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I got dealt five cards of the same suit and one of them was the right bower (Jack). Since I had 5 out of the 7 total trump, it seemed a good bet to call a loner. Twice in 80 hands I was dealt 5 trumps, and twice I was set. We thought the odds of this happening was very small.

If I have 5 trumps, what are the odds that one of my opponents has both the other two trumps?

Example: I am dealt the J♠, A♠, K♠, 10♠, 9♠. What are the odds that the J♣ AND the Q♠ are in the same opponent's hand?

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Probably not as rare as you think, around 13%

First calculate the chance of one of your opponents having the Left Bower.

They have 10 cards, and you know about five or six cards, so there are 18-19 left. You didn't mention whether you were dealer or not. For this, I'll assume not, so there are 18 cards left out of the deck. (24 card deck, less the five in your hand, and the one card turned over that nobody made trump. Obviously it isn't one of the two other trump.)

10/18 is ~ 56% chance that one of your opponents has the Left.

Next we need the chance of that specific opponent having the other trump. That opponent has four other cards, and one of them must be the remaining trump. There are now 17 other cards left.

4/17 is 23.5%

Multiply the two together to get our overall chance of 13.1% chance.

If you were the dealer and made trump, the odds drop to 11.7% (10/19 * 4/18).

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    We were playing with a 24 card deck so I think that would make it a 10/18 (55.6%) chance that my opponents have the Left. Then a 4/17 (23%) chance that the last trump is in that hand. So final odds would be 10/18 * 4/17 or about a 13% chance!
    – Boone80
    Dec 20, 2021 at 1:45
  • @Boone80 - oh geez, it's been a while since I've played Euchre, you're right, it's a 24 card deck. I'll redo the numbers.
    – Pat Ludwig ♦
    Dec 20, 2021 at 18:14

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