Is "A person from Group X was 3 times more likely than a person from Group Y to have the disease" an accurate gloss of an odds ratio of 3?

Question

All the interpretations I've seen have referred instead to the odds being 3 times higher in Group X. However, It is not clear to me that there is a meaningful difference between these formulations.

Edit: In thinking that it was OK to express things in this way, I was influenced by http://sphweb.bumc.bu.edu/otlt/sparta/html/32a.html, which states: "An odds ratio is interpreted as if it were a relative risk. In this case an odds ratio of 3.46 indicates that people who ate food from the bakery had 3.46 times the risk of developing hepatitis compared to people who did not eat food from the bakery."

Is that website perhaps wrong? Or am I perhaps interpreting it incorrectly?

bdeonovic · Accepted Answer · 2014-04-14T00:17:15.443

Odds ratios are in terms of odds, and not probabilities (or likelihoods, or risks, they all mean basically same thing).

If you have an odds ratio of 3 (where the odds ratio was constructed by comparing the odds of disease given you are in group X relative to odds of disease given you are in group Y) then the proper interpretation is that the odds of having the disease are 3 times higher in group X than in group Y, just like you said.

Odds and probability of disease are very closely related, just think of it as being on a different scale.

$odds = \dfrac{p}{1-p}$ where $p$ is the probability of disease. Probabilities are numbers in range from 0 to 1, but odds represent the same phenomenon but on a range from 0 to $\infty$.

If you have $p=0.5$, probability of disease is 50%, then the corresponding odds of disease is $0.5/(1-0.5)=1$.

If you have an odds of disease of 3, then probability of disease is $3/4=0.75$.

That is NOT the same as having an odds RATIO of 3. Since an odds ratio of 3 could be represented by (odds in group X is 3)/(odds in group Y is 1) or (odds in group X is 6)/(odds in group Y is 2). In the first scenario if you convert odds to probabilities you get $(0.75)/(0.5) = 1.5$ (A person from Group X was 1.5 times more likely than a person from Group Y to have the disease). In the second scenario you have $(6/7)/(2/3) \approx 1.28$ (A person from Group X was 1.28 times more likely than a person from Group Y to have the disease).

All this to show that odds and probability/likelihood/risk (although they both are a measure of the same thing) are not equal.

Thanks! I edited my post to add where I got this interpretation from. Is that website wrong? — user1205901 - Слава Україні, Apr 14 '14 at 00:39
When $p$ is low (rare disease) then the odds and the relative risk are very similar. I can't tell from your example, but I'm guessing that thats what the answer is trying to imply. See wikipedia — bdeonovic, Apr 14 '14 at 01:11
This might be a silly question for me to ask, but is the superficial-level answer to my question "No, this is not an accurate gloss"? — user1205901 - Слава Україні, Apr 15 '14 at 01:23
In general, the statement in the title is not correct. It is correct for rare events because OR is approximately equal to relative risk in that situation. — bdeonovic, Apr 15 '14 at 02:51
@Sasha If you found this answer useful you can select as correct :) Let me know if you are still confused. I know I am not the most eloquent writer. — bdeonovic, Apr 15 '14 at 18:41

Is "A person from Group X was 3 times more likely than a person from Group Y to have the disease" an accurate gloss of an odds ratio of 3?

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