Here is a card game. At the beginning of the game, you choose what rank of card you will be taking. Every round, 5 cards are dealt. Any that match the rank you chose are kept, the rest are shuffled back into the deck before the next round. You win when you get 3 cards of the chosen rank.
What is the distribution for how many turns it takes to win?
At first I was looking at a binomial distribution for each deal, but I am unsure how to account for the removal of successful card draws.
Here's my working so far: Let's imagine I choose rank 5 as the card I want and we only need 2 of them to win. If we assume a binomial distribution and we want to work out the odds of winning on round 2. Is it right to assume 2 hands of 5 cards = 1 hand of 10 cards? Below, for 10 cards we have
$$P(X\ge2)=1-P(X = 0)-P(X=1) =1-\frac{\binom{4}{0}\binom{48}{10}}{\binom{52}{10}}-\frac{\binom{4}{1}\binom{48}{9}}{\binom{52}{10}}\approx0.481$$
This assumes the 2nd hand dealt is independent of the 1st, which I don't think is true. So, alternatively, if we look at a hand of 5 on its own and note down some probabilities: $$P(Y=0)=\frac{\binom{4}{0}\binom{48}{5}}{\binom{52}{5}}\approx0.659$$ $$P(Y=1)=\frac{\binom{4}{1}\binom{48}{4}}{\binom{52}{5}}\approx0.299$$ Then, $$P(Y\ge2)=P(Y_1=0,\, Y_2\ge2\,\,||\,\,Y_1=1,\, Y_2\ge1\,\,||\,\,Y_1\ge2)$$ (Where $Y_1$ indicates the $1^{st}$ deal, $Y_2$ the $2^{nd}$) $$=0.659\cdot(1-0.659-0.299)+0.299\cdot(1-\frac{\binom{3}{0}\binom{48}{5}}{\binom{51}{5}})+1-0.659-0.299$$ $$\approx0.028+0.299\cdot0.271+0.042\approx0.151$$
These are vastly different results so I figure I've gone wrong somewhere, but I can't see it