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Old 07-30-2020, 05:48 PM   #1
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How to account for probability in this case of chances of disease

Situation: A man has a son. The man is diagnosed in his early 40s with a deadly genetic disease, where there is a 50% chance that the child will also be a carrier of it (and die from it as well).

Years later, the son is all grown up and is fifty years old. He is considering getting genetic testing to find out if he's a carrier. He does research on this disease and finds that for 80% of people who do have it, onset of symptoms happens between 30 and 50 years old.

Is there a way to derive just from that what the odds are that he's a carrier? It seems intuitively that it's no longer 50/50, since the odds that he's not a carrier has to be higher than the odds that he is a carrier, but falls in the 20% category.

How would you calculate the probability? Would it just be (50+20)/2= 35%?

I guess I'm trying to figure out how to factor in "circumstantial" probability.
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Old 07-30-2020, 06:11 PM   #2
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Re: How to account for probability in this case of chances of disease

Apply Bayes' Theorem: P(have it) = .2*.5 / (.2*.5 + .5) = 1/6
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Old 07-31-2020, 12:08 AM   #3
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Re: How to account for probability in this case of chances of disease

Quote:
Originally Posted by heehaww View Post
Apply Bayes' Theorem: P(have it) = .2*.5 / (.2*.5 + .5) = 1/6
I appreciate the response. I'm not that familiar with working with Bayes'Theorem but I think I get it, although I'm still not quite sure which pieces of info to fill into to which variables. What does the (.2*.5+.5) of the denominator represent?

Also, could you guys help me with one addition to this problem? (This is a personal thing happening, not for some exam or anything)

The 50 year old man dies suddenly soon after, so we never find out if he has the disease. He leaves behind two children. One of the children gets tested for the gene and is negative.

What can we now conclude about whether or not the dad had it?
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Old 07-31-2020, 06:06 AM   #4
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Re: How to account for probability in this case of chances of disease

Bayes Theorem: P(A given B) = P(A and B) / P(B) = P(A)*P(B given A) / P(B)

Applied here: P(have it | no symptoms before 50) = P(have it & no symptoms) / P(no symptoms)

Quote:
The 50 year old man dies suddenly soon after, so we never find out if he has the disease. He leaves behind two children. One of the children gets tested for the gene and is negative.

What can we now conclude about whether or not the dad had it?
We can start with the 1/6 and update it:

P(had it) = P(had it & kid negative) / P(kid negative) = (1/6)(1/2) / (1/12 + 5/6) = 1/11

Or had we not already calculated the 1/6:
P(had it | no symptoms and kid negative) = .5*.2*.5 / (.5*.2*.5 + .5)
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Old 07-31-2020, 11:30 PM   #5
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Re: How to account for probability in this case of chances of disease

Quote:
Originally Posted by Double Down View Post
I appreciate the response. I'm not that familiar with working with Bayes'Theorem but I think I get it, although I'm still not quite sure which pieces of info to fill into to which variables. What does the (.2*.5+.5) of the denominator represent?
If you are willing to forget formulas, its simple logic. The probability that you are positive and get to 50 is 50% x 20% or 10%. The probability you are negative and get to 50 is (if you assume that nothing else could kill you) is 50% x 100% or 50%. 40% of the time you don't reach 50. Given you do (60%), 10 of that 60 you will be positive.
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Old 08-01-2020, 12:31 AM   #6
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Re: How to account for probability in this case of chances of disease

Basically the range of possibilities has been narrowed by the info.

It's like, the chance of a spade is 1/4, but then if somehow you know that the card is black, that changes it to 1/2 because it's 13/26 instead of 13/52, or put another way, (1/4) / (1/2).

Same thing here. Initially, the chance of being positive and asymptomatic is 50% * 20% = 10%. But then we're given the info that he's asymptomatic, so the possibility of being symptomatic is eliminated, like ruling out an answer to a multiple-choice test question. The eliminated possibility had a chance of 50% * 80% = 40%, so now our denominator shrinks from 100% to (100-40)%
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Old 08-03-2020, 08:24 AM   #7
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Re: How to account for probability in this case of chances of disease

Quote:
Originally Posted by Double Down View Post
Is there a way to derive just from that what the odds are that he's a carrier?
In principle, you don't have enough information to answer. All the answers given had to make another assumption, which was not a given, and namely that the probability of having those symptoms is zero if you are not a carrier.

Suppose for instance that those symptoms are present even within the 80% of the healthy population. Having (or not having) them does not change the situation, and the man is still even money to be a carrier.

In another extreme situation, where 95% of the healthy people has those symptoms, actually not having them increases the chance that you are a carrier.

I illustrated the above examples just to let you understand what you need to properly answer and that you miss a bit of information: the probability of having (or not having) the symptoms if you are not a carrier.

I'd also add that making the assumptions that were made ITT could lead to totally irrational considerations were the question framed differently. I will point it out if there is any interest in it.

Last edited by nickthegeek; 08-03-2020 at 08:31 AM.
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Old 08-03-2020, 10:19 AM   #8
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Re: How to account for probability in this case of chances of disease

As usual, nickthegeek is right.

In school I had a math teacher that used to say, "When you assume, you make an ass out of u and me."
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Old 08-04-2020, 03:52 AM   #9
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Re: How to account for probability in this case of chances of disease

Thank you heehaww for your consideration (and btw I'm right much less often than "usually")!

Pretty recently I was interested in this work, in which the authors postulate that the capability of taking rational decisions is a separate ability of the intelligence that doesn't get tested with standard IQ tests. The kind of "problems" that they propose to measure a "rationality quotient" is very similar to the one proposed here. It's pretty common for our brain to jump to some conclusions in some instances and being wrong.

Just to outline how it's wrong to make assumptions (and how they mostly are unreasonable) let's consider these two variations.

1) You know that 80% of carriers support democracy. The man does not support democracy.
2) Same as the original question, but the man now does have symptoms.

It's pretty clear in variation 1) that you don't have enough information to estimate the probability of the man being a carrier, and making assumptions would be pretty silly. Notice that the information you got is totally equivalent to the original question.

More striking, in my opinion, is variation 2). Now, with the same assumption presented ITT, one should conclude that it's 100% sure, guarantee that the man is a carrier. I highly doubt that, if the question were presented as variation 2), anybody would assume that is 0 the probability of having symptoms if you are not a carrier.

Last edited by nickthegeek; 08-04-2020 at 03:58 AM.
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Old 08-05-2020, 05:21 AM   #10
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Re: How to account for probability in this case of chances of disease

Thank you for your input Nick. I actually think there are things that you're considering that aren't actually relevant factors in this scenario, but that's my fault for not being clear.

I would actually really love some help on this from any of you, it has real world implications. The situation:

My wife and I are pregnant (yay!) We're still in the beginning, about 10 weeks in and are doing genetic testing. Aside from the two most common varieties of testing, one for being high risk for standard conditions, and another to test for Jewish Ashkenazi conditions, we're considering getting a third, to test my wife for the Huntington's Disease gene that runs in her family. It's a pretty scary notion, one that for some reason I haven't really dealt with yet, but with the baby coming it's really hitting now.

For those of you who don't know, Huntington's is a degenerative disease like MS, MD, or Parkinson's where you slowly lose control of yourself physically, mentally, and behaviorally and you eventually die from it. ****ing awful.
The onset of symptoms can be any time from age 2-80, but onset in most cases is 30s-40s. And from the time of onset the average lifespan is another 21, 22 years. There is currently no cure. The real ***** of it is that if you have it, every one of your offspring has a 50% chance of inheriting it from you.

So when people decide whether to be tested, they have to weigh the risk they might have it and not know or risk knowing that you do and you essentially know how much longer you're going to live and how you'll die (best case scenario), which is frankly a part of my own mortality that I don't evenknow if I'd want to find out. I'm of course going to leave it up to my wife, especially since it's not going to change our plan anyway. We'll have the baby regardless, it'll just be a real kick in the life nuts.

So when we meet with a geneticist next week to have a "counseling" session, they're going to give us the odds that she has it and then we can weigh our options. But I'd like to make sure that everything is accounted for because I obviously want us to make the decision with the most and most accurate information.

So if you guys are willing to help, I'll list what I think is pertinent info, I'll tell you then how I applied it (using Bayes', thanks you guys) and if I've misstepped along the way, please chime in. I've been doing a lot of research about statistics on the commonness of the onset of symptoms based on age, so let's just use the numbers I provide. Also for the sake of this, we can assume that any symptoms displayed are to be interpreted that the person does have HD, so we don't have to worry about scenarios of if the person has symptoms but it might be something else.

Actually... none of the people (that we're concerning ourselves with) have had symptoms, so it's a totally moot point. OK, here it goes:

My wife's grandfather had HD, The onset was in his early 40's, which is typical (most common range is 30-40s) and died in his 60s. He had 3 children. His youngest daughter (my wife's aunt) had HD, and like her father, got it in her 40s and died about 15 years later.

Her father, and her uncle, have never been tested. Luckily, at the this point, her uncle almost certainly doesn't have it, because he's in his 70s and hasn't developed any symptoms (and to develop them now would put him in like 5% of cases that develop this late) plus he has 3 kids, all of whom are in their 40s, and haven't developed any symptoms either. None of this will have bearing on our calculations, just wanted to paint the whole picture.

My wife's father is where it gets tough, because he died over 20 years ago, from stomach cancer at age 54. By the time he died, he still hadn't developed any symptoms according to anyone in the family. They wouldn't have a reason to lie, and they would've been familiar with the symptoms. So I think it's a rational thing that we can assume they're right.

So here's where the calculating starts. There's a 50/50 chance he had it. But if he made it all the way to 54, it would put him in the 20% to get it past age 50 (not even going to include the little extra EV for getting to 54 specifically) So using Bayes' theorem, we get that 50% of the time he doesn't have it, and 80% of the other 50% of the time he does have it but already developed symptoms (which is 40% of the total) and then 20% of the 50% (so 10% of the total) he has it and has yet to develop symptoms.
So we take away the 40% from the denominator since those scenarios are impossible, and we get a 10%:50%, or a 1 out of 6 chance he had it. This is his new odds.

Now, my wife has two brothers. One of them had the genetic testing done years ago when he and his wife first got pregnant. It was negative. So now we can take it another step further to realize that there's even less of a chance her dad had it.

So if he's 1/6 to have it, the 5/6 of the time he doesn't. And half of the 1/6 that he does, her brother also has it (remember, each child inherits it 50% of the tie) so that's 1/12. and the other 1/12 is where the dad has it but the brother doesn't. We know that the former example couldn't have happened bc the son is negative, so we subtract that from the denominator. Now, the 10/12 of the time the dad doesn't have it is now 10/11, and the 1/12 is now 1/11.

So the dad's odds are down to 1/11. So far, this is EXACTLY what heehaww says. Is everything good so far or is there something we did wrong? Nick?

Ok, next, my wife had a brother who lived to 34 and then sadly died from suicide. Yes, my wife's family has dealt with a **** ton of tragedy.

But her brother never developed any symptoms. And to make it to 34 means he's already past 33% of people who develop symptoms by then.

So now we can deduce the father's chances again.
10/11 the dad is negative
1/2 of the 1/11, the dad is positive but the younger brother is negative
2/3 of that other half of the 11th, the dad and brother are positive, and he would have developed symptoms at some point
1/3 of the other half of the 1/11, the dad and brother are positive, and already developed symptoms (which we know is the scenario that didn't happen so we'll subtract this from the denominator)
Common denominator is now 66.
60/66 dad is negative
3/66 dad is positive, brother is negative
2/66 dad and bro positive, would have developed (possible)
1/66 dad and bro positive, already developed (not possible)

Subtract the 1, so
63/65 the dad was negative
2/65 dad was positive

Am I off my rocker here? Are these 3 pieces of info enough to reduce the dad's chances from a 50% all the way down to around 3%???

If everything is right so far, we can go even further. My wife, just like her deceased brother, is 34 and does not display symptoms. So she's also made it past the first 3rd.

So we can do the exact same calculations for him again based on this info.
Shorthand..
63/65 negative
1/65 he's pos, she's neg
2/3 of 1/65 he's pos, she's pos and will develop
1/3 of 1/65 he's pos, she's pos and already developed (not possible)

so 189/195 he's neg
3/195 he's pos, she's neg
2/195 he's pos, she's pos will develop
1/195 he's pos, she's pos already developed

so 192/194 he's neg
2/194 he's pos

Is this totally ridiculous? And if her chances are 50/50 of that (actually less since she's made it to 34 without symptoms) that means she's less than a 1/194 to be positive.

Can someone please verify all of this for me? If I'm right and the odds are just that low, I can maybe sleep at night again. Also, it's such a low chance that we'd probably not even get tested and just assume she's negative.

I would really love everyone's input. I know this is long but I appreciate anyone who's read it and willing to chime in. It's really been weighing on my mind.


Thank you.

-DD
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Old 08-06-2020, 04:21 AM   #11
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Re: How to account for probability in this case of chances of disease

Sorry OP for the situation you are facing. I wish you and your family all the best. Luckily, till now, everything is pointing towards the right direction and you are right in assuming that the chances that your wife has HD is very low.

Nonetheless, you also need to know the probability of having the symptoms for people with no HD. It's crucial because you want to have a better estimate and, more importantly, you don't want to jump to the conclusion that your wife has it if she shows the slightest of the symptoms. There is important evidence for the good side and we don't want to overestimate evidence for the wrong side.

To solve this kind of problems, I suggest the use of software implementing bayesian networks. For instance, the netica software is free if used for small networks (as in this case). You can build a network with nodes (your variables) and links (causal relations between variables). In your case, a network might look like something as in figure.



With the data you provided, it looks like that your wife is 98.9% not a carrier. You might have some mistake in the calculations and/or I might not have entered the values correctly. However, if you download Netica it should be simple for you to create the bayesian network, to put the correct values and to see the results.

Last edited by nickthegeek; 08-06-2020 at 04:35 AM.
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Old 08-06-2020, 11:01 AM   #12
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Re: How to account for probability in this case of chances of disease

Quote:
Originally Posted by nickthegeek View Post
Sorry OP for the situation you are facing. I wish you and your family all the best. Luckily, till now, everything is pointing towards the right direction and you are right in assuming that the chances that your wife has HD is very low.

Nonetheless, you also need to know the probability of having the symptoms for people with no HD. It's crucial because you want to have a better estimate and, more importantly, you don't want to jump to the conclusion that your wife has it if she shows the slightest of the symptoms. There is important evidence for the good side and we don't want to overestimate evidence for the wrong side.

To solve this kind of problems, I suggest the use of software implementing bayesian networks. For instance, the netica software is free if used for small networks (as in this case). You can build a network with nodes (your variables) and links (causal relations between variables). In your case, a network might look like something as in figure.



With the data you provided, it looks like that your wife is 98.9% not a carrier. You might have some mistake in the calculations and/or I might not have entered the values correctly. However, if you download Netica it should be simple for you to create the bayesian network, to put the correct values and to see the results.
Nick thank you for responding. The 1.1% chance you give her definitely puts me at ease, but it is still higher than the 1/194 chance I came to in my post right before.

Do you happen to know where the errors were in my calculations to lead me to a different answer than the the software you used?
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Old 08-07-2020, 09:08 AM   #13
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Re: How to account for probability in this case of chances of disease

I agree with Nick: I calculate a 1/91 chance that he had it (and 90/91 is 98.9%).

Quote:
10/11 the dad is negative
1/2 of the 1/11, the dad is positive but the younger brother is negative
2/3 of that other half of the 11th, the dad and brother are positive, and he would have developed symptoms at some point
1/3 of the other half of the 1/11, the dad and brother are positive, and already developed symptoms (which we know is the scenario that didn't happen so we'll subtract this from the denominator)
Common denominator is now 66.
60/66 dad is negative
3/66 dad is positive, brother is negative
2/66 dad and bro positive, would have developed (possible)
1/66 dad and bro positive, already developed (not possible)

Subtract the 1, so
63/65 the dad was negative
2/65 dad was positive
1/11 * 1/2 * 2/3 = 1/33
1/33 / (1/33 + 10/11) = 1/31

The next calculation is similar:
1/31 * 1/2 * 2/3 = 1/93
1/93 / (1/93 + 30/31) = 1/91

I, too, hope you can fade this runner-runner if no other!
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Old 08-07-2020, 12:21 PM   #14
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Re: How to account for probability in this case of chances of disease

Quote:
Originally Posted by heehaww View Post
I agree with Nick: I calculate a 1/91 chance that he had it (and 90/91 is 98.9%).
Nick was saying that the 98.9% negative was my wife's chances, not her father's. Do you think he misspoke?

I hope so, and that you're right. Because if it's the dad's chance that it's 1/91, then my wife would be slightly less than half of THAT. So we'd be well under 1% then which would be a big relief.

In fact, that's low enough that we might then opt to not even take the test.
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Old 08-07-2020, 05:16 PM   #15
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Re: How to account for probability in this case of chances of disease

Oops I tried to rush, forgot some scenarios, then fell victim to confirmation bias when I saw Nick's coincidental %.

Hm, now I'm getting 5/77 for the father and 2/77 for your wife (so 1/77 for the baby). No sense posting my math since it might change again. I'll try again later.
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Old 08-07-2020, 06:40 PM   #16
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Re: How to account for probability in this case of chances of disease

Quote:
Originally Posted by heehaww View Post
Oops I tried to rush, forgot some scenarios, then fell victim to confirmation bias when I saw Nick's coincidental %.

Hm, now I'm getting 5/77 for the father and 2/77 for your wife (so 1/77 for the baby). No sense posting my math since it might change again. I'll try again later.
No need to post the math, but can you speak to what aspects of the scenario you're accounting for differently than you did before?
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Old 08-07-2020, 09:46 PM   #17
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Re: How to account for probability in this case of chances of disease

Still getting 5/77 and 2/77, so I'll show my math and maybe someone will find my mistakes.

I did it with and without splitting the father calculation into two parts, with the same result, so I'll show the two-part version.

P(father had it | 1st kid asymptomatic) = P(had it & kid asymptomatic) / P(1st kid asymptomatic)

P(had it & asymp) = (1/11)[1/2 + (1/2)(2/3)] = 5/66
P(asymp) = 1 - 1/11/6 = 65/66
Therefore, P(father had it | 1st kid asymptomatic) = 1/13
(Btw, @OP, your math actually agreed with this, but then you turned your 60 into a 63.)

P(father had it | 2nd kid asymptomatic) = (1/13)(5/6) / (1 - 1/13/6) = 5/77


P(wife has it | both asymp) = P(father had it & passed it to her & both asymptomatic) / P(both asymp)

numerator = (1/11)(1/2)(2/3)(5/6)
denominator: 1 - (1/11)(1 - 25/36)
Therefore, P(wife has it) = 2/77
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Old 08-08-2020, 12:00 AM   #18
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Re: How to account for probability in this case of chances of disease

So.... I only somewhat understand the calculations there (TOTALLY on me, I'm sure your breakdown is perfectly clear to a person better versed in all of this) but it does seem like the one scenario you didn't take into consideration was reducing the father's odds based on my wife's so far asymptomaticness, the way you did with the 2 brothers.

It looks like the steps you took was:

1. Odds of dad having it, based on older son testing negative

Then

2. Odds of dad having it based on dad not being symptomatic after 55 years

Then

3. Odds of dad having it based on the other son still being asymptomatic after 34 years and then finally

4. Odds of my wife having it based on the dad's odds.

But there was one more calculation that you skipped. Step 4 should be to calculate dad's odds based on my wife having been asymptomatic for 34 years (similar to the 2nd brother). And THEN we can calculate her true odds based on that.
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Old 08-08-2020, 06:15 AM   #19
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Re: How to account for probability in this case of chances of disease

I didn't see your detailed calculation, but let's use this post to agree on the data. There is no way my calculation is wrong (since it's made by a sw), but there is the chance that I input the wrong probabilities. So, let's agree on that.

P(dad has HD a priori) = 0.5
P(a child has HD if a parent has it) = 0.5
P(a child has HD if no parent has it) = 0
P(having some symptom before age 54 if you have HD) = 0.8
P(having some symptom before age 54 if you don't have HD) = 0
P(having some symptom before age 34 if you have HD) = 0.67
P(having some symptom before age 34 if you don't have HD) = 0

Next you have this bit of information.

- Dad is 50/50 a priori to have HD and has 3 child
- Dad didn't have symptoms in his life and died at 54
- 1 child is assured 100% not having it because of a test
- 1 child didn't have symptoms till death, which happened at 34
- OP's wife, the 3rd child, didn't experience symptoms and she is currently 34

Is the above the correct statement of the problem?


Just in case one wants to use Netica. After download, copy/paste the following in a file with a .dne extension and load it into Netica through the File->Open menu.

Code:
// ~->[DNET-1]->~


bnet tptHD {
AutoCompile = TRUE;
autoupdate = TRUE;
whenchanged = 1596881098;

visual V1 {
	defdispform = BELIEFBARS;
	nodelabeling = TITLE;
	NodeMaxNumEntries = 50;
	nodefont = font {shape= "Arial"; size= 9;};
	linkfont = font {shape= "Arial"; size= 9;};
	ShowLinkStrengths = 1;
	windowposn = (22, 22, 1126, 482);
	resolution = 72;
	drawingbounds = (1412, 720);
	showpagebreaks = FALSE;
	usegrid = TRUE;
	gridspace = (6, 6);
	NodeSet Node {BuiltIn = 1; Color = 0x00e1e1e1;};
	NodeSet Nature {BuiltIn = 1; Color = 0x00f8eed2;};
	NodeSet Deterministic {BuiltIn = 1; Color = 0x00d3caa6;};
	NodeSet Finding {BuiltIn = 1; Color = 0x00c8c8c8;};
	NodeSet Constant {BuiltIn = 1; Color = 0x00ffffff;};
	NodeSet ConstantValue {BuiltIn = 1; Color = 0x00ffffb4;};
	NodeSet Utility {BuiltIn = 1; Color = 0x00ffbdbd;};
	NodeSet Decision {BuiltIn = 1; Color = 0x00dee8ff;};
	NodeSet Documentation {BuiltIn = 1; Color = 0x00f0fafa;};
	NodeSet Title {BuiltIn = 1; Color = 0x00ffffff;};
	PrinterSetting A {
		margins = (1270, 1270, 1270, 1270);
		};
	};

node Dad_Has_It {
	discrete = TRUE;
	states = (yes, no);
	kind = NATURE;
	chance = CHANCE;
	parents = ();
	probs = 
		// yes          no           
		  (0.5,         0.5);
	whenchanged = 1596783726;
	belief = (0.5, 0.5);
	visual V1 {
		center = (492, 168);
		height = 1;
		};
	};

node Brother1 {
	discrete = TRUE;
	states = (yes, no);
	kind = NATURE;
	chance = CHANCE;
	parents = (Dad_Has_It);
	probs = 
		// yes          no            // Dad_Has_It 
		  (0.5,         0.5,          // yes        
		   0,           1);           // no         ;
	whenchanged = 1596783722;
	belief = (0.25, 0.75);
	visual V1 {
		center = (288, 300);
		height = 2;
		};
	};

node Wife {
	discrete = TRUE;
	states = (yes, no);
	kind = NATURE;
	chance = CHANCE;
	parents = (Dad_Has_It);
	probs = 
		// yes          no            // Dad_Has_It 
		  (0.5,         0.5,          // yes        
		   0,           1);           // no         ;
	whenchanged = 1596636266;
	belief = (0.25, 0.75);
	visual V1 {
		center = (492, 294);
		height = 4;
		};
	};

node Brother2 {
	discrete = TRUE;
	states = (yes, no);
	kind = NATURE;
	chance = CHANCE;
	parents = (Dad_Has_It);
	probs = 
		// yes          no            // Dad_Has_It 
		  (0.5,         0.5,          // yes        
		   0,           1);           // no         ;
	whenchanged = 1596636806;
	belief = (0.25, 0.75);
	visual V1 {
		center = (774, 294);
		height = 3;
		};
	};

node Symptoms_54 {
	discrete = TRUE;
	states = (yes, no);
	kind = NATURE;
	chance = CHANCE;
	parents = (Dad_Has_It);
	probs = 
		// yes          no            // Dad_Has_It 
		  (0.8,         0.2,          // yes        
		   0,           1);           // no         ;
	whenchanged = 1596783719;
	belief = (0.4, 0.6);
	visual V1 {
		center = (774, 168);
		height = 6;
		};
	};

node Symptoms_34_w {
	discrete = TRUE;
	states = (yes, no);
	kind = NATURE;
	chance = CHANCE;
	parents = (Wife);
	probs = 
		// yes          no            // Wife 
		  (0.67,        0.33,         // yes  
		   0,           1);           // no   ;
	whenchanged = 1596783721;
	belief = (0.1675, 0.8325);
	visual V1 {
		center = (492, 402);
		height = 7;
		};
	};

node Symptoms_34 {
	discrete = TRUE;
	states = (yes, no);
	kind = NATURE;
	chance = CHANCE;
	parents = (Brother2);
	probs = 
		// yes          no            // Brother2 
		  (0.67,        0.33,         // yes      
		   0,           1);           // no       ;
	whenchanged = 1596783718;
	belief = (0.1675, 0.8325);
	visual V1 {
		center = (774, 396);
		height = 5;
		};
	};
ElimOrder = (Brother1, Symptoms_54, Symptoms_34_w, Wife, Dad_Has_It, Brother2, Symptoms_34);
};

Last edited by nickthegeek; 08-08-2020 at 06:29 AM.
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Old 08-08-2020, 07:42 AM   #20
heehaww
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Re: How to account for probability in this case of chances of disease

Quote:
Originally Posted by nickthegeek View Post
P(having some symptom before age 34 if you have HD) = 0.67
My understanding is that it's the opposite. Hopefully this the only source of our difference.

Quote:
Just in case one wants to use Netica. After download, copy/paste the following in a file with a .dne extension and load it into Netica through the File->Open menu.
When I paste your code, it says it's 50/50 that the dad had it and 25/75 that the wife has it. Is there some evaluation button I have to click? (If i can get it working, I'll swap the 2/3 and 1/3 to see what it says.)

Quote:
Originally Posted by Double Down View Post
but it does seem like the one scenario you didn't take into consideration was reducing the father's odds based on my wife's so far asymptomaticness, the way you did with the 2 brothers.
I started from 1/11 which already had the 55yo brother factored in. Then I factored one of the 34yo's being asymptomatic, which changed the dad's chance to 1/13. Then I factored in the other 34yo which changed your dad's chance to 5/77. (If instead I calculate with both of the 34yo's in a combined step, I get the same result.)

For your wife, I didn't use the 5/77 (which I'm pretty sure would be a no-no); I calculated separately from the initial 1/11.
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Old 08-08-2020, 10:05 AM   #21
nickthegeek
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Re: How to account for probability in this case of chances of disease

Quote:
Originally Posted by heehaww View Post
When I paste your code, it says it's 50/50 that the dad had it and 25/75 that the wife has it. Is there some evaluation button I have to click? (If i can get it working, I'll swap the 2/3 and 1/3 to see what it says.)
Click on the "evidence" you got. Click on "no" for Brother1, Symptoms_34_w, Symptoms_54 and Symptoms_34 nodes. You see that the probabilities will update as long as you add new evidence, so you can check also your intermediate results.

To change 67 with 33, you have to double click on each Symptom_34* node, go to Table and change the values accordingly.
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Old 08-08-2020, 10:36 AM   #22
heehaww
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Re: How to account for probability in this case of chances of disease

Thanks. In the code I swapped the two instances of .67 and .33, but added more precision: .33333 and .66667. Then per your suggestion I right-clicked the nodes and used "Enter Finding" to plug in the No's.

Now it agrees with me:
Dad = 6.49 ≈ 5/77
Wife = 2.60 ≈ 2/77
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Old 08-08-2020, 02:02 PM   #23
Double Down
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Re: How to account for probability in this case of chances of disease

I just figured out my mistake. After factoring in for the younger brother, the 2/65 I came to, I mistook for the dad's chances but it was actually the brother. Yes, it's 5/65 which is 1/13. And then we do the same with her, and we get that the dad's true odds are 5/77, and my wife's is 2/77.

But this isn't her true odds! Now that we have the most accurate odds for the dad at 5/77, we can more accurately calculate her odds.

By my calculations, we get 10/457, which is a little less than 2/77.
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Old 08-08-2020, 02:20 PM   #24
Double Down
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Re: How to account for probability in this case of chances of disease

Hm... I just recalculated just the scenario in a vacuum of my wife having it with the dad being positive (asnif that were the first step), and I'm getting 40%.

3/6 she's neg
1/6 she's pos already symptomatic
2/6 she's pos, will be symp

Remove the 1/6 and we have 2/5.

So now it does make sense that if he's 5/77, she'd be 2/77.

Where is my mistake in doing it the other way?
5/77 he's positive, let's make the common denominator 462 so 30/462.

15/462 she's negative
5/462 she's pos, already has symptoms
10/462 she's pos, will develop

Remove the 5 from the denominator

So I have her odds at 10/457.

What is the reason for the discrepancy??? I know we're only talking .4% of a percentage point difference between 2/77 and 10/457 but I'm trying to figure out my mistake.


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Old 08-08-2020, 05:44 PM   #25
heehaww
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Re: How to account for probability in this case of chances of disease

Quote:
Originally Posted by me
For your wife, I didn't use the 5/77 (which I'm pretty sure would be a no-no)
I said that without actually trying it, because I felt like circular logic would be involved. But I stand corrected below:
Quote:
Originally Posted by Double Down
3/6 she's neg
1/6 she's pos already symptomatic
2/6 she's pos, will be symp

Remove the 1/6 and we have 2/5.
That's valid and obv way more efficient than what I showed.

However:
Quote:
Where is my mistake in doing it the other way?
5/77 he's positive, let's make the common denominator 462 so 30/462.

15/462 she's negative
5/462 she's pos, already has symptoms
10/462 she's pos, will develop

Remove the 5 from the denominator

So I have her odds at 10/457.
Great question, and this shows that my suspicion wasn't completely unfounded. What's wrong is that you have to use 1/13 instead of 5/77. The 1/13 only factors in her sibling being asymptomatic. This way, her being asymptomatic isn't redundantly factored in.

1/13/3 / (1 - 1/13/6) = 2/77

Let's bring both solutions up a level of abstraction for a wider view of what's happening.

First way you showed:
P(wife pos) = P(dad pos & wife asy)*P(wife pos | dad pos & wife asy)
P(dad pos & wife asy) is simply 5/77 because her being asy was already factored into the 5/77. This is key.
P(wife pos | dad pos & wife asy) = 2/5 for the reason you said

2nd way:
P(wife pos | asy) = P(pos & asy) / P(asy)
P(pos & asy) = P(dad pos & dad passed it & wife asy)
P(asy) = 1 - P(dad pos & dad passed it & wife symp)

We can't have overlap between "dad pos" and "wife asy/symp". Setting P(dad pos)=5/77 causes overlap, whereas 1/13 avoids it. With 1/13, the "wife asy" part only builds on the info using new info.
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