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Let's say that I have 4 of a card in my 60 card deck.
What are my odds of drawing at least 1 of that card in my initial 7 card hand?
If I discard those 7, what are the odds of drawing 1 of the 4 in another 7 card draw? (I guess you could think of a Wheel of Fortune type situation)
Since no one has provided the correct answer, at least not they current way you have it worded, I will take a stab at it.
What are my odds of drawing at least 1 of that card in my initial 7 card hand?
Answer: 1-Hypergeometric Distribution(population=60,successes in population=4,sample size=7,successes in sample=0) = 1 - .600500 = 39.95%
Everyone got this right. I personally like corsiKa's answer, because if your aren't interested in drawing at least 1 success then it is easier to do the calculation (especially by hand) by figuring the odds of not drawing any cards and subtracting from 1.
If I discard those 7, what are the odds of drawing 1 of the 4 in another 7 card draw? (I guess you could think of a Wheel of Fortune type situation)
This is where everyone mucks it up. First realize that this question is fundamentally different than any of the following
What you actually asked for was the bolded question above. So, what that means is that at that moment, you have removed 7 cards from the deck that weren't successes, leaving you 53 cards. The odds of drawing exactly 1 card can easily be found, HGD(53,4,7,1) = 36.29% or of getting at least 1 is (1 - HGD(53,4,7,0) = 44.27%).
Now that you have finally clarified what it is that you want. The answer that you seek is above. It is the last answer, which is = all possibilities - (chance of missing first 7 cards * chance of missing again after replacement) = 1 - (.6005 * .6005) = 63.94%
In general, you either multiply probabilities (when determining how likely two independent events will occur in a row), or add them together (when determining if either of two mutually exclusive events will occur). As for when to "add" these probabilities together and when to "multiply" them, it is very difficult to know when to do which.
In your above question, you might realize that there are 4 independent mutually exclusive events for drawing 7 cards, replacing those 7 cards, shuffling and drawing 7 cards again. From the answer from the first question, you know that you have about a 40% chance of drawing at least one copy, and a 60% chance of drawing zero copies. (0.60 + 0.40 = 1 these mutually exclusive events equal 100%, so that makes sense)
Your first and second (draw 7 cards and see if you succeed) events are independent of each other, because you are replacing the cards you drew back into the deck. Neither event has any effect on the results of the other, so you can multiply the first and second events together to see how likely it is to get a particular result twice in a row. Finally, because you are interested in the chances of succeeding at least once, you can either add all the mutually exclusive results above that succeed one or more times, or just subtract 1 from the chance of missing twice. (realizing that those two values should be the same)
The chance of drawing the card will be 1 - (chance of not drawing). I know that sounds silly, but bear with me.
You get to draw 7 cards. The chance of avoiding it on your first draw is 56 / 60 because there's 56 cards that are not that card. The next card is a chance 55 / 59. This carries on for 7 cards.
56 * 55 * 54 * 53 * 52 * 51 * 50
--------------------------------
60 * 59 * 58 * 57 * 56 * 55 * 54
You can cancel out some of those values (56, 55, 54)
53 * 52 * 51 * 50 7,027,800
----------------- = ---------- = .6005
60 * 59 * 58 * 57 11,703,240
So there's a 60.05% chance of NOT getting the card on the first go. That means there's a 39.95% chance of getting it.
You can carry this out to 14 cards to represent your wheel. As in the last one, we can keep the 60, 59, 58, and 57 on the bottom, and the last four on top (46, 45, 44, 43). The rest will cancel out
46 * 45 * 44 * 43 = 3,916,440
----------------- = ---------- = .3346
60 * 59 * 58 * 57 11,703,240
So you have a 33.46% chance of avoiding your target card, which means a 66.54 chance of getting it.
Of course, the answer to the first part of your question is here, and describes to some degree what a hypergeometric distribution is.
For the first part of your question, your odds are about 39.95% of drawing a particular card in your opening 7 card hand given 4 in a 60 card pool.
If you are in a situation where you draw 7 then draw 7 more without replacement, you are effectively drawing 14 cards instead, then your probability changes to 66.53%
