Ever wonder what the probability of certain scenarios are in the game of Euchre? Euchre is a trump game played with 24 cards (9 through Ace.) 5 cards are dealt to each player leaving a kitty of 4 cards. The top card is flipped over and is called the upcard. The upcard is important because everyone must reject the upcard as the trump suit before bidding on an alternate trump suit.

Well here are your answers:

5Pn * 18P6-n * 6Cn

23P6

What is the probability of getting exactly cards of the same suit as the upcard?
%
But keep in mind that there are four players. So the expected number of occurrences per hand is 4 times this value.

What this means is that if there is a 30.312% probability that you will get 2 cards suited to the upcard, then an average hand will see 1.21 players have exactly 2 cards suited to the upcard.

The following averages would be expected per hand (also assuming 12 hands per game):

1.69 players will have 1 card suited to the upcard (every 0.59 hands or 20 times per game)
1.21 players will have 2 cards suited to the upcard (every 0.83 hands or 14 times per game)
0.32 players will have 3 cards suited to the upcard (every 3.10 hands or 3.88 times per game)
0.03 players will have 4 cards suited to the upcard (every 33 hands or every 2.75 games)
0.0007 players will have 5 cards suited to the upcard (every 14285 hands or every 1200 games)

6-nCn' * 18-(6-n)C5-n' 6-nCn' * 6-n-n'Cn' * 18-(6-n)C5-n' * 18-(6-n)-(5-n')C5-n'



18C5 18C5 * 13C5

Where n is the number of upcard suited cards which will be dealt to you, and
n' is the number of upcard suited cards which will be dealt to at least one of your opponents.
The subtrahend (the subtracting term; the probability that both opponents have exactly n' upcard suited cards) only applies when n + (2 * n') is 6 or less -- since including the upcard, there are only 7 bowers available.
 
When you have exactly upcard suited cards,
what is the probability that at least one of your opponents will have exactly upcard suited cards (including the left bower)?
%
These percentages account for the fact that you face two opponents, and each one might of course have exactly n cards suited to the upcard.


I received a query from someone interested in this page. The point of interest was about the probability of being dealt a hand worthy of a redeal. I use the rule, "No ace, no face, no trump." But there are other rules which only require no aces or face cards (only 9's and 10's.)

The calculation for no aces or face cards is a bit easier:
8C5     =   0.13175%

24C5

The chance of a player receiving only 9's or 10's is 0.13175%, or one out of every 759 hands. Since there are four players, and each has exactly the same probability of receiving such a hand at the beginning of the deal, the probability that any given Euchre hand must be redealt is 0.52701%, or one out of every 189.75 hands, or about once every 15.8 games.


So what is the probability of getting no aces, no faces, and no trump? The specific rule says you have to wait until trump has been decided, but that cannot be calculated (since it relies on human bidding.) But the calculation that you have no aces, no faces, and no upcard suited cards is as follows:
6C5 * ( 6 / 19 )     =   0.00446%

24C5

The ( 6 / 19 ) expression is the probability that the upcard is the absent suit. In order for a hand to qualify, when you are dealt only 9's and 10's, the upcard must be the absent suit. The absest suit will have all 6 cards available for the remaining card positions, and there are only 19 cards remaining.

The chance of a player receiving only 9's or 10's without any of them being the upcard suit is 0.00446%, or one out of every 22433 hands. Since there are four players, and each has exactly the same probability of receiving such a hand at the beginning of the deal, the probability that any given Euchre hand must be redealt is 0.01783%, or one out of every 5608 hands, or about once every 467 games. (Or about once every 6 weeks if you play 8 hours a day, non-stop.)


How many possible Euchre hands are there?

24P20 * 4C1     =   62,336,074,312,512

(5P5)4 * ( 2P2 * 2P2 * 2P2 )

This equation answers the question, how many unique Euchre situations are there? There are over 62.3 quadrillion. The number of hands is reduced by the fact that the exact same hand can come up with the suits interchanged. Since bowers apply to the opposite suit of the same colour, we cannot simply say 4P4, but rather 2P2 for the reds, and 2P2 for the blacks, and then 2P2 because the reds and the blacks can themselves be interchanged. Since the play starts based on who dealt, player positioning is important.

You can instead use the equation:

24C4 * 19C4 * 14C4 * 9C4 * 4C1     =   62,336,074,312,512

2P2 * 2P2 * 2P2


Hopefully these percentages jive with your experience. I've been playing Euchre on Yahoo! and I thought I was getting bad cards more than anyone else. So I actually calculated and determined, that for whatever reason, I was correct, and Yahoo! does not shuffle the cards fairly -- for whatever reason. Which explains why I seem to be a better Euchre player on Playsite than on Yahoo! -- Go figure!


Due to some errors in calculations that I discovered, this page was last updated on September 23rd, 2002.