The Cryptography made simple (page 207, under Fig 11.12)(Nigel Smart) say that adversary's advantage of IND-PASS Game is $$Adv1 = 2\times|Pr[b=b']-\frac{1}{2}|$$.
The reason for multiplying by 2 is to normalize advantage from $$[0,\frac{1}{2}]$$ to $$[0,1]$$.

But in this paper (page 5, line 9), the advantage of IND-CKA Game is $$Adv2 = |Pr[b=b']-\frac{1}{2}|$$ which is not normalized and scale is $$[0,\frac{1}{2}]$$.
And this value is used with the advantage of pseudo random function
$$Adv3 = |Pr[A(f)=0]-Pr[A(g)=0]|$$
(f is pseudo random function. g is random function)
which scale is $$[0,1]$$.

Does $$Adv2$$ need not be normalized to $$[0,1]$$ for use with $$Adv3$$?
Or do I usually not need to be aware of normalization of "advantage"?

• Welcome to Cryptography.SE. We have $\LaTeX$/MathJax enabled on our site. You can also provide the page number of the book. It is free to download that I've given the link. Nov 10, 2021 at 12:36
• thank you! Added page number. (It is 207 pages in print, but it looks 212 page in Google Chrome) Nov 10, 2021 at 14:51
• The "value" of the advantage often depends on the context and agreed-upon conventions as long as they express something meaningful. Consequently, the expression for a CPA game may look different than the one for a hash collision game. For a specific game, the scale should not really matter as long as the goal is understood. I would say it is still good to be aware of the different conventions even though they express the same thing. Nov 11, 2021 at 14:40