# How to Calculate 5 to 11-Card Poker Straights When Dealt 11 Cards from 8-Deck Shoe

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!

• There is often very little point in keeping track of the number of decks used in the shoe when using multiple decks. You can approximate the results just fine by using an infinite shoe instead, allowing for far easier calculations by effectively treating every draw as being completely independent from each other. Further, since you don't care about suits here, it can be further simplified to effectively be drawn from a 13 card deck with replacement each time. Sep 17, 2020 at 1:26
• That said, even after the simplifications, there are still around $2$ trillion possibilities where order matters ($13^{11}\approx 1.792\times 10^{12}$) so looping over all of these is infeasible. This is far more than the $\binom{52}{5}\approx 2.6\times 10^6$ different 5-card hands from a traditional deck. You could satisfy yourself with a random sampling, picking just a mere million or so at random and using the ratio of those seen using some introductory programming. Sep 17, 2020 at 1:31

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

$$10\cdot 32^{5}{411\choose 6} - 9\cdot 32^{6}{410\choose 5}+ 8\cdot 32^{7}{409\choose 4} - 7\cdot 32^{8}{408\choose 3} + 6\cdot 32^{9}{407\choose 2} + 5\cdot 32^{10}{406\choose 1} - 4\cdot 32^{11}$$

• Why is the combination representing the random card for the 10-card straight (406 C 1) when the combinations for the lesser straights follow a pattern of 415, 414, 413, 412, to 411? I presume this was a typo and it's supposed to be (416 C 1)? Sep 17, 2020 at 10:22
• Brain fart... the series should be declining numbers of free cards as the straights get bigger. Sep 17, 2020 at 17:37
• Ahh, that makes more sense... however, with closer inspection these calculations aren't working out... I have these plotted out in an Excel sheet and the numbers are wrong. The sum percentage exceeds 100%, as the sum of the numbers of each type of straight is larger than the calculated number of total 11-card poker hands dealt from an 8-deck shoe, (416 C 11) = ~1.415E21. Even calculating your 5-card straight count example verbatim gives ~1.503E21, meaning that there's a 106.23% chance of drawing a 5-card straight in an 11-card stud hand from an 8-deck shoe, which is obviously very wrong. Sep 18, 2020 at 23:53
• Before trying your 5-card straight example verbatim, I was calculating the number of each type of possible straight in an 11-card hand slightly differently, but I thought equally valid. The math for my version also doesn't work, though it isn't quite as extremely out-of-bounds. I have the total number of hands as COMBIN(416, 11). Then I calculate the number of 11-card straights as 4*(32^11). Then 10-card straights as 5*(32^10)*COMBIN(416-10,1) - C3 (the previous count of 11-card straights) Then 9-card straights as 6*(32^9)*COMBIN(416-9,2) - SUM(C3:C4) (the sum of 11 and 10 straight counts) Sep 19, 2020 at 0:01
• I continue this pattern, using your formulas to count the base number of each type of straight, then subtracting the sum of the previous counts of the higher types of straights, e.g., so that for 7-card straights, it's 8*(32^7)*COMBIN(416-7,4) - the sum of the previous values for 11, 10, 9, and 8-card straights. This makes more sense to me and is easier to write out, however the results are still incorrect, with the total probability for each type of straight 11 to 5 being 1.52. This is less zany than when I plugged in your 5-card count example, yielding a probability of 1.06 for just them Sep 19, 2020 at 0:03