VP probability challenge

A couple of Double Double Bonus VP probability questions:

Does anyone know the post-draw ratio of the royal flush to 8-Q and 9-K straight flushes for Double Double Bonus VP playing optimal strategy?

What fraction of all flushes (not including straight flushes) would have a span of 7 cards or fewer (ex 3-4-5-6-9 suited, max span)? This scenario also assumes optimal play.

Lastly, does this conditional probability statement look close: P(4 Aces| quads) = 0.27 ? And could that be extrapolated to, say, P(3 Aces over any pair | full house) = 0.135? My guess is no because the preferential contingencies for keeping aces would affect each outcome slightly differently.

Thank you!

For the flushes question, if I am counting right, there are C(13,5) x 4 = 5148 possible flushes, of which 40 are straight flushes.
If you count, say, A-2-3-5-6 as a flush of span 6, then:
There are 9 possible ranks for the low card in a span-6 flush (A-6, 2-7, ..., 9-A); for each one, there are C(4,3) = 4 choices for the middle three cards, so there are 9 x 4 x 4 suits = 144 flushes of span 6.
There are 8 possible ranks for the low card in a span-7 flush; for each one, there are C(5,3) = 10 choices for the middle three cards, so there are 8 x 10 x 4 suits = 320 flushes of span 7.
Thus, there are a total of 464 flushes of span 7 or less that are not straight flushes, out of 5108 flushes that are not straight flushes. This reduces to the fraction 116/1277, or about 9.1%.

Thanks! That looks entirely accurate to me.

Now this probability would hold for a stud game, but for a perfectly played draw game, the number of tighter range flushes would presumably increase significantly because of the preferential holding of high suited cards and suited cards in general separated by 5 or less. Not sure how one would compute that effect though... My guess is the probability would end up somewhere between 0.15 and 0.2.

Quote: Gorillanoah

Thanks! That looks entirely accurate to me.

Now this probability would hold for a stud game, but for a perfectly played draw game, the number of tighter range flushes would presumably increase significantly because of the preferential holding of high suited cards and suited cards in general separated by 5 or less. Not sure how one would compute that effect though... My guess is the probability would end up somewhere between 0.15 and 0.2.
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I see what you are asking now. You want to know what fraction of the flushes you end up with have span 6 or 7.

Right. After a draw round. This seems quite onerous without some computing power.

Seems like you’d have to account for all potential flush hands on the deal, partitioning them into subtypes of equipotential flush spreads to eliminate redundancies, and then churn out some probability chains.

Same logic seems like it would apply to computing the real-world ratio of ‘Aces over trips Full Houses’ to all Full Houses.

But perhaps there’s a more streamlined means here…