# Odds of this happening in euchre

Last night in euchre, 4 times in a row around the table we all turned up the Jack of spades. All good shufflers and we had 2 splits and 2 no splits. I know there are many variances in determining the odds of a given situation in a euchre game, but, is there anyone whiz out there that can break it down? What are the odds of 4 different dealers flipping up the same card consecutively?

• Wikipedia say "[Euchre] is played with a deck of 24, 28, or 32 standard playing cards" You'll need to say how many cards you were playing with and if the deck included jokers. – StartPlayer Dec 27 '20 at 20:58

Assuming a Euchre deck with 24 cards (9, 10, J, Q, K, A of each suit, no joker) and perfect shuffling, the chance that the 21st card in the deck (the card that gets turned up) is a particular card is 1/24. Thus, if you play 4 deals, there is a (1/24)^4 = 1/331776 = 0.0003% that all four deals will turn up the Jack of Spades.

There's much higher chance of all four deals turning up the same card: (1/24)^3 = 1/13824 = 0.007%. That's because there's a 100% chance that the first deal turns up some card, and each subsequent deal has a 1/24 chance of turning up the same card as the previous deal.

If you play N deals over the course of the night, the probability that at some point all four deals turn over the same card consecutively is (N - 4) * (1/24)^3 (the sequence can start on any but the last 3 deals, and the probability of the sequence is (1/24)^3). This works out to about 0.1% if you play 20 hands and 0.7% if you play 100.

The odds of turning up a given card that occurs once in the deck are 1 in N, where N is the size of the deck. Since you say the shuffles were all good, this means the trials are all independent. So the odds of it happening four times are

(1/N)^4

If N is 24, this comes out to 1 in 331776.

If N is 32, this comes out to 1 in 1048576.

If you don't restrict it to the Jack of spades, then, obviously, you're reducing the number of trials by one: the first draw sets the card and is not a trial itself. So for drawing 4 of the same card, whatever the card, you use (1/N)^3 to calculate the odds of the subsequent three draws matching the first one.

• Worth pointing out that with a standard deck of cards (N=52), the answer is 1 in 7,311,616...not as unlikely as you might think! Across America, say every family carefully shuffles a deck of cards then selects a card at random. They write the card value down eg. `9D`. Repeat this another 3 times and compare the card values. Dozens of families will see the "miracle" of all four cards matching. – AlainD Dec 28 '20 at 15:55
• As the other answer points out, this is correct ONLY iof yiu specify the card in advance. If the question is the chanve of getting the same card four times i a row from a deck mof N cards it is (1/N)^3 not ^4, which makes a significant difference. – David Siegel Dec 28 '20 at 22:18
• Yes. The explanation of the odds can be used in many different cases, sure. – L. Scott Johnson Dec 29 '20 at 13:39