Thursday, July 06, 2006

Euchre strategy - Analysis of a euchre hand

There’s a lively debate going on over at the Euchre Science pages. The question is what you should do about the following hand.

Jack of Hearts
Jack of Diamonds
Ace of Diamonds
Jack of Clubs
10 of Clubs

The Ace of Hearts was turned down, and you are sitting in seat 1.

Should you go alone or take your partner along? Should you lead trump or lead the loser? Should the score matter in your decision?

Let’s look at the first question.

Alone or with partner?

In case you don’t much care about all the math, here is the bottom line to this answer first.

Bottom line: Unless your team’s score is 8 points, you should go alone on this hand. If you have 8 points, you should bring your partner along with you, lead 2 trump, and hope for her to have the highest club.

Now the details…
This question is the source of much debate. First, consider all the ways that your partner could possibly help you sweep or march.

A. They have 4 trump and an off-suit winner
B. They have the highest club
C. They are void in clubs and have a trump


A. They have 4 trump.
There are a total of 8568 possible hands that your partner can have. This is found by taking the total number of cards left unseen (18) and figuring out how many ways they can be sorted into groups of 5. Combination of 18 things taken 5 at a time is 8568.

Of these possible hands only 14 will have 4 trump and only 3 of those will have a guaranteed off-suit winner. That means partner will only have a 4 trump helping hand 0.05% of the time. Those aren’t good odds.

B. They have the highest club.
If your partner has the highest club and you’ve saved one of your winner trumps, you can sweep for 2 points. To keep it simple we’ll consider only hands in which your partner has the Ace of clubs. Of the 8568 possible hands, your partner will have the Ace of clubs 2380 times or 27.8%. You’ll have to lead 2 of your trump first and your partner will have to save the Ace but you will sweep under these conditions nearly 30% of the time. Which also means you won’t sweep 70% of the time.

C. They are void in clubs, have a trump.
Complicated, yes. But a situation like this is not simple. Of the 8568 hands, only 2002 will be void in clubs (23.4%), and 1750 of those will also have a trump. That puts the chances of this sweep at 20%

Unfortunately, none of these situations can occur simultaneously. And the way you play will depend on which of these hand holdings your partner has. Based on the scenarios above, your best play would be option B which gives you a 27.8% chance of a sweep.

However, what are the chances of you making a sweep by going alone?
The chances that the 3 clubs are buried (either in partner's hand or in the uh kitty) and you make your loner is 6.86%. That was found by the following simple combination equation.

1. Number of unknown cards = 18. (24 - your 5 cards - the turned down card)
2. Number of cards that can foil your loner = 3 (A, K, & Q of clubs)
3. Number of possible opponent distribution of unknown cards = 43,758 (combination of 18 things taken 10 at a time)

4. Number of possible opponent distributions that don't contain any of the cards that can foil you = 3003 (combination of 15 things taken 10 at a time)

5. 3003 / 43,758 = 6.86%

Now, I believe we can also add in the times when the opponents have a lone Queen of clubs because after you've played your 3 trumps they probably dumped the lone Queen in favor of other cards. This improves your chances as follows.

6. Number of possible opponent distributions that are either void of clubs or could contain the lone Queen of Clubs = 8008. (combination of 16 things taken 10 at a time)

7. 8008 / 43,758 = 18.3%

So you should expect to make a loner with this hand a little more than 1 out of every 6 times.

Game theory comparison.
Expectation alone = 4 * 18.3% + 1 * 81.7% = 1.55 points
Expectation with partner = 2 * 27.8% + 1 * 72.2% = 1.28 points

Over the long run, going alone will get you more points and that’s what you should play.

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