[Q] If only 2 out of 10 people are chosen. Is your chance 2/10 which is 20% or 1/10 + 1/9 which is 21.11%?

Here's where I think you're going wrong (I'm a student, so I may be wrong):

Thinking about this as randomizing a deck, it intuitively makes sense that the probability of any individual card in the deck, from position 1 to 52, has a 1/52 chance of being the card of interest. But once you know that the first card isn't the card of interest, the probabilities change. The situation you're considering:

If you draw a card randomly from a deck, look at it, and it is an ace of spades, and then you draw another card randomly from the deck, and don't look at it, what is the chance that that the second card is ace of hearts? Is it 1/51 because now there are only 51 cards in the deck? No, it is still 1/52

excludes the possibility that the first card is the card of interest. What you're asking is P(Ace_Hearts_Second|Ace_Spades_First). The conditional probability should be

P(A|B) = P(AnB) / P(B)

where (AnB) is the probability of drawing the ace of hearts after drawing that specific card, or (1/52)(1/51), divided by P(B) which is 1/52 is equal to 1/51.

Or, where B is equal to NOT drawing the ace of hearts first, it would be (51/52)(1/51)/(51/52). Or, out of the sample space of every situation where the ace of hearts is not drawn first, the likelihood of drawing it second is 1/51.

When you say

Because you didn't intentionally select the ace of spades in the first case, and the option to select the ace of hearts was open, the chance of selecting the ace of hearts is still 1/52 after the ace of spades is withdrawn, because on average 1/52 times you do the first draw you will withdraw the ace of hearts, and can not even move on to the second half of the experiment.

This is where you lose me. Originally you were talking about conditional probability, but if you're saying that the option to select the ace of hearts was open then it's an entirely different question, to which the answer would be 1/52 (the likelihood that the ace of hearts is in any single location in a randomized 52-card deck). Once you have more information, that the first card was not the ace of hearts, you need to update the model.

/r/statistics Thread Parent