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This happens to be about poker but I would like to ask it in a general way

There are 47 cards remaining in the deck with 2 cards to come
Of the 47 cards 9 make my hand - I only need 1 but 2 two also makes my hand

$$\frac { \binom{9}{1} \binom{38}{1} + \binom{9}{2} } { \binom{47}{2} } = 0.349676226 $$

Pretty sure that is the correct number as it is posted in poker odds

Tried to get that same number with fractions

The first card is 9/47 = 0.191489362

Second card
Two possibilities - first card hit or not

  • first card hit: (9/47)*(8/46) = 0.03330249

  • first card did not hit: (1 - 9/47)*9/46 = 0.158186864

Add those up 0.191489362 + 0.033302498 + 0.158186864 = 0.382978723

That is not the same number as combinations 0.382978723 != 0.349676226
My question why are those not the same result?

With fractions this matches 9/47 + (1 - 9/47)*9/46 = 0.349676226
But that does not make sense to me as it seems to me that should be the chance of 1 card hitting excluding the chance of 2 cards hitting

If I go at it 1 - not then it matches
1 - ((47-9)/47*(46-9)/46)
= 1 - (1 - 9/47)*(1 - 9/46)
= 1 - 1 + 9/47 + 9/46 - 9*9/47/46
= 9/47 + 9/46(1 - 9/47)

From the answer I found my mistake - this matches combinations

  • first card hits and second does not (9/47)(46-8)/46 = 0.158186864
  • first and second (9/47)(8/46) = 0.033302498
  • first card does not hit and second does ((47-9)/47)(9/46) = 0.158186864
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Your method (the one that produces the wrong result) consists of summing the following three probabilities

  • The first card hits
  • The first card hits and the second card hits
  • The first card does not hit and the second card hits

But these events are not mutually exclusive so the probability of their union (probability that at least one of them occurs) is not obtained by summing the probabilities of the individual events. You are double-counting the possibility of both cards hitting, since it is contained both in the first term and the second term.

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