I just came back from Vegas. I was playing Deuces Wild Triple Play. Was dealt 3 deuces on bottom hand, of course i held the deuces and hit the draw button. To my amazement all 3 hands filled in the 4th deuce !!!! I would like to know the odds of this happening or what the probability is of this happening. Thanks in advance.

If I did this correctly the probably of getting a deuce in all three rows given you already got three deuces is 0.0077%.

Yes, as others have said, converting any dealt trips to quads on all 3 lines is: (2/47)^3 ~ 0.000077 ~ 1 in 12,978 Nice pull! And I'm glad you seem a lot more excited than the only time I saw it (via a neighbor). He pulled quad Jacks on DDB on each of the 3 lines, doesn't react at all, and immediately moves onto the next hand. He really sucked the joy out of that moment.

Correct me if I'm wrong, but isn't it (1/47+1/46)^3 ~ 0.000080? Wouldn't 2/47 be the odds of getting one of two specified cards drawing one card, where (1/47+1/46) is the odds of getting a specific card drawing twice? And of course to the third power for three different hands.

If you save three cards (the three deuces), there are two cards to be drawn. To get four deuces, either the first of two cards has to be a deuce or the second has to be deuce. So the probability is (1/47) x (46/46) + (46/47) x (1/46) = 2/47. Since the three lines are independent, the final answer is (2/47)^3 = 0.000077.

A lot of times calculations like this do what you describe with card removal, but when you are looking at it the way you're describing, you are double counting during the times you hit the quad on the first card. Probability of getting the quad on the first card: 1/47 Probability of getting the quad on the second card if you didn't get the quad on first card: 1/46 Probability of getting the quad on the second card if you got the quad on the first card: 0 since there are only 4 cards of that rank in the deck. So using your method, the probability of getting a quad drawing two cards is like this: Probability of getting quad on 1st card + Probability of not getting quad on 1st card x Probability of getting quad on 2nd card 1/47 + (46/47) x (1/46) = 1/47 + 1/47 = 2/47 Alternatively you can find the probability this way...many multi card draws are calculated this way. Find the probability of it failing and subtract from 1. Probability of Quad = 1 - (Probability of not getting quad on 1st card) x (Probability of not getting quad on the 2nd card) = 1 - (46/47) x (45/46) = 1 - 45/47 = 2/47

Ah, got it! Thanx! It makes sense to me now because if you draw 47 times, the chances of getting the 4OAK should be 1 = 1/47 + 1/47 +1/47 + ... (47 times). So the odds of getting a 4OAK from a 3OAK is the same as getting a straight flush from an outside straight flush draw?