Why bridged bicyclic compound isn't a preferred product over fused bicylic compound in Robinson Annulation reaction?

**Why a doubly-fused cyclohexanone is formed instead of bridged cyclohexanone like product which would be more flexible (entropically-favoured)?**


**Also, the OH group not being at the bridgehead carbon unlike former compound, shouldn't it have a better chance of forming stable alkenone after dehydration as the preferred product formed in Robinson annulation?** The following image elaborates my doubt.


[![Doubt regarding the possible alternative solution to Robinson annulation](https://i.stack.imgur.com/sxdov.jpg)](https://i.stack.imgur.com/sxdov.jpg)


[![Baldwin rule for enolate](https://i.stack.imgur.com/eYWVv.png)](https://i.stack.imgur.com/eYWVv.png) (source : <https://en.wikipedia.org/wiki/Baldwin%27s_rules#Rules_for_enolates>)


As per the Baldwin rule for enolate, the enol-endo reaction and enol-exo reaction is favoured for both exo-tet and exo-trig for six/seven-membered chain (from the excerpt is provided as above image). 


Both the alternatives follow six-membered exo-trig which is favoured according to above Baldwin rule. (Refer to the image where the carbon chain of both the alternatives is numbered.) 
The bridged-bicycle compound has 3 faces, where each face has a 6-membered ring when it was visualized through ChemDraw 3D. 
So, it seems both the alternatives as the product of Robinson annulation is favoured according to modified Baldwin rules.


