I am trying to solve a probability problems. I know that probability of getting flush (contains five cards all of the same suit) in 5 card poker is 0.196. But what if we have the option to replace a card with a new one on second turn?
Suppose 'X' is dealt five cards. That contains 4 hearts and another card of different suit. He discarded the card of different suit and draw a new card from remaining deck. What is the probability that the new card will also be a heart?
This is what I tried:
Probability of 4 heart and one from diffent suit is=$$\binom{4}{1}\binom{13}{4}\binom{3}{1}\binom{13}{1}\bigg/\binom{52}{5}$$ After discard the card from different suit, probability of choosing a heart would be =9/47. So the probability of getting flush would be , $$ \frac{\binom{4}{1}\binom{13}{4}\binom{3}{1}\binom{13}{1}\bigg/\binom{52}{5}}{9/47} $$ Am I on the right track? If not how can I solve it?
I didn't understand your calculation. Why are you dividing with ${52 \choose 5}$ after you have taken 4 cards only?
Also, the terms in the numerator is confusing. Even if those were correct, its not correct to divide with $(9/47)$.
This is what I think:
The probability of getting 4 cards of hearts and one from a different suite.
$(\frac{13}{52}) * (\frac{12}{51}) * (\frac{11}{50}) * (\frac{10}{49}) * (\frac{39}{48}) = 0.00214585834$
Now, discard the last card, which was of different suite. In the remaining deck, there are 9 hearts and 47 cards in total. So, now separately, the probability of choosing a hearts is $9/47$.
And the probability that the flush would be completed(considering the whole situation, meaning from the start of the dealing) is simply the multiplication of the previous value and $9/47$.
So: $0.00214585834 * 0.191489362 = 0.00041090904$