Puzzle/logic question

[CoinOperatedBoy]

Originally posted by: Chiropteran

Originally posted by: CoinOperatedBoy
There is no implication that flipping a random coin will always achieve this result.

YES THERE IS. Can't you read?

You pick a random coin from the jar and flip it 4 times, and each time it comes up heads.

The result happens. It doesn't say *if* it comes up heads 4 times, it simply says it comes up as heads.

If it wasn't supposed to imply that the coin will always come up as heads 4 times, it would have been worded like so:


You pick a random coin from the jar and flip it 4 times, and each time it comes up heads or tails.

Awesome. You almost had me convinced you were legit. Well played.

 

[her209]

Let's modify the the OP's question:

You have 20 coins. One of the coins is a trick coin that has both sides heads. The rest have heads/tails. Someone adds one trick coin that has both sides tails to the jar. 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?

Since the question assumes that you picked a coin and flipped it 4 times and got heads, can we essentially ignore the coin with both sides tails from our calculations? In other words, the probability of picking the coin with both sides heads is 1/20 and not 1/21 since picking the coin both sides tails invalidates the question?

 

[TallBill]

Originally posted by: Chiropteran

Originally posted by: CoinOperatedBoy

Is this a joke? If you don't understand, leave it at that and avoid statistics like the plague.

This is not a statistics issue at all, it is a grammar/linguistics issue. And I'm not convinced you are correct

edit

Originally posted by: CoinOperatedBoy
There is no implication that flipping a random coin will always achieve this result.

YES THERE IS. Can't you read?

You pick a random coin from the jar and flip it 4 times, and each time it comes up heads.

The result happens. It doesn't say *if* it comes up heads 4 times, it simply says it comes up as heads.

If it wasn't supposed to imply that the coin will always come up as heads 4 times, it would have been worded like so:


You pick a random coin from the jar and flip it 4 times, and each time it comes up heads or tails.

Usually when its a large group of people against one, the one person is wrong; unless of course they are incredibly smart, which you are not.

 

[Dragula22]

Originally posted by: Chiropteran


You pick a random coin from the jar and flip it 4 times, and each time it comes up heads.

The result happens. It doesn't say *if* it comes up heads 4 times, it simply says it comes up as heads.

If it wasn't supposed to imply that the coin will always come up as heads 4 times, it would have been worded like so:


You pick a random coin from the jar and flip it 4 times, and each time it comes up heads or tails.

I don't understand the confusion either. The problem is pretty clear.

You're right, the result happens. 4 heads came up. It could have been 3 heads/1tail or 2heads/2tails. But nope, it was 4 heads. That's the current situation.

From that, you have to deduce the likelihood you are holding the trick coin and calculate the odds.

 

[CoinOperatedBoy]

Q: You have a bag with a bunny and a squirrel inside. You blindly reach in and pull out the bunny. What is left in the bag?
Chiropteran: I JUST TRIED TO REPRODUCE THIS IN REALITY AND I PULLED OUT THE SQUIRREL AND THE SQUIRREL BIT ME. THIS IS ABSURD!

 

[TallBill]

Originally posted by: CoinOperatedBoy
Q: You have a bag with a bunny and a squirrel inside. You blindly reach in and pull out the bunny. What is left in the bag?
Chiropteran: I JUST TRIED TO REPRODUCE THIS IN REALITY AND I PULLED OUT THE SQUIRREL AND THE SQUIRREL BIT ME. THIS IS ABSURD!

 

[Engineer]

Originally posted by: CoinOperatedBoy
Q: You have a bag with a bunny and a squirrel inside. You blindly reach in and pull out the bunny. What is left in the bag?
Chiropteran: I JUST TRIED TO REPRODUCE THIS IN REALITY AND I PULLED OUT THE SQUIRREL AND THE SQUIRREL BIT ME. THIS IS ABSURD!

:laugh:

 

[tokie]

Originally posted by: CoinOperatedBoy
Q: You have a bag with a bunny and a squirrel inside. You blindly reach in and pull out the bunny. What is left in the bag?
Chiropteran: I JUST TRIED TO REPRODUCE THIS IN REALITY AND I PULLED OUT THE SQUIRREL AND THE SQUIRREL BIT ME. THIS IS ABSURD!

nice.

someone should really do something about Chiropteran's trolling.

 

[silverpig]

Originally posted by: Chiropteran

Originally posted by: CoinOperatedBoy
It's a statistics question hinging on a single possible scenario. It's not implied that this scenario will occur every time you hypothetically choose a coin from the jar.

The way I read it, it is implied.

It says you can pick a coin at random, and that coin flips heads 4 times in a row. If it didn't have the "pick a coin at random part", it would make more sense. for example "you are given a particular coin out of the 20". If it was worded like that, then there would be no implication that every coin is going to show the same result. The "at random" part, to me, implies that you could pick *any* coin and it's going to come up as heads 4 times in a row. Which in turn implies that if you repeat the process you can pick each and every coin and every one of them will flip as heads 4 times in a row.

Dude just stop. While you weren't the only one in the thread to misunderstand the problem, you're the only one who doesn't understand it after having it explained in multiple ways by many people.

 

[Chiropteran]

I'm not trolling, but it's pointless to discuss this further.

If you would actually read my posts, you could see that I have zero issues with the math or logic used to come up with the answer. The issue is that the OP is ambiguous and the answer doesn't give us any useful information in reality.


As a counter example, look at the "3 prizes" puzzle. The announcer offers you one of 3 prizes. You pick one, then he reveals a prize you didn't pick to be junk. Is it worth switching? It's a very similar puzzle, but unlike this one it makes sense because you can actually run through it in reality.

 

[Schfifty Five]

Originally posted by: Chiropteran
I'm not trolling, but it's pointless to discuss this further.

If you would actually read my posts, you could see that I have zero issues with the math or logic used to come up with the answer. The issue is that the OP is ambiguous and the answer doesn't give us any useful information in reality.


As a counter example, look at the "3 prizes" puzzle. The announcer offers you one of 3 prizes. You pick one, then he reveals a prize you didn't pick to be junk. Is it worth switching? It's a very similar puzzle, but unlike this one it makes sense because you can actually run through it in reality.

That would be the "Monty Hall" question and the answer is YES, you switch everytime because your chances go from 33% -> 66%.

 

[CoinOperatedBoy]

Originally posted by: Chiropteran
I'm not trolling, but it's pointless to discuss this further.

If you would actually read my posts, you could see that I have zero issues with the math or logic used to come up with the answer. The issue is that the OP is ambiguous and the answer doesn't give us any useful information in reality.

The OP is in no way ambiguous. It's just an illustration of a principle -- that probability can change conditionally based on given events. Your puzzle is in no way more useful.

You can test this in reality. GodlessAstronomer created a program to do exactly that, but given enough time and proper conditions, you could come up with the same answer with an actual collection of coins.

 

[Chiropteran]

Originally posted by: CoinOperatedBoy

You can test this in reality. GodlessAstronomer created a program to do exactly that, but given enough time and proper conditions, you could come up with the same answer with an actual collection of coins.

In reality, when I pick a coin at random, and flip it 4 times, the vast majority of the time it's NOT going to end up as heads 4 times in a row. If anything other than heads X4 results, the puzzle fails.

That is the problem with it.



The "Monty Hall" puzzle is NEVER invalidated by real life, while this puzzle would be invalidated the majority of the time. You can repeat the "Monty Hall" puzzle a million times and the overall results will be the exact same.


If you even try to repeat the coin toss puzzle a couple times, you will almost assuredly get a result other than heads X4, invalidating the entire puzzle.

 

[CoinOperatedBoy]

Originally posted by: Chiropteran
If you even try to repeat the coin toss puzzle a couple times, you will almost assuredly get a result other than heads X4, invalidating the entire puzzle.

No shit. The question specifically applies only to the situation when you did get four heads in a row, which is obviously one of the possible outcomes. If you want to reproduce the conditions of the puzzle, you pick a coin and flip four times. If you don't get four heads, put the coin back and randomly pick again until you do. The puzzle doesn't have to spell that out.

 

[Chiropteran]

Originally posted by: CoinOperatedBoy

Originally posted by: Chiropteran
If you even try to repeat the coin toss puzzle a couple times, you will almost assuredly get a result other than heads X4, invalidating the entire puzzle.

No shit. The question specifically applies only to the situation when you did get four heads in a row, which is obviously one of the possible outcomes. If you want to reproduce the conditions of the puzzle, you pick a coin and flip four times. If you don't get four heads, put the coin back and randomly pick again until you do. The puzzle doesn't have to spell that out.

Which negates the first line of the puzzle. If you are throwing out coins that flip tails, you are NOT picking a coin at random, you are picking a coin that is much more likely to flip heads 4 times in a row.

 

[CoinOperatedBoy]

Originally posted by: Chiropteran

Originally posted by: CoinOperatedBoy

Originally posted by: Chiropteran
If you even try to repeat the coin toss puzzle a couple times, you will almost assuredly get a result other than heads X4, invalidating the entire puzzle.

No shit. The question specifically applies only to the situation when you did get four heads in a row, which is obviously one of the possible outcomes. If you want to reproduce the conditions of the puzzle, you pick a coin and flip four times. If you don't get four heads, put the coin back and randomly pick again until you do. The puzzle doesn't have to spell that out.

Which negates the first line of the puzzle. If you are throwing out coins that flip tails, you are NOT picking a coin at random, you are picking a coin that is much more likely to flip heads 4 times in a row.

Ban.

 

[Chiropteran]

Originally posted by: CoinOperatedBoy
Ban.

If you are going to be a baby about it just quit posting.

 

[Engineer]

Damn guys, you still debating this, lol???!??

The answer has been given. The OP was slightly worded "funny" but once explained, it makes perfect sense. I don't even have to know the formulas to understand why it's above 50% (near 72%).

 

[her209]

Originally posted by: her209
Let's modify the the OP's question:

You have 20 coins. One of the coins is a trick coin that has both sides heads. The rest have heads/tails. Someone adds one trick coin that has both sides tails to the jar. 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?

Since the question assumes that you picked a coin and flipped it 4 times and got heads, can we essentially ignore the coin with both sides tails from our calculations? In other words, the probability of picking the coin with both sides heads is 1/20 and not 1/21 since picking the coin both sides tails invalidates the question?

Bump!

 

[Chiropteran]

Originally posted by: her209

Originally posted by: her209
Let's modify the the OP's question:

You have 20 coins. One of the coins is a trick coin that has both sides heads. The rest have heads/tails. Someone adds one trick coin that has both sides tails to the jar. 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?

Since the question assumes that you picked a coin and flipped it 4 times and got heads, can we essentially ignore the coin with both sides tails from our calculations? In other words, the probability of picking the coin with both sides heads is 1/20 and not 1/21 since picking the coin both sides tails invalidates the question?

Bump!

I'm pretty sure the answer is yes. If you pick the double tails coin you flip it once, don't get heads, and start over. It's effectively not among the choices.

 

[TuxDave]

Originally posted by: CoinOperatedBoy
Q: You have a bag with a bunny and a squirrel inside. You blindly reach in and pull out the bunny. What is left in the bag?
Chiropteran: I JUST TRIED TO REPRODUCE THIS IN REALITY AND I PULLED OUT THE SQUIRREL AND THE SQUIRREL BIT ME. THIS IS ABSURD!

That is surprisingly an accurate analogy to this question.

 

[PlasmaBomb]

Originally posted by: CoinOperatedBoy
Q: You have a bag with a bunny and a squirrel inside. You blindly reach in and pull out the bunny. What is left in the bag?
Chiropteran: I JUST TRIED TO REPRODUCE THIS IN REALITY AND I PULLED OUT THE SQUIRREL AND THE SQUIRREL BIT ME. THIS IS ABSURD!

:laugh:

 

[GodlessAstronomer]

Originally posted by: Chiropteran

Originally posted by: CoinOperatedBoy

Originally posted by: Chiropteran
If you even try to repeat the coin toss puzzle a couple times, you will almost assuredly get a result other than heads X4, invalidating the entire puzzle.

No shit. The question specifically applies only to the situation when you did get four heads in a row, which is obviously one of the possible outcomes. If you want to reproduce the conditions of the puzzle, you pick a coin and flip four times. If you don't get four heads, put the coin back and randomly pick again until you do. The puzzle doesn't have to spell that out.

Which negates the first line of the puzzle. If you are throwing out coins that flip tails, you are NOT picking a coin at random, you are picking a coin that is much more likely to flip heads 4 times in a row.

Holy facepalm. Did you eat paint chips as a kid?

 

[Chiropteran]

Originally posted by: GodlessAstronomer
Did you eat paint chips as a kid?

No, but I eat them as an adult. Why?

 

[CoinOperatedBoy]

Originally posted by: her209

Originally posted by: her209
Let's modify the the OP's question:

You have 20 coins. One of the coins is a trick coin that has both sides heads. The rest have heads/tails. Someone adds one trick coin that has both sides tails to the jar. 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?

Since the question assumes that you picked a coin and flipped it 4 times and got heads, can we essentially ignore the coin with both sides tails from our calculations? In other words, the probability of picking the coin with both sides heads is 1/20 and not 1/21 since picking the coin both sides tails invalidates the question?

Bump!

Yes, the result will be the same, 51/70. You can essentially omit the tails/tails coin when considering the scenario of four heads flips. Here is the calculation if you include it.

Event TH = You choose the trick coin that is only heads
Event TT = You choose the trick coin that is only tails
Event N = You choose a normal coin
Event X1 = You flip the coin once and get heads
Event X4 = You flip the coin 4 times and get heads each time
Event X5 = You flip the coin 5 times and get heads each time

P(TH) = 1/21
P(TT) = 1/21
P(N) = 19/21

P(X4|TH) = 1
P(X4|TT) = 0
P(X4|N) = (1/2)(1/2)(1/2)(1/2) = 1/16
P(X4|TH') = P(X4|TT)P(TT) + P(X4|N)P(N) = (0)(1/21) + (1/16)(19/21) = 19/336

P(X4) = P(X4|TH)P(TH) + P(X4|TH') = (1)(1/21) + 19/336 = 5/48

P(TH|X4) = (P(X4|TH)P(TH)) / P(X4) = ((1)(1/21)) / (5/48) = 16/35
P(TT|X4) = (P(X4|TT)P(TT)) / P(X4) = 0
P(N|X4) = (P(X4|N)P(N)) / P(X4) = (1/16)(19/21)) / (5/48) = 19/35

P(X5) = P(TH|X4)P(X1|TH) + P(TT|X4)P(X1|TT) + P(N|X4)P(X1|N)
= (16/35)(1) + (0)(0) + (19/35)(1/2) = 51/70 = 0.7286


This makes sense logically, because as soon as you see heads you know you don't have the tails/tails coin and it ceases to have any effect on the probability of future flips.