## Frequency of Flannery vs. Weak 2

I played Flannery for the first time yesterday. My bridge guru had always loathed the convention, so I had naturally loathed it as well. Recently I looked at his convention card for the US Team Trials and saw it on there. "What's with this?" I asked. He responded with this story.

So I decided to give it a try. Guess what: the first time I played it, it came up three times in 27 boards. Many of the Flannery fans at the club said things like, "That's typical -- you'll get several hands in a session and then it won't come up for months."

Mathematically speaking, that can't be literally true -- at least, not more so than any other convention. The reason is probably psychological: when you make an unusual or artificial bid, it sticks in your head, and so when it happens to come up a few times in a session, that really sticks in your head.

Anyway, how often does Flannery actually come up? It's easy to calculate the likelihood of being dealt a Flannery shape and a weak 2.

• Number of hands with 6 diamonds: $$\binom{13}{6}\binom{13}{7}=26,393,687,892$$
• Chance of being dealt this shape: $$26,393,687,892/635,013,559,600\approx 4.16\%$$
• Number of hands with 4 spades and 5-6 hearts: $$\binom{13}{4}\binom{13}{5}\binom{26}{4} + \binom{13}{4}\binom{13}{6}\binom{26}{3} = 16,947,108,750$$
• Chance of being dealt this shape: $$16,947,108,750/635,013,559,600\approx 2.67\%$$

In other words, you are about half again as likely to be dealt a weak-two shape as a Flannery shape. Notice that I didn't consider point count -- that calculation would be MUCH more complicated. But I'm guessing that restricting the weak-two to 5-9 HCP and Flannery to 11-15 HCP (or insert your own ranges here), while it would reduce the number of possible hands, would have little effect on the relative frequency of the two hand-types. It should actually make Flannery relatively less likely, since 5-9 HCP is more common than 11-15 (if it weren't, bridge players wouldn't complain so much about holding bad cards).

Let's do a quick and dirty approximation. I don't want to work out the probabilities of being dealt each possible number of HCP, so I'll Google it instead and hope that this table is accurate.

• Chance of being dealt 5-9 HCP: $$5.19+6.55+8.03+8.89+9.36 = 38.02\%$$
• Chance of being dealt 11-15 HCP: $$8.94+8.03+6.91+5.69+4.42 = 33.99\%$$

Yep, as I suspected. So your actual chance of being dealt a weak 2 hand is closer to $$4.16\%\times 38.02\%=1.58\%$$, while Flannery comes out to $$2.67\%\times 33.99\%=0.91\%$$. The ratio of these probabilities is more like 1.75 to 1.

Why did I write "closer to" just now? Because the point-count distribution of hands with six diamonds is not the same as it is for all hands. (In probability language, the events "I hold six diamonds" and "I have 5-9 HCP" are not independent.) To consider an extreme case, the table says that the probability of being dealt 10 HCP is 9.41%; is that still true if you know that the hand contains thirteen clubs? In the case we're considering, the effect is likely much less drastic, but we should be aware that these last couple of numbers are really approximations to avoid much more icky calculations.