Maybe this is a hard question, maybe not, but the first subscriber to answer it correctly in the comments will get a free signed copy of one of my books.
For those who are interested, here's the calculation. There are 52 cards in the deck; 7 cards are showing; 45 cards are left. The other player needs one of 3 Qs or one of 3 Js as the next card. The chances of this are 6/45. If the player got a Q, he needs one of the remaining 3 Js from the remaining 44 cards. Similarly, if he got a J, he needs one of the remaining 3 Qs out of the remaining 44 cards. Either way, the probability is 3/44. To get the chances of both cards coming, you need to multiply the probabilities = 6/45 x 3/44 = .0091 probability (out of 1). In terms of odds, they were in my favor .9909 to .0091 or 109 to 1.
Like Don below your odds were clearly 92+% and the only way you lose would be a QJ runner runner.
Not unheard of, but still a truly bad beat story
Larry W.
Your odds seem to be 99.17%, assuming he won by hitting a Queen and a Jack and I'm not missing some other avenue.
-Don
You got it, Don! Let me know which book you want.
For those who are interested, here's the calculation. There are 52 cards in the deck; 7 cards are showing; 45 cards are left. The other player needs one of 3 Qs or one of 3 Js as the next card. The chances of this are 6/45. If the player got a Q, he needs one of the remaining 3 Js from the remaining 44 cards. Similarly, if he got a J, he needs one of the remaining 3 Qs out of the remaining 44 cards. Either way, the probability is 3/44. To get the chances of both cards coming, you need to multiply the probabilities = 6/45 x 3/44 = .0091 probability (out of 1). In terms of odds, they were in my favor .9909 to .0091 or 109 to 1.