OT - Five card flush puzzle

[Vaj]

As for the pill problem, I think the point of confusion involves Grunion's maintining of the a priori probabilities of the color of each pill, and then adjusting the probabilities of the color of the second pill drawn based on the additional knowledge provided.

There are 30 possible permutations of pills that can be drawn. Let's list them:

R1-R2, R1-B1, R1-B2, R1-G1, R1-G2
R2-R1, R2-B1, R2-B2, R2-G1, R2-G2
G1-G2, G1-B1, G1-B2, G1-R1, G1-R2
G2-G1, G2-B1, G2-B2, G2-R1, G2-R2
B1-B2, B1-R1, B1-R2, B1-G1, B1-G2
B2-B1, B2-R1, B2-R2, B2-G1, B2-G2

Clearly, there are 8 fatal permutations here. Also, each pill has an equal probability of being the first one selected.

Now, let's list the possible permutations after we know that a red/green combination was not drawn:

R1-R2, R1-B1, R1-B2
R2-R1, R2-B1, R2-B2
G1-G2, G1-B1, G1-B2
G2-G1, G2-B1, G2-B2
B1-B2, B1-R1, B1-R2, B1-G1, B1-G2
B2-B1, B2-R1, B2-R2, B2-G1, B2-G2

Again, 8 fatal permutations, only now there are only 22 possible permutations. Grunion's calculation reduces to 8/22.5. I don't see how we can have 22.5 possible permutations.

Alternatively, we can use Grunion's method of calculating a red/blue combination consisitently applying the posterior probabilitites. The posterior probability that the first pill is red is 6/22. Given that the first pill is red, the posterior probability that Fred croaks is 2/3. The posterior probability that the first pill is blue is 10/22. Given that the first pill is blue, the posterior probability
that Fred dies is 2/5. If the first pill chosen is green, Fred lives, so I'll ignore the resultant zero term. Thus, the probability that Fred dies, given that he did not choose a red and green pill, is (6/22)*(2/3) + (10/22)*(2/5) = 12/66 + 20/110 = 12/66 + 12/66 = 24/66 = 8/22.

[QuikSand]

Here is my best re-statement of the semantic differences between Grunion and QuikSand at this point:

Grunion states his preferred wording of my original puzzle thusly:

What are the odds of drawing a flush given that your hand contains at least one ace.

And it appears that he and I differ in how we might "elucidate" on this phrase's intended meaning.

I would state it this way:

Given only the knowledge that your hand does, in fact, include at least one ace (and absent any knowledge about the method used to generate the hand), what is the likelihood that your hand is a flush?

And I'd answer it 0.223%.

He would (I believe) state it something like this:

Given that you have a hand that was created by a method ensured to generate a hand that includes at least one ace, what is the likelihood that your hand is a flush?

And he'd answer 0.189%.

---

Since this seems to pretty rapidly dissolve into a Clintonesque debate over the meanings of specific words, I'm not sure of there is a way to "resolve" this. But at this point, I don't think there are any real mathematical differences between us-- the differences are in the translation of sentences and phrases into the mathematical constructs needed to solve them.

Incidentally, Grunion, I'm hoping to provide a fair recap here-- if you would prefer to re-state your position in the context above in some manner, I'd be more than happy to re-write it in your words. I think I fairly understand our difference, but I don't seek to put improper words into your mouth.

[Grunion]

However, when we're given no set of conditions about the distribution of the original set of hands (as we were not
in the original flush problem) I argue that it is invalid to use the conditions that are observed, and to then use any process to generate anything other than an even distribution of hands.

Ahhh, but we are given information about the distribution about the original set of hands. We are being told that we have at least one ace, which is information relevant to the distribution of hands.

[Grunion]

Again, 8 fatal permutations, only now there are only 22 possible permutations. Grunion's calculation reduces to 8/22.5. I don't see how we can have 22.5 possible permutations.

8/22.5=16/45=(2*2*2*2)/(5*3*3)

So, in actuality, my solution factors quite nicely.

Alternatively, we can use Grunion's method of calculating a red/blue combination consisitently applying the
posterior probabilitites. The posterior probability that the first pill is red is 6/22. Given that the first pill is red, the
posterior probability that Fred croaks is 2/3. The posterior probability that the first pill is blue is 10/22. Given that
the first pill is blue, the posterior probability that Fred dies is 2/5. If the first pill chosen is green, Fred lives, so I'll ignore the resultant zero term. Thus, the probability that Fred dies, given that he did not choose a red and green pill, is (6/22)*(2/3) + (10/22)*(2/5) = 12/66 + 20/110 = 12/66 + 12/66 = 24/66 = 8/22.

Vaj,

Debate fairly or don't debate at all. You can't solve a problem using an initial set up which I have clearly indicated I disagree with, and call it Grunion's method. Then use "Grunion's" method to achieve QuikSand's result, implying that QuikSand is correct.

[Grunion]

I would state it this way:

Given only the knowledge that your hand does, in fact, include at least one ace (and absent any knowledge about the method used to generate the hand), what is the likelihood that your hand is a flush?

And I'd answer it 0.223%.

But we do have knowledge about the method used:

For purposes of this puzzle, all the probabilities are as they seem-- the hands described are the product of some purely random selection process-- there is no trickery involved, just pure probability.

So in essence, are disagreement can be also looked at this way.
Given a condition, does a random model apply the constraint prior to the generation of combinations or afterwards?

What I disagree with is your bastardization of Bayes Theorem, which is a tool to calculate posterior probability by
using prior probabilities. You've created a very elegant little Bayesian problem here, whoch lends itself quite nicely to a Bayesian analysis. However, in doing so, you are not properly using prior probabilities... specifically the prior probability that p(B) =Fred gets a red pill = 10/12.

Your own words seem to indicate that you believe it to be more appropriate to apply the constraint prior to the generation of combinations.
My method clearly does this, yours clearly does not.

[Grunion]

QuikSand,

I believe it would help us bring this to a conclusion if you could look at the "full" process for the flush problem I posted on 12/22 at 3:04 PM, and identify specifically what you believe to be improper with it, and why.

[Vaj]

I'm sorry you took offense to my post, Grunion. I was just trying to show how you would have arrived at what I feel is the correct answer to the question at hand by weighting the probabilities of Fred croaking given the color of the first pill ingested by the probability of that first pill being selected. I was too lazy to type that all out, so I used "Grunion's method" as a shorthand description. I won't do that again!

All I was trying to show was that not only does the posterior probability of the color of the second pill chosen change with additional knowledge, but that the posterior probability of the color of the *first* pill must change as well.

Suppose we were told, after the fact, that neither pill Fred picked was green. What would be the posterior probability of the first pill being green? Can we agree that it's not 1/3, but 0? Perhaps more relevant questions are in my 7:43 post. I really shouldn't post before eating breakfast.


Incidentally, I still can't see how, when you start with 30 *discrete* permutations, a correct answer can compare the 8 fatal permutations to a winnowed down universe of possible permutations to 22.5.


[This message has been edited by Vaj (edited 12-23-2000).]

[QuikSand]

Responding to my comments:

However, when we're given no set of conditions about the distribution of the original set of hands (as we were not
in the original flush problem) I argue that it is invalid to use the conditions that are observed, and to then use any process to generate anything other than an even distribution of hands.

- - -

Grunion replied:

Ahhh, but we are given information about the distribution about the original set of hands. We are being told that we have at least one ace, which is information relevant to the distribution of hands.

I disagree, but again, let me be clear about my semantics here. When I say "distribution" I mean "the likelihood of appearance of the various 106million+ given hands." I'm pretty sure that is/was clear, but I wanted to make sure.

I would argue that since the puzzle simply calls for the "flush probability" *of the subset* that we must use the simple exercise if counting the elements of the subset, then counting its elements that meet the condition (are a flush), then divide. No other choice.

I believe I understand your interpretation, but I would re-word it: "If we have one hand that we know is from that subset (the hands with aces) then we'll use some information about a/the process that would be used to generate all those hands in that subset, employ the resulting probabilities from that process that this might be certain hands (more likely than others), and then we can calculate the overall resulting likelihood that this particular hand is a flush."

I think this is a fair statement of your method. I disagree with the interpretation that gets from the original puzzle to this method, but I understand your method and as of yet, have found nothing else with which I disagree (your math, to my reckoning, has been on target).

Another way I would articulate our differences is that I am calculating the probability of the set, while you are calculating the probability of a hand drawn from the set, using an uneven probability for various hands within the set. We both have argued our side ad infinitum, but the questions remains-- which is in keeping with the original question?

[QuikSand]

Regrettably (or perhaps not), I am walking out the door to embark on holiday travels. Perhaps it comes at a good time for this debate, I'm not sure.

Anyway, I'll probably be away for a few days.

Grunion, I've truly enjoyed this back-and-forth, and I apologize for the times when I may have slipped into a tone that suggested otherwise. I suspect you've reached a point where you are saying (about me) "he seems smart and able to understand this stuff, why can't he see he's wrong?" I feel the same way (with an emphasis on the former portions).

I hope you and your family have a happy and safe holiday season.

[Vaj]

Here's another pill question. Given the same 6 pills, and given that a red/green combination was not drawn, what is the probability of two green pills being drawn? Is this different from the probability of two blue pills being drawn?

I submit there the two possible ways to select two green pills are G1-G2 and G2-G1, and that there are 22 possible permutations (the original 30 less the 8 red/green). Thus, P(2 green|!red/green)=2/22. I would go through an analogous procedure to arrive at a 2/22 chance of 2 blue pills being selected (B1-B2, B2-B1) given that one red pill and one green pill weren't chosen.

Now I'll go through this exercise using my understanding of your approach. The probability of the first pill being red, blue, or green are all 1/3. If the first pill is green, the probability of the second pill being green is 1/3, and the probability of the second pill being blue is 2/3, as we know green/red is not allowed. Thus, the probability of two pills being green is (1/3)*(1/3), or 1/9. Now, if the first pill is blue, the probabilities of the second pill being red, green, or blue are 2/5, 2/5, and 1/5, respectively. So the probability of two blue pills being selected, given that one red and one green pill weren't selected, is (1/3)*(1/5), or 1/15.

Is there really a higher probability of choosing two green pills than two blue pills? Absent any knowledge about the two pills chosen, I think we would agree that the chances of choosing 2 blue pills and 2 green pills are, in fact, 2 in 30. However, once we know that a red/green combination wasn't selected, your method implies that suddenly it's more likely that 2 green pills were selected than 2 blue pills. I disagree.

I apologize in advance if I've misrepresented your methodology. It's not my intent at all.

[Grunion]

Vaj,

Suppose we were told, after the fact, that neither pill Fred picked was green. What would be the posterior probability of the first pill being green? Can we agree that it's not 1/3, but 0? Perhaps more relevant questions are in my 7:43 post. I really shouldn't post before eating breakfast.

Given that Fred does not have a green pill, Fred has a 0% percent chance of having a green pill.

It may seem like splitting hairs, but this is not the same as:

Fred picks two pills at random. If Fred has a green pill, the result does not count. If Fred does not have a green pill, the result counts. What is the probability that a counted result will have a green pill? (Answer is 0%)

Incidentally, I still can't see how, when you start with 30 *discrete* permutations, a correct answer can compare the 8 fatal permutations to a winnowed down universe of possible permutations to 22.5.

Once again:

Remembering that order is not important:
P(1) has a 1/3 chance of being red, a 1/3 chnce of being green and a 1/3 chance of being blue.

If P(1) is red, P(2) has a 2/3 chance of being blue and a 1/3 chance of being red (as a red/green combo is impossible)

If P(1) is blue, P(2) has a 2/5 chance of being red, a 2/5 chance of being green and a
1/5 chance of being green.

If P(1) is green, P(2) has a 1/3 chance of being green and a 2/3 chance of being blue (as a red/green combo is impossible)

P(red/blue) = .33(.67)+.33(.4)+ .33(0)= 35.5%

As you can see, the factors in my denominators are all threes and fives.
=1/3(2/3)+(1/3)(2/5)+(1/3)(0).

No offense taken on any of your posts Vaj, I enjoy the exchange.

[Grunion]

QuikSand,

Anyway, I'll probably be away for a few days.

Grunion, I've truly enjoyed this back-and-forth, and I apologize for the times when I may have slipped into a tone that suggested otherwise. I suspect you've reached a point where you are saying (about me) "he seems smart and able to understand this stuff, why can't he see he's wrong?" I feel the same way (with an emphasis on the former portions).

I hope you and your family have a happy and safe holiday season.

I feel exactly the same way. Best wishes over the holiday.

[Grunion]

Vaj,

Here's another pill question. Given the same 6 pills, and given that a red/green combination was not drawn, what is the probability of two green pills being drawn? Is this different from the probability of two blue pills being drawn?

Ok, given: a red/green combo was not drawn.

Is there really a higher probability of choosing two green pills than two blue pills? Absent any knowledge about the two pills chosen , I think we would agree that the chances of choosing 2 blue pills and 2 green pills are, in fact, 2 in 30. However, once we know that a red/green combination wasn't selected, your method implies that suddenly it's more likely that 2 green pills were selected than 2 blue pills. I disagree.

I contend that the given stated as a constraint in your problem gives us knowledge about the two pills chosen, which is if one is red the other can not be green.

I have not verified your math or your calculations, but what your result is in this exercise is not really at issue.

Our disagreement, it seems, is that I believe the contraint of a given must be applied before the randomization process, while you contend that the constraint of a given must occur after a result which may or may not be in agreement with the stated given, and all non-conforming results thrown out. This is exactly what my discussion with QuikSand boils down to, and unfortunately I believe we (myself and QuikSand) are at an impasse.
Our arguements have many, many issues raised, and when all is said and done they all boil down to the above disagreement. Unfortunately, our disagreement is conceptual. We can't resolve it because:

While I think we have both demonstrated we are fairly adept at math and logic, most likely neither one has the knowledge that will allow us to resolve our issue, as it is a theoretical one.

Wignasty said it best earlier, both of us are right. This is true, given that each of our methodologies are sound, both of our processes are right. Unfortunately, one of our methodologies are unsound and it proving difficult to prove that either one is incorrect.

QuikSand mentioned something at one point about getting together, and settling the issue through experimental means. Then we both quickly realized, that we would never be able to agree on the rules to conduct the experiment.

Vaj,
I see you are on the same page as QuikSand (which is fine). If you have time, go back through my posts to get a better understanding of my point of view. I'm not asking you to do so because I think it will change your mind, I am asking you to do so in order to fully understand the one issue which QuikSand and myself disagree on.

[Vaj]

I suspect you're right about us being at an impasse. In the pill problem, I was viewing each permutation as a unit (say a piece of paper in a hat), and, when we know that a red/green combination was not picked, I would remove the 8 pieces of paper with a red/green combination (R1-G1, R1-G2, etc.), and then choose a permutation at random. I actually see this as applying the constraint before the randomization process.

I see your procedure as providing the constraint *during* the randomization process. My last problem [P(2 green) and P(2 blue)] was an example of what I feel is a nonsensical result caused by changing the randomization process midstream.

FWIW, another way I would solve these would be as follows. Given no red/green combinations, the probabilities of pill 1 being red, green, or blue are 6/22, 6/22, and 10/22, respectively. I do agree with your breakdown of the probabilites of the color of the second pill, given the color of the first. Specifically, P(2=G|1=G)=1/3, and P(2=B|1=B)=1/5. I would then calculate P(2 greens) as (6/22)*(1/3) = 6/66, and P(2 blues) as (10/22)*(1/5) = 10/110, which equals 6/66.

I will review your posts on this again to make sure I'm not misunderstanding you. Again, please correct me if I misrepresented/misunderstood how you would solve these last problems.

Considering this thread is almost longer than the religion discussion, I think declaring an impasse is a good idea Happy holidays.

[Grunion]

QuikSand, Vaj, etc.

Unless you guys want to break the post record (I think after this we will be four behind the religion discussion), I'm going to throw in the towel.

I believe I am correct, however I do not know I am correct. I'm sure both of you can make the identical statement.

I would much rather know that my premise is incorrect than believe that my premise is valid, without absolute proof. Unfortunately, this is not the case at this time.

Of course, if anybody comes across anything which could possibly resolve this one and for all please post it.

Anyway QuikSand, I believe I monopolized enough of your time, its been a while since your last OT challenge.

[QuikSand]

Retreived from the archives and bumped for the nearly sixth anniversary of the granddaddy of all FOFC puzzle debates. Not for the weak of heart, I'll warn you.

[albionmoonlight]

A classic.

If Monte Hall was the "Son House" of FOFC Puzzles, then this is the "Robert Johnson."

[Dutch]

Apparently I was a lot smarter in December 2000 than now. When I looked at the start of this thread, I thought, "Who fucking knows that!?" then six posts down I am stunned to see myself trying to give the answer.

[Maple Leafs]

Any truth to the rumor than Grunion disappeared from the board because he spent the next six years in a cabin in the woods repeatedly dealing himself five-card hands to prove he was right?

[Arctus]

Originally Posted by Maple Leafs

Any truth to the rumor than Grunion disappeared from the board because he spent the next six years in a cabin in the woods repeatedly dealing himself five-card hands to prove he was right?

None whatsoever.

He stopped visiting this board (and a bunch of others) due to some very demanding out of town disaster recovery assignments.

Years later, after his existance returned to normal, he rediscovered this board on May 10, 2005. After spending 1/2 hour trying to remember his old password, he decided to just reregister under a different username instead.

[MJ4H]

Originally Posted by Arctus

None whatsoever.

He stopped visiting this board (and a bunch of others) due to some very demanding out of town disaster recovery assignments.

Years later, after his existance returned to normal, he rediscovered this board on May 10, 2005. After spending 1/2 hour trying to remember his old password, he decided to just reregister under a different username instead.

You sound like his mom or something


(yes i get it)

[Passacaglia]

Originally Posted by Dutch

Apparently I was a lot smarter in December 2000 than now. When I looked at the start of this thread, I thought, "Who fucking knows that!?" then six posts down I am stunned to see myself trying to give the answer.

Same here. Although, I guess I was just out of college, then.

[QuikSand]

fun times

[Marc Vaughan]

Interesting question ...

[Dutch]

Damn, I lived in Los Angeles way back then when you started this....and I was 29 years old...

[QuikSand]

came across this old post while searching for another one... fun fun