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Author Topic: A question in probability  (Read 34980 times)

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Watno

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Re: A question in probability
« Reply #100 on: December 02, 2015, 01:00:27 pm »
+1

Just to make sure: Can you check what the result is if you always look at the first 3 non-Ts?
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silverspawn

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Re: A question in probability
« Reply #101 on: December 02, 2015, 01:01:51 pm »
+1

I'm pretty sure randomizing which subste we look at is redundant. Since the order is not fixed, every arrangement is equally likely. The probability of any particular 6-element subset to fulfill the conditions is the same, so it doesn't matter which one we look at.

Yes, it is. The result is the same. But I changed it anyway for the same of the argument.

PPE: I had it like that before.

silverspawn

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Re: A question in probability
« Reply #102 on: December 02, 2015, 01:04:58 pm »
+2

Man. I really wish I had not been confused halfway. What I was thinking initially was correct, and then for some reason I wrote a simulation that has the exact same flaw that I've been trying to explain all along and that made me think Haddock was right after all. Now, instead of being the guy who was right all along, I'll just be the guy who flip-flopped  :(

But regardless, I'm back to being completely sure that it's as simple as I've been saying since the beginning. 10 numbers out of 50. 20%.

XerxesPraelor

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Re: A question in probability
« Reply #103 on: December 02, 2015, 01:41:48 pm »
+1

What matters is that you ended up being right, and you helped other people be right. Flip-flopping should happen much more than it does.
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scott_pilgrim

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Re: A question in probability
« Reply #104 on: December 02, 2015, 03:40:20 pm »
+1

So I was asleep for this whole discussion, but I just finished reading through it now.  I meant to say in my post last night that there might be a way of phrasing the problem to get Haddock's answer, and as I was thinking about it this morning, I realized that there was, so I semi-take back what I said in the other thread, because the problems I addressed there were ambiguous (my solutions were correct only for a particular understanding of the problems).

As for the actual game, I am pretty certain that silverspawn's simulation is correct (and also that randomizing the order is unnecessary and should yield the same result as looking at the first three non-T letters, though it looks like that has been confirmed by the simulation too).  So let me see if I can get the actual reasoning correct:

The three T's have no effect on anything, so really we're just looking at four letters.
We know that a random two letters are P and a random one is K.  But again, we can ignore these, because we only care about the one that we don't already know.
The probability of getting whatever letter we wanted (I forgot which one) given that it's not a T is 1/5.

So yeah, this is the same as my understanding of the two 5-sided dice problem.  The way in which we get the information is not to re-roll until we get a 5 (Haddock's interpretation of the problem), but instead to look at what we already rolled, look at a random one of the dice (this is the same as always looking at the first die), and see that it happened to be a 5.  The same thing happened in the game.  We found out first that there are exactly three T's.  Of the remaining four letters, we looked at a random three of them, and saw their results; then asked what the fourth one showed.

For Haddock's argument to be correct, we would have needed to re-roll the set-up over and over until we got a situation where we had 2+ P's and 1+ K, and then ask what the other one showed.  This is subtly different, because it's not taking into account that we stumbled upon that information randomly.  This would have been the correct solution if the situation was something like "any time 2 (or more) P's and 1 (or more) K is rolled, the corresponding roles will claim", and we wanted to find the probability of the non-claimer representing a P from there.
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Haddock

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Re: A question in probability
« Reply #105 on: December 02, 2015, 05:40:58 pm »
+2

Just to make sure: Can you check what the result is if you always look at the first 3 non-Ts?
Yes and so it should.  Jeez I'm thick.  Yes this randomised-looking procedure should absolutely yield 20% that's obvious.  I'm an idiot.  And in that setup order doesn't matter at all, you can just fix the frame, no problem.

Scott, you're right, I'm sorry.  The way to phrase the question so that my answer is correct I have finally figured out.  It's the following. The setup is: you know from the start that the info you will get is simply yes/no answer to the 3 questions, namely:
"Are there exactly 3 Ts?" "Are there 2 or more Ps?"  "Is there at least one K?"
If you know from the start that the information you will get is just answers to these three questions, then when you're given three yes answers the resulting probability is 2/23.

I agree that this is not a good model for what would happen in a mafia game; although it is probably the most likely way for a problem like this to be phrased if it were just an academic exercise, which is why I was so subconsciously attached to it.

In fact the randomised frame model probably isn't great as a mafia approximate either, but it's definitely closer.

You guys are right, I'm clearly going crazy.  :-[

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M Town Wins-Losses (6-2, 75%): 71, 72, 76, 81, 83, 87 - 79, 82.  M Scum Wins-Losses (2-1, 67%): 80, 101 - 70.
RMM Town Wins-Losses (3-1, 75%): 42, 47, 49 - 31.  RMM Scum Wins-Losses (3-3, 50%): 33, 37, 43 - 29, 32, 35.
Modded: M75, M84, RMM38.     Mislynched (M-RMM): None - 42.     Correctly lynched (M-RMM): 101 - 33, 33, 35.       MVPs: RMM37, M87

Hyphen-ated

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Re: A question in probability
« Reply #106 on: December 02, 2015, 06:29:10 pm »
+2

It's the difference between:

"I flip two coins, at least 1 is a head, what is the probability that both are heads?"
and
"I flip two coins, the first is a head, what is the probability that the 2nd is too?"

Completely different questions with completely different answers, depending only on KNOWING which coin you're missing information about.
Noooooo, this is wrong. Both of those problems are ill-posed because the information distribution procedures in question are unspecified! The different wordings between them are causing you to have different intuitions about what the procedures would have been, and then solving as if the procedures you intuited are actually stated in the problem.

It always matters how you know what you know.
« Last Edit: December 02, 2015, 06:31:03 pm by Hyphen-ated »
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liopoil

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Re: A question in probability
« Reply #107 on: December 02, 2015, 08:44:07 pm »
+6

This thread: the most civil debate about (relatively basic, though not easy) probability on the internet ever.
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Hyphen-ated

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Re: A question in probability
« Reply #108 on: December 02, 2015, 08:52:29 pm »
+2

The different wordings between them are causing you to have different intuitions about what the procedures would have been
Specifically, the procedures you're intuiting are probably these:

#1: Flip two coins. If there's at least one heads, reveal "there's at least one heads". Otherwise reveal that it was TT. With this procedure, given that you know there's at least one heads, the chance of HH is 1/3.

#2: Flip two coins. Reveal the first. With this procedure, given that the first is heads, the chance of HH is 1/2.

You can come up with OTHER procedures that might reveal "there's at least one heads" or "the first coin is heads". These other procedures can give you different probabilities of HH at the end. Here are two examples:

#1.b: Flip two coins. If they are HH, reveal "there's at least one heads". Otherwise reveal "there's at least one tails". With this procedure, given that you know there's at least one heads, the chance of HH is 100%.

#2.b: Flip two coins. If they are HH, reveal the first. Otherwise reveal the second. With this procedure, given that you know the first is heads, the chance of HH is 100%.

Moral of the story: you cannot analyze these problems independently from the information procedure.
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GendoIkari

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Re: A question in probability
« Reply #109 on: December 02, 2015, 11:48:16 pm »
+5

This thread: the most civil debate about (relatively basic, though not easy) probability on the internet ever.

That's not even close to true you idiot.
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Titandrake

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Re: A question in probability
« Reply #110 on: December 03, 2015, 12:13:54 am »
0

This thread: the most civil debate about (relatively basic, though not easy) probability on the internet ever.

That's not even close to true you idiot.

Not sure if joke or serious. Poe's Law please.
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liopoil

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Re: A question in probability
« Reply #111 on: December 03, 2015, 12:31:38 am »
+1

This thread: the most civil debate about (relatively basic, though not easy) probability on the internet ever.

That's not even close to true you idiot.

Not sure if joke or serious. Poe's Law please.
He's probably serious in that I am of course completely wrong, but he's an idiot for thinking I can take a joke.
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Haddock

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Re: A question in probability
« Reply #112 on: December 03, 2015, 05:57:56 am »
0

The different wordings between them are causing you to have different intuitions about what the procedures would have been
Specifically, the procedures you're intuiting are probably these:

#1: Flip two coins. If there's at least one heads, reveal "there's at least one heads". Otherwise reveal that it was TT. With this procedure, given that you know there's at least one heads, the chance of HH is 1/3.

#2: Flip two coins. Reveal the first. With this procedure, given that the first is heads, the chance of HH is 1/2.

You can come up with OTHER procedures that might reveal "there's at least one heads" or "the first coin is heads". These other procedures can give you different probabilities of HH at the end. Here are two examples:

#1.b: Flip two coins. If they are HH, reveal "there's at least one heads". Otherwise reveal "there's at least one tails". With this procedure, given that you know there's at least one heads, the chance of HH is 100%.

#2.b: Flip two coins. If they are HH, reveal the first. Otherwise reveal the second. With this procedure, given that you know the first is heads, the chance of HH is 100%.

Moral of the story: you cannot analyze these problems independently from the information procedure.
I got there eventually.  See my earlier post, where I finally figure out what the information procedure is that leads to my original answer.

I'm crazy embarrassed now.  In my defence, I think the information procedure that gives my answer IS the one that would be used if this were just an abstract exercise in probability, while the randomised procedure is somewhat closer to the real situation.
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The best reason to lynch Haddock is the meltdown we get to witness on the wagon runup. I mean, we should totally wagon him every day just for the lulz.

M Town Wins-Losses (6-2, 75%): 71, 72, 76, 81, 83, 87 - 79, 82.  M Scum Wins-Losses (2-1, 67%): 80, 101 - 70.
RMM Town Wins-Losses (3-1, 75%): 42, 47, 49 - 31.  RMM Scum Wins-Losses (3-3, 50%): 33, 37, 43 - 29, 32, 35.
Modded: M75, M84, RMM38.     Mislynched (M-RMM): None - 42.     Correctly lynched (M-RMM): 101 - 33, 33, 35.       MVPs: RMM37, M87

Haddock

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Re: A question in probability
« Reply #113 on: December 03, 2015, 06:00:49 am »
0

But regardless, I'm back to being completely sure that it's as simple as I've been saying since the beginning. 10 numbers out of 50. 20%.
Bah.  Just as I'd agreed with you. 
I think this whole argument (between intelligent and well-qualified people) proves that this is really not as simple as that.

The fact that the answer is 20% is a result of something much more complex than just "10 numbers out of 50".

But yes 20% is probably a pretty good approximation to what the answer actually was in the context of the game. (as previously mentioned we'll never get an exact answer because of the human element.)
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The best reason to lynch Haddock is the meltdown we get to witness on the wagon runup. I mean, we should totally wagon him every day just for the lulz.

M Town Wins-Losses (6-2, 75%): 71, 72, 76, 81, 83, 87 - 79, 82.  M Scum Wins-Losses (2-1, 67%): 80, 101 - 70.
RMM Town Wins-Losses (3-1, 75%): 42, 47, 49 - 31.  RMM Scum Wins-Losses (3-3, 50%): 33, 37, 43 - 29, 32, 35.
Modded: M75, M84, RMM38.     Mislynched (M-RMM): None - 42.     Correctly lynched (M-RMM): 101 - 33, 33, 35.       MVPs: RMM37, M87

Asper

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Re: A question in probability
« Reply #114 on: December 11, 2015, 08:30:28 pm »
0

Huh, that reminds me of a gameshow puzzle. I hope nobody minds if i describe it here, because it seems applicable to some degree. I also apologize for not having read this thread completely before:

In a gameshow we have three doors. One contains a prize, the other two don't. I choose a door. Now the show master opens one of the doors that does not have the prize and is not the door i chose. He asks me whether i want to change my mind and choose the other door. Should i or does it not matter?

The answer: I should. I have 1/3 probability to choose the right door in the beginning. Only in that case will he open one of two bad doors, and only in that case would switching lose me the game. On the contrary, if i chose wrongly, with a probability of 2/3, he will choose the other bad door, and switching would win me the game. So it's double likely to win if i switch. Of course, the analogy isn't perfect. A real game master, who knows the door, might in fact refuse to open the bad door and just leave me with my bad door, or open the good one and just plain tell me i lost, or act in another way. But i tried this. Maybe somebody finds it halfway applicable and it helps understand the problem.

Because i don't. I find it incredible how easily i will be absolutely sure to be right on probability matters, just to be proven wrong. That teached me a lesson.

Edited because i originally wrote "price" instead of "prize". English is hard.
« Last Edit: December 12, 2015, 05:39:33 am by Asper »
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heron

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Re: A question in probability
« Reply #115 on: December 11, 2015, 08:56:18 pm »
0

Actually the best strategy is to keep the same door to avoid having to pay a price.
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eHalcyon

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Re: A question in probability
« Reply #116 on: December 11, 2015, 10:06:30 pm »
+2

Huh, that reminds me of a gameshow puzzle. I hope nobody minds if i describe it here, because it seems applicable to some degree. I also apologize for not having read this thread completely before:

In a gameshow we have three doors. One contains a price, the other two don't. I choose a door. Now the show master opens one of the doors that does not have the price and is not the door i chose. He asks me whether i want to change my mind and choose the other door. Should i or does it not matter?

The answer: I should. I have 1/3 probability to choose the right door in the beginning. Only in that case will he open one of two bad doors, and only in that case would switching lose me the game. On the contrary, if i chose wrongly, with a probability of 2/3, he will choose the other bad door, and switching would win me the game. So it's double likely to win if i switch. Of course, the analogy isn't perfect. A real game master, who knows the door, might in fact refuse to open the bad door and just leave me with my bad door, or open the good one and just plain tell me i lost, or act in another way. But i tried this. Maybe somebody finds it halfway applicable and it helps understand the problem.

Because i don't. I find it incredible how easily i will be absolutely sure to be right on probability matters, just to be proven wrong. That teached me a lesson.

This is the Monty Hall problem, and I think it was referenced earlier in this thread.

If you find the Monty Hall problem tough to intuit, it may be easier if you exaggerate the conditions.  There are 100 doors and 1 prize.  When the host offers the switch, he first opens 98 of the dud doors.
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Awaclus

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Re: A question in probability
« Reply #117 on: December 12, 2015, 05:22:30 am »
0

If you find the Monty Hall problem tough to intuit, it may be easier if you exaggerate the conditions.  There are 100 doors and 1 prize.  When the host offers the switch, he first opens 98 of the dud doors.

I don't think this really helps, there's still only two doors remaining so if you're not focusing on the process of opening the dud doors, I'm not sure how it will change it. What originally helped me understand the problem was emphasizing the fact that the host won't open your chosen door ever, regardless of its contents.
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eHalcyon

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Re: A question in probability
« Reply #118 on: December 12, 2015, 11:07:59 am »
+2

If you find the Monty Hall problem tough to intuit, it may be easier if you exaggerate the conditions.  There are 100 doors and 1 prize.  When the host offers the switch, he first opens 98 of the dud doors.

I don't think this really helps, there's still only two doors remaining so if you're not focusing on the process of opening the dud doors, I'm not sure how it will change it. What originally helped me understand the problem was emphasizing the fact that the host won't open your chosen door ever, regardless of its contents.

It helps because it's easier to see how much less likely you were to choose the right door in the first place.
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liopoil

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Re: A question in probability
« Reply #119 on: December 12, 2015, 11:24:58 am »
+2

The one that got me was linearity of expectation. Once I had learned all of these subtle things about conditional probabilities, I had a lot of trouble believing the answer to this question:

I put the numbers 1,2,3,4,5 in a hat, then draw them out randomly one at a time. How many total times will the nth number drawn be n?

The expected number of times this occurs per trial: 1 (moat). Moreover, this holds for arbitrarily large values of 5
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scott_pilgrim

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Re: A question in probability
« Reply #120 on: December 12, 2015, 03:00:02 pm »
+1

If you find the Monty Hall problem tough to intuit, it may be easier if you exaggerate the conditions.  There are 100 doors and 1 prize.  When the host offers the switch, he first opens 98 of the dud doors.

I don't think this really helps, there's still only two doors remaining so if you're not focusing on the process of opening the dud doors, I'm not sure how it will change it. What originally helped me understand the problem was emphasizing the fact that the host won't open your chosen door ever, regardless of its contents.

I've used that explanation a lot to help people understand the problem, and mostly I've found it to be the most helpful explanation.  I think the fact that people find it helpful shows that their problem is not with understanding the set-up itself, but rather with understanding the intuition behind the solution.  It sounds like your difficulty with the problem was not the same as most people's (I think it was the same as mine), which was actually thinking about how the host decided which door(s) to open.

I guess what I'm saying is, the fact that it wouldn't have helped you personally doesn't mean it wouldn't help anyone.  I've had a lot of success convincing people about it by using the 100 doors argument.
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silverspawn

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Re: A question in probability
« Reply #121 on: December 12, 2015, 03:05:14 pm »
0

What originally helped me understand the problem was emphasizing the fact that the host won't open your chosen door ever, regardless of its contents.

the same is true for me

blueblimp

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Re: A question in probability
« Reply #122 on: December 12, 2015, 04:54:20 pm »
+2

The one that got me was linearity of expectation.
Linearity of expectation is the best. It makes all sorts of expected value calculations unexpectedly easy.
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Re: A question in probability
« Reply #123 on: December 12, 2015, 05:35:58 pm »
+1

It makes all sorts of expected value calculations unexpectedly easy.

>:(
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Titandrake

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Re: A question in probability
« Reply #124 on: December 12, 2015, 06:27:52 pm »
0

It makes all sorts of expected value calculations unexpectedly easy.

>:(

I wouldn't be mad, doesn't look like an intended pun.
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