Setup
During a game of cutthroat euchre (3 players instead of 4), my girlfriend won the game by playing as her last card, the 9$\spadesuit$. Given that euchre is played with a deck consisting of 9s, 10s, Js, Qs, Ks, and Aces, and trump was a different suit, this was one of the 3 lowest cards in the deck.
That's a pretty awesome way to win the game, but it gets better. In addition to playing the 9$\spadesuit$, she also took all 5 tricks in that hand and therefore got two points to win the game instead of the standard one point you get for only making your bid.
Adding on to the improbable nature of this game, the 9$\spadesuit$ was turned up five times as the bid card -- she turned it up three times, her father turned it up once, and I turned it up once.
The game ended with scores of 10, 8, and 3, which meant, since hers was the only two point hand, we played 19 hands.
I refer those who are unfamiliar with Euchre to the rule set here.
Goal
I'd love to figure out the probability of the 9$\spadesuit$ turning up 5 times as the bid card and then extend that probability to include the fact that the the 9$\spadesuit$ was also the card used to win the game by taking all tricks in the final hand.
Start of a Solution
Because you shuffle between each hand, I believe the probability of any one 9 (since we would have been just as flabbergasted with any 9) coming up 5 times as the bid card in a game of 19 hands is:
$$ P_{\text{particular card coming up 5 times}} = \frac{4}{24} \cdot \left(\frac{1}{24}\right)^{4} = \frac{1}{1,990,656} $$
But here is where my stats fail me. How do I model the probabilities of the hand being played? If I want to model the probability of the 9$\spadesuit$ being the card to win the last hand, do I have to include the probability of the 9$\spadesuit$ not being the last card during the other hands? I'm thinking something like this pseudo equation:
$$ P_{\text{dealt the right 9}} \cdot P_{\text{playing the 9 last}} \cdot P_{\text{taking all tricks}} \cdot P_{\text{something about winning the game?}} \cdot P_{\text{something else?}} $$
I believe we can calculate the first probability via:
$$ P_{\text{dealt the right 9}} = 1 - (P_{\text{not being dealt the right 9}}) = 1 - (\frac{23}{24} \cdot \frac{22}{23} \cdot \frac{21}{22} \cdot \frac{20}{21} \cdot \frac{19}{20}) = \frac{5}{24} $$
Can I calculate the second probability via this? This result is non-intuitive to me.
$$ P_{\text{playing the 9 last}} = 1 - P_{\text{not playing the 9 last}} = 1 - (\frac{4}{5} \cdot \frac{3}{4} \cdot \frac{2}{3} \cdot \frac{1}{2}) = \frac{4}{5} $$
Could I use a pre-calculated probability for the fact that she took all 5 of the tricks on the last hand? If I found a resource online that gave the probability of taking all 5 tricks in a Euchre hand, how would I integrate that?