produce heads) + P(A'|B) P(flipping the normal coin will produce heads)
(0.457142)(1) + (1-0.457142)(0.5) = 0.728571

 

[her209]

Originally posted by: gdextreme

Originally posted by: her209
Probability of picking a normal coin * probability of flipping heads:
19/20 * 1/2

Probability of picking the abnormal coin * probability of flipping heads:
1/20 * 1

19/40 + 2/40 = 21/40 = 52.5%

Here you're ignoring that the coin already gave 4 heads earlier. I got to ask this question to my teacher for a confirmation.

The results of the previous flips don't matter because the questions asks for "the chances that the next flip will come up heads".

 

[Mo0o]

Originally posted by: her209

Originally posted by: gdextreme

Originally posted by: her209
Probability of picking a normal coin * probability of flipping heads:
19/20 * 1/2

Probability of picking the abnormal coin * probability of flipping heads:
1/20 * 1

19/40 + 2/40 = 21/40 = 52.5%

Here you're ignoring that the coin already gave 4 heads earlier. I got to ask this question to my teacher for a confirmation.

The results of the previous flips don't matter because the questions asks for "the chances that the next flip will come up heads".

They do because it gives information about the likelihood of you having selected teh trick coin. If you replace "4 heads" with "1 million heads", i think intuitively you know the chance of another head coming up should be well above 52.5%

 

[CoinOperatedBoy]

Originally posted by: her209

Originally posted by: gdextreme

Originally posted by: her209
Probability of picking a normal coin * probability of flipping heads:
19/20 * 1/2

Probability of picking the abnormal coin * probability of flipping heads:
1/20 * 1

19/40 + 2/40 = 21/40 = 52.5%

Here you're ignoring that the coin already gave 4 heads earlier. I got to ask this question to my teacher for a confirmation.

The results of the previous flips don't matter because the questions asks for "the chances that the next flip will come up heads".

No. Any single flip has a 52.5% chance of being heads. However, the more heads you get in a row, the higher the likelihood that you chose the trick coin, and thus the higher the chance the next flip will also be heads.

 

[Mo0o]

Originally posted by: CoinOperatedBoy

Originally posted by: her209

Originally posted by: gdextreme

Originally posted by: her209
Probability of picking a normal coin * probability of flipping heads:
19/20 * 1/2

Probability of picking the abnormal coin * probability of flipping heads:
1/20 * 1

19/40 + 2/40 = 21/40 = 52.5%

Here you're ignoring that the coin already gave 4 heads earlier. I got to ask this question to my teacher for a confirmation.

The results of the previous flips don't matter because the questions asks for "the chances that the next flip will come up heads".

No. The first flip has a 52.5% chance of being heads. However, the more heads you get in a row, the higher the likelihood that you chose the trick coin, and thus the higher the chance the next flip will also be heads.

corrected a bit, but you have the right idea

 

[GodlessAstronomer]

Thanks for the detailed explanation CoinOperatedBoy, I learned something today

 

[CoinOperatedBoy]

Originally posted by: Mo0o

Originally posted by: CoinOperatedBoy

Originally posted by: her209

Originally posted by: gdextreme

Originally posted by: her209
Probability of picking a normal coin * probability of flipping heads:
19/20 * 1/2

Probability of picking the abnormal coin * probability of flipping heads:
1/20 * 1

19/40 + 2/40 = 21/40 = 52.5%

Here you're ignoring that the coin already gave 4 heads earlier. I got to ask this question to my teacher for a confirmation.

The results of the previous flips don't matter because the questions asks for "the chances that the next flip will come up heads".

No. The first flip has a 52.5% chance of being heads. However, the more heads you get in a row, the higher the likelihood that you chose the trick coin, and thus the higher the chance the next flip will also be heads.

corrected a bit, but you have the right idea

I guess you could say "any single flip, disregarding the results of any previous flip," but this is all pretty unnecessarily semantic.

 

[Mo0o]

Originally posted by: CoinOperatedBoy

Originally posted by: Mo0o

Originally posted by: CoinOperatedBoy

Originally posted by: her209

Originally posted by: gdextreme

Originally posted by: her209
Probability of picking a normal coin * probability of flipping heads:
19/20 * 1/2

Probability of picking the abnormal coin * probability of flipping heads:
1/20 * 1

19/40 + 2/40 = 21/40 = 52.5%

Here you're ignoring that the coin already gave 4 heads earlier. I got to ask this question to my teacher for a confirmation.

The results of the previous flips don't matter because the questions asks for "the chances that the next flip will come up heads".

No. The first flip has a 52.5% chance of being heads. However, the more heads you get in a row, the higher the likelihood that you chose the trick coin, and thus the higher the chance the next flip will also be heads.

corrected a bit, but you have the right idea

I guess you could say "any single flip, disregarding the results of any previous flip," but this is all pretty unnecessarily semantic.

I paranoid now, just incase someone latches on to some semantic mistake and renews this retarded debate

 

[silverpig]

I'm sensing a plane and a treadmill in this thread...