Probability & miscarraige - help a non-maths person :)


I'm trying understand this paper on probability, parental age and miscarriage. I have no training in statistics. I'd really appreciate any help coming my way.

There's a table within the article, or here:; results of a multicentre European study

The table shows that the probability of miscarriage for couples where paternal age is 40+ and maternal age is 35+ is 6.73.

I realise this is a very basic question, but if the standard risk is probability 1 for the low risk group of under thirties, and this translates to 8% of those couples having a miscarriage (I'm getting this figure from other research), what is the risk for the older group as a percentage? It isn't just 6.73 multiplied by 8?

Thanks a lot for any help on this one! :)


Omega Contributor
The table presents the odds ratios for miscarriage for the two oldest groups versus the youngest groups (as reference) while controlling for other covariates.

So when both partners are in the oldest groups, their odds of miscarriage is 6.7 times higher than the reference (young) groups odds, when controlling for covariates.

An 8% probability is equal to odds of 0.087 or 0.08/0.92. So making a very very crude example not taking into account the results came from a model controlling for other variables. Then given your 8% probability the high risk group would have an odds of 0.5829 (or also 0.5829/1.5829 = probability of 0.368), so 0.5829/0.087 = 6.7 times greater odds.

Though, whenever making generalizations you have to keep in mind the sample and its original target population. The table is for European patients in a particular study in the 1990s. So the 8% probably may be irrelevant to them or not.
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Thanks so much for that, I really appreciate you taking the time to answer. Sorry in advance if this is a stupid question, but can the odds for the older group then be expressed as a percentage probability?