Loner Round 1 Seat 1?

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Richardb02
Posts: 748
Joined: Mon Dec 10, 2018 8:57 pm
Location: Florida

Loner Round 1 Seat 1?

Unread post by Richardb02 » Sat May 23, 2020 5:39 pm

R1 S1 Score 0-0 1st game ever with these players
(Card_Q-S) Up
(Card_J-C) (Card_A-D) (Card_9-S) (Card_J-S) (Card_9-H) Your hand in S1
Would you order?
Would you go Aone?



Wes (aka the legend)
Posts: 1541
Joined: Wed Jun 13, 2018 3:03 pm

Unread post by Wes (aka the legend) » Tue May 26, 2020 9:02 pm

I would go alone in the first round.

Now if I had this hand as the dealer after I picked up an upcard:

(Card_J-C) (Card_A-D) (Card_9-S) (Card_J-S) (Card_9-H)

I would just call. I have seen some mathematical evidence from Eric Zalas' book strongly suggesting that if you have a 9 loser your team will accumulate more points in the long run if you just call vs going alone.

HOWEVER, this does not hold for S1 loners. Now the 9H loser isn't quite as bad becuz you have the lead, and with that lead you deny your opponents any chance to see which suit doesn't work making it more likely they will throw away the wrong cards.

E.G if you were the dealer going alone and S1 leads a club, your opponents would immediately learn you have no clubs and thus not hold onto them in the end which makes them significantly more likely to keep a heart.

When you have the lead from S1, your opponents never get this critical information--that clubs are no good--until it's too late, so now they're forced to guess more and they can guess wrongly. Eric Zalas' work also supports the idea that going alone with a 9 loser is better than calling or at least no worse than calling from S1.

If I did not go alone, I'd pass hoping to trap the dealer and if the dealer passed, I'd call Next for a relatively easy point. For those wondering how often the dealer will call if you pass such a strong hand, a decent approximation would be around 20% if we assume the dealer is always picking up with 3 trump or 2 trump+2 aces:

How often the dealer will have 3 trump:

(3C2 x 15C3)/18C5 = 1365/8568 = 15.93%

How often the dealer will have 2 trump + 2 aces:

(3C1 x 2C2 x 16C2)/18C5 = 360/8568 = 4.20%

4.20 + 15.93 = 20.13%

Wes (aka the legend)
Posts: 1541
Joined: Wed Jun 13, 2018 3:03 pm

Unread post by Wes (aka the legend) » Wed May 27, 2020 5:34 pm

Wes (aka the legend) wrote:
Tue May 26, 2020 9:02 pm
If I did not go alone, I'd pass hoping to trap the dealer and if the dealer passed, I'd call Next for a relatively easy point. For those wondering how often the dealer will call if you pass such a strong hand, a decent approximation would be around 20% if we assume the dealer is always picking up with 3 trump or 2 trump+2 aces:

How often the dealer will have 3 trump:

(3C2 x 15C3)/18C5 = 1365/8568 = 15.93%

How often the dealer will have 2 trump + 2 aces:

(3C1 x 2C2 x 16C2)/18C5 = 360/8568 = 4.20%

4.20 + 15.93 = 20.13%
I made some mistakes with the number crunching. I fixed this post below. The blue numbers represent my corrections:
Wes (aka the legend) wrote:
Tue May 26, 2020 9:02 pm
If I did not go alone, I'd pass hoping to trap the dealer and if the dealer passed, I'd call Next for a relatively easy point. For those wondering how often the dealer will call if you pass such a strong hand, a decent approximation would be around 18-19% if we assume the dealer is always picking up with 3 trump or 2 trump+2 aces:

How often the dealer will have 3 trump:

(3C2 x 15C3)/18C5 = 1365/8568 = 15.93%

How often the dealer will have 2 trump + 2 aces:

(3C1 x 2C2 x 13C2)/18C5 = 234/8568 = 2.73%

2.73 + 15.93 = 18.66%

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