• #### What are the odds against a 'Yarborough' rummy hand?

From James Dow Allen@21:1/5 to All on Thu Jun 25 02:57:29 2020
I played 3-handed rummy yesterday, starting with 9 cards
randomly dealt from an ordinary 52-card deck. I was surprised
to see that my hand was a Yarborough -- No pairs, no suited
connections (e.g. 2-3 of clubs), not even a gutshot connection
(e.g. 2-4 of clubs). I did have a double-gutshot 2-5 of clubs,
but let's ignore those. (By the way, Ace is always high: K-A
of the same suit is a connection, but not A-2.)

I was so surprised to see this "Yarborough" that I took a photo of it!
But estimating the odds just now, I see that it isn't really that rare.

I usually play 2-handed rummy, starting with ten cards instead of
nine. Now the probability of a Yarborough is much less.

What are those probabilities?

James Dow Allen

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• From Ilan Mayer@21:1/5 to James Dow Allen on Sun Jun 28 10:44:09 2020
On Thursday, 25 June 2020 05:57:30 UTC-4, James Dow Allen wrote:
I played 3-handed rummy yesterday, starting with 9 cards
randomly dealt from an ordinary 52-card deck. I was surprised
to see that my hand was a Yarborough -- No pairs, no suited
connections (e.g. 2-3 of clubs), not even a gutshot connection
(e.g. 2-4 of clubs). I did have a double-gutshot 2-5 of clubs,
but let's ignore those. (By the way, Ace is always high: K-A
of the same suit is a connection, but not A-2.)

I was so surprised to see this "Yarborough" that I took a photo of it!
But estimating the odds just now, I see that it isn't really that rare.

I usually play 2-handed rummy, starting with ten cards instead of
nine. Now the probability of a Yarborough is much less.

What are those probabilities?

James Dow Allen

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In order to not have pairs each card in the hand must have a different value. From these hands, in order to avoid suited connections, each combination of values must checked (with values sorted):
Card 1 has 4 possibilities
Card n has 2 possibilities if values differs by 1 from card n-1 and by 2 from card n-2, 3 possibilities if value differs by 1 or 2 from card n-1 and by more than 2 from card n-2, and 4 possibilities otherwise.
The products for all combinations are then added up.
They total 6,820,068 for a 9 card hand and 4,262,880 for a 10 card hand.
The total numbers of hands are 52!/(43!*9!) = 3,679,075,400 for 9 cards and 52!/(42!*10!) = 15,820,024,220 for 10 cards.
The probabilities therefore are:
9 cards - 6,820,068 / 3,679,075,400 = 0.00185374510128 (about 1 in 539)
10 cards - 4,262,880 / 15,820,024,220 = 0.000269461028676 (about 1 in 3,711)