# Riddle Me This

Every week the geeks at FiveThirtyEight.com post a couple of riddles for their followers to solve.  One of last week’s riddles was about Texas Holdem poker, proposing the you only want to make the best possible hand after all community cards are dealt – the nuts!  What two starting cards are mostly likely to end up as the absolute nuts?

I posed this question once or twice at poker games during the week, and the answers strongly clustered around suited connectors and suited aces.  Can JTs make the nuts more ways than 87s?  Is AKs better because almost every flush is a nut flush?  What about ATs or A5s… the flushes are still nut flushes, plus there are a few more straight flushes using only the T?

The first thing to do here is determined whether the combinatorics favor nut flushes or nut straights.  JTs can make a number of different straights, all of which are the nuts (unless somehow counterfeited).  If the community cards include AKQ, KQ9, Q98, or 987, the resulting straight is the nuts.  Start with AKQ… there are 4 aces, 4 kings, and 4 queens, making  4 x 4 x 4 = 64 combinations of these three cards.  This would be further multiplied by the number of combinations for the other two cards (not including cards that pair the board or counterfeit the J or T), but we can defer that for now.  With 64 combinations of four different straights, we have 256 core combos.  The only nut flushes using JTs are straight flushes.

Contrast that to AKs and the number of nut flush combos.  With 13 cards of each suit and two of them in our hand, we need one of the other 11 cards of that suit, then one of the remaining 10, and finally one of the remaining 9 outs.  11 x 10 x 9 = 990 combinations.  After removing the number of these combinations that create potential straight flushes for another player, the remainder should still far exceed the number of straight flush combos for a hand like JTs (or T9s or 87s or any other suited connectors).

In summary so far, AKs should make the absolute nuts more different ways than any suited connectors without an ace.  My back of the napkin style analysis doesn’t tell us the definitive correct answer, but it does strongly suggest that suited aces will make the nuts more ways than suited connectors without an ace.

I’ll fast forward from here, as the 538 geeks have now posted the answer, which is ATs.

ATs will hit all the nut flushes that AKs (or AQs or AJs) will. ATs also makes more straight flushes than AKs (KKing David note:  there are a few hands that will make a straight flush using only the T, surrounded by the same suit QJ98 or J987 or 9876.)  And ATs guards against other hands in a way that AKs does not. For example, if the board includes the 6, 7, 8 and 9 of spades (with some useless fifth card), AK of spades would not be the nuts because JT of spades would beat it (flush vs. straight flush). But the JT of spades can’t exist in our ATs scenario — the 10 of spades would be in our hand. Further discussion about the intuition behind this solution broke out last week on the /r/poker subreddit.

The difference between ATs and AKs or AQs or AJs is very small, nevertheless it does exist and tips the answer in favor of ATs despite it not being the overall stronger hand to play.

For purposes of the riddle, the worst hand is 62o, making the nuts less often than any other hand including 72o.

If you want a simpler riddle, try this one:  A boy and a doctor went fishing.  The boy was the doctor’s son, but the doctor was not the boy’s father.  Who is the doctor?

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