20210613, 12:58  #1 
"University student"
May 2021
北京
2^{2}×3^{2} Posts 
Is the "Combined" factoring probability actually wrong on mersenne.ca?
The "Combined" probability just add the trial factoring probability and the P1 factoring probability, but that makes no sense.
Take 108071849 for example: It has been trial factored to 2^77, with 64.9351% chance of finding a factor. Also, the chance of finding a factor in P1 with b1=755000, b2=21551000 assuming no factor below 2^77 is 3.5775% (a conditional probability). Let A denote the event of "finding a factor below 2^77" B denote the event of "finding a factor in P1 with b1=755000, b2=21551000" then P(A) = 64.9351%, P(B(!A)) = P(!A && B) / P(!A) = 3.5775% So the total probability of finding a factor should be P(A  B) = P(A) +P(!A && B) = P(A) + P(B(!A)) * P(!A) = P(A) + P(B(!A)) * (1P(!A)) = 0.0649351 + 0.035775 * (10.649351) = 66.1895% Which makes sense. This formula can also avoid having over 113% probability (???) on exponents such as 1277. (PS: I'm only a freshman student studying probability. If my formula is wrong, please point out :) Last fiddled with by Zhangrc on 20210613 at 12:59 
20210614, 00:52  #2  
"James Heinrich"
May 2004
exNorthern Ontario
6535_{8} Posts 
Quote:
If others agree that your implementation makes sense I'm happy to revise the site accordingly. But I'll need a little more explanation in simple terms what you mean with the logical combinations of probabilities, such as P(!A && B) or P(B(!A)) 

20210614, 03:25  #3  
"University student"
May 2021
北京
24_{16} Posts 
Quote:
Last fiddled with by Zhangrc on 20210614 at 03:52 

20210614, 03:53  #4  
"University student"
May 2021
北京
2^{2}×3^{2} Posts 
Quote:


20210614, 04:01  #5 
"James Heinrich"
May 2004
exNorthern Ontario
3421_{10} Posts 
Unfortunately that means about the same to me as this.

20210614, 04:03  #6 
"James Heinrich"
May 2004
exNorthern Ontario
D5D_{16} Posts 

20210614, 04:07  #7 
"University student"
May 2021
北京
2^{2}×3^{2} Posts 
It seems so. But you'd better wait for another one who really knows the stuff and states that the formula is correct.

20210614, 06:03  #8  
"Mihai Preda"
Apr 2015
2533_{8} Posts 
Quote:
A) the chances of being eaten by the tiger is: "60% of 10%" == 0.6 * 0.1 == 6% (because, if the aligator gets him first, the tiger is out of luck). Overall being eaten is: 90% + 6%, 96%. B) considering the complement: surviving the whole trip means: not being eaten by the crocodile (10%), AND not being eaten by the tiger (40%). Surviving = 10% * 40% == 4%. The complement of surviving thus is 1  4% == 96%, same as above. Last fiddled with by preda on 20210614 at 06:10 Reason: spelling 

20210614, 12:52  #9 
"James Heinrich"
May 2004
exNorthern Ontario
D5D_{16} Posts 

20210614, 14:44  #10 
If I May
"Chris Halsall"
Sep 2002
Barbados
2627_{16} Posts 

20210614, 15:42  #11  
Sep 2009
2^{2}×523 Posts 
Quote:
In the normal case where both TF and P1 have only a few% chance of finding a factor adding the probabilities would be nearly right. Which is probably why it's not been noticed until now. Chris 

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