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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?

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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: $0.00214585834 * 0.191489362 = 0.00041090904$

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  • $\begingroup$ wouldn't it be "given getting 4 cards of hearts and one from a different suite, what is the probability of discard that last card(not a heart, different suit from the other 4) and get a heart"? I'm thinking there would be a conditional probability in this question though.. $\endgroup$ – Liz Sugar Sep 16 '16 at 20:14
  • $\begingroup$ i think the discarding part is certain. that's why i didn't include it in the calculations. $\endgroup$ – Hasan Iqbal Sep 17 '16 at 1:41
  • $\begingroup$ wouldn't the other four cards being certain as well? I'm thinking the answer is 9/47, but it's just too straight forward to be true.... lol $\endgroup$ – Liz Sugar Sep 18 '16 at 5:57

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