How to compose easily two ORs and CI95% I need to compose ORs and CIs. I have an OR/CI95% for treatment A vs treatment B and an OR/CI95% for treatment B vs treatment C, and I want to get the OR of treatment A vs treatment C with the CI95.
It seems I can multiply ORs and CIs, but I should get a CI99.7%. How can I get the CI95 in cargo cult mode ?
Thanks !

It sounds like you want to perform a network meta-analysis and you do not have the estimated standard errors, only confidence limits.  Typically a confidence interval for an odds ratio is constructed by inverting a Wald test with a log link function, i.e.
$$\text{exp}\Bigg[\text{log}\{\hat{\theta}_{AB}\}\pm1.96\cdot\hat{\text{se}}_{AB}\Bigg],$$
where $\hat{\theta}_{AB}$ is the estimated odds ratio and $\hat{\text{se}}_{AB}$ is the estimated standard error of $\text{log}\{\hat{\theta}_{AB}\}$.  If you are given the  point estimate and confidence limits you can solve for the standard error, i.e.
$$\hat{\theta}^U_{AB}=\text{exp}[\text{log}\{\hat{\theta}_{AB}\}+1.96\cdot\hat{\text{se}}_{AB}]\implies \hat{\text{se}}_{AB}=[\text{log}\{\hat{\theta}^U_{AB}\}-\text{log}\{\hat{\theta}_{AB}\}]/1.96 $$
Let $\hat{\theta}_{BC}$ be the estimated odds ratio between groups $B$ and $C$.  Then the estimated odds ratio between $A$ and $C$ is $\hat{\theta}_{AB}\cdot\hat{\theta}_{BC}$, a product of odds ratios.  The corresponding 95% confidence interval from inverting a Wald test with a log link function is
$$\text{exp}\Bigg[\text{log}\{\hat{\theta}_{AB}\cdot\hat{\theta}_{BC}\}\pm1.96\cdot\hat{\text{se}}\Bigg],$$
where $\hat{\text{se}}=\sqrt{\hat{\text{se}}^{2}_{AB} + \hat{\text{se}}^{2}_{BC} }$ is the estimated standard error of $\text{log}\{\hat{\theta}_{AB}\cdot\hat{\theta}_{BC}\}$ based on the delta method and $\hat{\text{se}}^{2}_{BC}$ is the esitmated standard error of $\text{log}\{\hat{\theta}_{BC}\}$.  This is for the case when $\hat{\theta}_{AB}$ and $\hat{\theta}_{BC}$ are estimated from independent samples.
Based on the figure linked to in the comments section I get 0.67 (0.44, 1.02) as the point estimate and 95% CI, which matches what the original poster calculated in the comments section.
NOTE: In my original solution I incorrectly provided inference on $\hat{\theta}_{AB}/\hat{\theta}_{BC}$ instead of $\hat{\theta}_{AB}\cdot\hat{\theta}_{BC}$.
