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bonanova

Probability 101

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Plato: Good morning, class, today's lesson is on probability.

Aristotle: Fantastic.
I'm headed to Vegas this weekend, and I can use some pointers.

P: Curb your enthusiasm kid, this is serious stuff.
Here, roll this pair of dice, but don't look at the the result. OK, good.
Now without looking, tell me the probability that you rolled a seven.
If you're going to play craps this is important.
By the way, I can tell you that one of your dice is a four.

A: Hmm...
So I could have rolled 41 42 43 44 45 46 14 24 34 54 or 64,
all with equal likelihood, with 34 and 43 making seven.
That's a probability of 2/11.

P: You're on a roll kid, now let's do it again. Great.
Again without looking, what's the probability you rolled a seven?
By the way, I can tell you one of your dice is a one.

A: So I could have rolled 11 12 13 14 15 16 61 51 41 31 or 21,
with 16 and 61 making seven. Hmm... it's the same as before - 2/11.

P: And if I had told you one of your dice was a five?

A: Well ... I guess it really doesn't matter what number you tell me.
It will always come out the same. The probability will be 2/11.

P: So what can we deduce from that?

A: That the probability of two dice making seven is ... 2/11.
But wait... Hey, you're not really Professor Plato, are you?

P: No. I'm an insurance salesman.

So ... what exactly is the probability of making seven?

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6 hours ago, CaptainEd said:

 

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The use of language is tricky here. The OP begins by saying “one of the dice is a 4”. We understand that to mean “at least one of the dice is a 4”, leading to the answer 2/11.

But later, in Bonanova’s response, a paragraph begins “one of his dice has a particular value”, and continues as if we are speaking of one particular die.

Let’s rephrase the original discussion in terms of two distinct dice: a red die and a white die.

Now Aristotle rolls the dice, and Plato says “the red one is a 4”. What is the probability of a seven total? 1/6. 

So if we are speaking of one specific die, the answer does not change : 1/6, and Thalia’s answer is right. 

The phrase “in each particular case he [Aristotle] reasons the probability of seven changes to a new value” shows that Aristotle has been swindled.

Thank you, Bonanova, this led me to think about the linguistic difficulties faced by Pascal and Fermat, as they initiated the study of probability.

 

 

 

Spoiler

Instead of red and white, you could just use "the die Plato said was a 4" and "the other one."

 

5 hours ago, CaptainEd said:

How did all those 2/11 become Thalia’s answer?

 

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draw a 6x6 table, and tally one in each cell each time Plato asks about another number (11 cells per number). When we are done, all 6 of the diagonal elements (11,22,33,44,55,66,) have one tally each, while all of the off diagonal cells have two tallies. Therefore, Aristotle can’t simply average the 2/11s. Once they have been normalized, we (Thalia, Pascal,  Fermat and others) see 36 cases, of which 6 have coordinates adding to 7.

 

 

 

Spoiler

First, Thalia just says the answer is 1/6 without the extra "i can tell you one die is X", so has nothing to do with 2/11.  I assume you mean Aristotle.

But why does there need to be a normalization at all?  What went wrong such that it was necessary?

Spoiler

 

On 6/27/2019 at 2:52 AM, bonanova said:

A: Hmm...
So I could have rolled 41 42 43 44 45 46 14 24 34 54 or 64,
all with equal likelihood, with 34 and 43 making seven.
That's a probability of 2/11.

It is true that P(11)=P(12)=...=P(66).  Each possible dice roll can happen with equal probability.

But once Plato says that "one die is a 4," it doesn't merely prune out the ones without 4's.  Using Bayes theorem we can see this.

P(41 | one is 4) = P(one is 4 | 41) * P(41) / P(one is 4)

Obviously P(41) is 1/36.  P(one is 4) and P(one is 4 | 41) need an assumption on how Plato decides which value to tell.  Assuming he would uniformly random choose a die to report on, we can see P(one is 4)=1/6 since all values are equally likely and P(one is 4 | 41)=1/2 since he has two options randomly chosen from.

P(41 | one is 4) = (1/2) * (1/36) / (1/6) = 1/12

This is the same for 42,43,45,46,14,24,34,54,64.

But since 44 only has one value possible to report, the value is different.

P(44 | one is 4) = 1 * (1/36) / (1/6) = 1/6.  So given that one is a 4, the roll 44 is twice as likely as the other 10 possibilities.

To finish finding the answer...

P(sum is 7 | one is 4) = P(34 | one is 4) + P(43 | one is 4) = 1/12 + 1/12 = 1/6.

Of course, if Plato always decides to tell the smaller value shown on the dice or anything other than uniformly random, the probability distribution could be wildly different.

 

 

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If you don't know any of the numbers, it's 1/6. 6x6=36 possible rolls, 6 of them adding up to 7.

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11 hours ago, Thalia said:

 

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And so thought poor Aristotle until today's class.

He knew the probability of two dice making seven, until his teacher told him the value of one of his dice. Then he reasoned it to be different. Ah, the magic of conditional probability, he thought. But then he reasoned further that it was not the knowledge of which value one of his dice had, for it did not matter whether that value was 1 2 3 4 5 or 6. It was seemingly only that it had a value. But what kind of conditional probability is that? Was he not already aware of that?

One of his dice has a particular value. Six values engender six cases. In each particular case he reasons the probability of seven changes to a new value. Worse, there are no other cases. Therefore in every case it changes to a new value. How then could it have the value he originally imagined?

So there's the question that lurks within the flavor text. Beneath the surface perhaps, but now fully revealed for all to ponder.

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The use of language is tricky here. The OP begins by saying “one of the dice is a 4”. We understand that to mean “at least one of the dice is a 4”, leading to the answer 2/11.

But later, in Bonanova’s response, a paragraph begins “one of his dice has a particular value”, and continues as if we are speaking of one particular die.

Let’s rephrase the original discussion in terms of two distinct dice: a red die and a white die.

Now Aristotle rolls the dice, and Plato says “the red one is a 4”. What is the probability of a seven total? 1/6. 

So if we are speaking of one specific die, the answer does not change : 1/6, and Thalia’s answer is right. 

The phrase “in each particular case he [Aristotle] reasons the probability of seven changes to a new value” shows that Aristotle has been swindled.

Thank you, Bonanova, this led me to think about the linguistic difficulties faced by Pascal and Fermat, as they initiated the study of probability.

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How did all those 2/11 become Thalia’s answer?

draw a 6x6 table, and tally one in each cell each time Plato asks about another number (11 cells per number). When we are done, all 6 of the diagonal elements (11,22,33,44,55,66,) have one tally each, while all of the off diagonal cells have two tallies. Therefore, Aristotle can’t simply average the 2/11s. Once they have been normalized, we (Thalia, Pascal,  Fermat and others) see 36 cases, of which 6 have coordinates adding to 7.

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18 hours ago, EventHorizon said:

 

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Instead of red and white, you could just use "the die Plato said was a 4" and "the other one."

 

 

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First, Thalia just says the answer is 1/6 without the extra "i can tell you one die is X", so has nothing to do with 2/11.  I assume you mean Aristotle.

But why does there need to be a normalization at all?  What went wrong such that it was necessary?

  Hide contents

 

It is true that P(11)=P(12)=...=P(66).  Each possible dice roll can happen with equal probability.

But once Plato says that "one die is a 4," it doesn't merely prune out the ones without 4's.  Using Bayes theorem we can see this.

P(41 | one is 4) = P(one is 4 | 41) * P(41) / P(one is 4)

Obviously P(41) is 1/36.  P(one is 4) and P(one is 4 | 41) need an assumption on how Plato decides which value to tell.  Assuming he would uniformly random choose a die to report on, we can see P(one is 4)=1/6 since all values are equally likely and P(one is 4 | 41)=1/2 since he has two options randomly chosen from.

P(41 | one is 4) = (1/2) * (1/36) / (1/6) = 1/12

This is the same for 42,43,45,46,14,24,34,54,64.

But since 44 only has one value possible to report, the value is different.

P(44 | one is 4) = 1 * (1/36) / (1/6) = 1/6.  So given that one is a 4, the roll 44 is twice as likely as the other 10 possibilities.

To finish finding the answer...

P(sum is 7 | one is 4) = P(34 | one is 4) + P(43 | one is 4) = 1/12 + 1/12 = 1/6.

Of course, if Plato always decides to tell the smaller value shown on the dice or anything other than uniformly random, the probability distribution could be wildly different.

 

 

The part that I turned red in the quote bugs me a little bit.

Spoiler

If you're given that one of the numbers is four, then saying that the probability of rolling a four and a one (in that order) is 1/12 seems at odds with there being eleven possible rolls with a four. But it does make sense if you assume that Plato will randomly pick one of the two dice and announce its number, since he's twice as likely to say that you rolled a four if you rolled a double-four than if you rolled a four and any other number, so you can count that possibility twice and bring the denominator to 12.

Suppose instead that Plato picked a number N before the dice were rolled and would plan to say either "You rolled an N" or "You didn't roll an N" as if he were in the Donny Hall puzzle?

 

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16 hours ago, plasmid said:

The part that I turned red in the quote bugs me a little bit.

  Reveal hidden contents

If you're given that one of the numbers is four, then saying that the probability of rolling a four and a one (in that order) is 1/12 seems at odds with there being eleven possible rolls with a four. But it does make sense if you assume that Plato will randomly pick one of the two dice and announce its number, since he's twice as likely to say that you rolled a four if you rolled a double-four than if you rolled a four and any other number, so you can count that possibility twice and bring the denominator to 12.

Suppose instead that Plato picked a number N before the dice were rolled and would plan to say either "You rolled an N" or "You didn't roll an N" as if he were in the Donny Hall puzzle?

 

 

Spoiler

The part you turned red follows directly from the assumption preceding it.  Perhaps I didn't make this point explicit enough in my post.  P(sum is 7)=1/6, but P(sum is 7 | rolled an N) is dependent upon how the information "rolled an N" is determined and the probability distribution can vary wildly based on it.

I assumed a method that left P(sum is 7 | rolled an N) = 1/6, but as your example shows it is not necessarily the case.

Another example, Lets say Plato always reported the smallest even number rolled, or just the smallest number if no evens are rolled.  It is easy to see that P(sum is 7 | rolled a 1) = P(sum is 7 | rolled a 3) = P(sum is 7 | rolled a 5) = 0 because if the other die is the even number necessary to make 7, it would have been reported instead.

I guess what I'm trying to say is that P(sum is 7 | rolled an N) says as much or more about the information "rolled an N" than the probability the dice sum to 7.

To liken it to the Monty Hall problem, the best action depends upon Monty himself.  If he would have always opened a door you didn't pick (which doesn't have the prize) and lets you switch, you should switch.  If he only makes it seem like he'd always have opened a door without the prize and given you a chance to switch, but simply opens the door you picked if you had picked wrong... you should stay.  The problem is, you don't know what is in Monty's head.

addendum:

I didn't come up with this myself.  Monty Hall was asked about the Monty Hall problem, and his response was such.

Edited by EventHorizon
Addendum added

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I agree completely with EventHorizon, and will try to summarize in a way that addresses the OP and deconvolutes the paradox at its heart:

Spoiler

If Plato were to pick a number, say 4, and decide that he would tell you "You rolled a 4" or "You didn't roll a 4", then Aristotle would actually be correct. There's an 11/36 chance that he would roll a 4, and among those outcomes there are 2 that sum to 7, so if Plato says "you rolled a 4" then there really is a 2/11 chance that the sum is 7. Alternatively, there's a 25/36 chance that he doesn't roll any 4s, and among those outcomes there are 4 that sum to 7, so if Plato says "you didn't roll a 4" then there's a 4/25 chance that the sum is 7. So Aristotle is slightly more likely to have rolled a 7 if Plato says he rolled a 4 (or any other number). The reason for this is that if Aristotle rolls doubles then it obviously won't sum to 7, and Plato is slightly less likely to say that Aristotle rolled any given number if he rolled doubles (1/6 chance of rolling the number Plato picked) than if he rolled two different numbers (1/3 chance of rolling the number Plato picked).

On the other hand, if Plato picks one of the two numbers that are rolled and calls it out, the math is different. If Aristotle rolls two different numbers then there's a 1/2 chance that Plato will call out either of the two numbers, but if he rolls doubles then there's a 1/1 chance that Plato will call out that number. So there are 11 possible rolls that could lead to Plato calling out 4, but one of those possibilities (rolling double 4s) is twice as likely to lead Plato saying that he rolled a 4 as the others, so the odds of rolling a sum of 7 should be 2/12 instead of 2/11. This fits with our intuition that if Plato simply picks one of the numbers to call out then it shouldn't affect the probability of whether the sum is 7.

The apparent paradox arises because you're doing the math assuming that Plato's acting as described in the first paragraph, but really he's acting as described in the second paragraph.

 

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@EventHorizon said,

But once Plato says that "one die is a 4," it doesn't merely prune out the ones without 4's

That’s the crux of it. 

EH is correct that this reads on Monty Hall. But more directly on the long-running Teanchy-Beanchy post (one of his two kids is a girl, what’s the probability he has two girls.) First people said has to be 1/2. Then others (including me) said (all that matters is that) it can’t be BB so it’s 1/3. Both wrong. 

so I made this one up to show that the informant’s algorithm has to be known. 

Nicely explained. 

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