# 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. Dec 27, 2020 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. Dec 28, 2020 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. Dec 28, 2020 at 22:18
• Yes. The explanation of the odds can be used in many different cases, sure. Dec 29, 2020 at 13:39