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 07-30-2020, 05:48 PM #1 Double Down veteran   Join Date: Oct 2003 Location: Studio City, CA Posts: 2,991 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.
 07-30-2020, 06:11 PM #2 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,825 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
07-31-2020, 12:08 AM   #3
Double Down
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Re: How to account for probability in this case of chances of disease

Quote:
 Originally Posted by heehaww 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.

07-31-2020, 06:06 AM   #4
heehaww
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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?

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

P(had it | no symptoms and kid negative) = .5*.2*.5 / (.5*.2*.5 + .5)

07-31-2020, 11:30 PM   #5
David Sklansky

Join Date: Aug 2002
Posts: 15,554
Re: How to account for probability in this case of chances of disease

Quote:
 Originally Posted by Double Down 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.

 08-01-2020, 12:31 AM #6 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,825 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)%
08-03-2020, 08:24 AM   #7
nickthegeek
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Join Date: Sep 2011
Posts: 270
Re: How to account for probability in this case of chances of disease

Quote:
 Originally Posted by Double Down 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.

 08-03-2020, 10:19 AM #8 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,825 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."
 08-04-2020, 03:52 AM #9 nickthegeek journeyman   Join Date: Sep 2011 Posts: 270 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.
08-06-2020, 11:01 AM   #12
Double Down
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Location: Studio City, CA
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Re: How to account for probability in this case of chances of disease

Quote:
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?

Yesterday, 09:08 AM   #13
heehaww
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Join Date: Aug 2011
Location: Tacooos!!!!
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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:
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!

Yesterday, 12:21 PM   #14
Double Down
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Join Date: Oct 2003
Location: Studio City, CA
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Re: How to account for probability in this case of chances of disease

Quote:
 Originally Posted by heehaww 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.

 Yesterday, 05:16 PM #15 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,825 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.
Yesterday, 06:40 PM   #16
Double Down
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Join Date: Oct 2003
Location: Studio City, CA
Posts: 2,991
Re: How to account for probability in this case of chances of disease

Quote:
 Originally Posted by heehaww 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?

 Yesterday, 09:46 PM #17 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,825 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
 Today, 12:00 AM #18 Double Down veteran   Join Date: Oct 2003 Location: Studio City, CA Posts: 2,991 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.
 Today, 06:15 AM #19 nickthegeek journeyman   Join Date: Sep 2011 Posts: 270 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; Today at 06:29 AM.
Today, 07:42 AM   #20
heehaww
Pooh-Bah

Join Date: Aug 2011
Location: Tacooos!!!!
Posts: 4,825
Re: How to account for probability in this case of chances of disease

Quote:
 Originally Posted by nickthegeek 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 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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