I'm looking for how to calculate the odds of a 5-card straight poker hand formed from 11 cards dealt from an 8-deck shoe, and then the odds for a 6-card straight out of those 11 cards, and then the same for 7, 8, 9, 10, and 11-card straights, all formed from 11-cards randomly dealt from a shuffled 8-deck shoe of 52-card poker decks (no jokers).
I understand how to calculate the odds of a straight formed from 5 cards dealt from a single 52-card deck. There are plenty of resources for calculating poker hand probabilities dealt from a single-deck shoe (i.e., a single deck of 52 cards), but I cannot for the life of me find any resources online regarding poker probability math from a multi-deck shoe, whether that's 2 decks, 4 decks, 8 decks as in my case, etc.
This problem also has the compounding factor that it's not just five card stud, but 11 cards dealt.
This seems a relatively trivial problem, but I'm afraid I'm going to mess up my calculations without realizing my error. I'd love any assistance, thanks!
The number of 11 card hands from your 8 deck shoe. ${416\choose 11}$
This will be our denominator for all that follow
And 11 card straight can have a $J,Q,K$ or $A$ high. Each values can be associated with $32$ cards in the shoe that might have that value.
$4\cdot 32^{11}$ for the numerator.
A 10 card straight can have a 5 different high values, plus one random card. But we don't want to count the possibility that the random card gives you the 11 card straight.
$5\cdot 32^{10}{406\choose 1} - 4\cdot 32^{11}$
And so we work down to the 5 card straight applying inclusion-exclusion along the way
$10\cdot 32^{5}{411\choose 6} - 9\cdot 32^{6}{410\choose 5}+ 8\cdot 32^{7}{409\choose 4} - 7\cdot 32^{8}{408\choose 3} + 6\cdot 32^{9}{407\choose 2} + 5\cdot 32^{10}{406\choose 1} - 4\cdot 32^{11}$