Combos in a deck of cards: probability I draw containing either an ace and a 2 or else a king and a queen

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Seja

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How would I calculate the probability of the first 5 cards I draw containing either an ace and a 2 or a king and a queen in a classic deck of 52 playing cards?

I'm having a hard time learning the hypergeometric calculator ?
The probability of successfully drawing one of the combos is 10.02%
I saw a Reddit post saying that to figure this out I should calculate the probabilities of each combo separately and then simply add them together. I double checked this by calculating the probability of drawing one card.
The probability of seeing an ace in your first five draws is 34.12%
If there were only two aces in your deck of 52 cards the probability would be 18.48%
18.48 + 18.48 is 36.96 not 34.12 so addition is certainly not the correct way to do this.
 
Seja said:
How would I calculate the probability of the first 5 cards I draw containing either an ace and a 2 or a king and a queen in a classic deck of 52 playing cards?

I'm having a hard time learning the hypergeometric calculator ?
The probability of successfully drawing one of the combos is 10.02%
I saw a Reddit post saying that to figure this out I should calculate the probabilities of each combo separately and then simply add them together. I double checked this by calculating the probability of drawing one card.
The probability of seeing an ace in your first five draws is 34.12%
If there were only two aces in your deck of 52 cards the probability would be 18.48%
18.48 + 18.48 is 36.96 not 34.12 so addition is certainly not the correct way to do this.
Click to expand...
I don't think the Reddit post applies to your problem. The probabilities of two events are added together only when the events do not intersect. I.e., when the probability of both events is 0, which is not your case. Do you know how to compute the probability of a union of two events in the general case?
 
Seja said:
How would I calculate the probability of the first 5 cards I draw containing either an ace and a 2 or a king and a queen in a classic deck of 52 playing cards?

I'm having a hard time learning the hypergeometric calculator ?
The probability of successfully drawing one of the combos is 10.02%
I saw a Reddit post saying that to figure this out I should calculate the probabilities of each combo separately and then simply add them together. I double checked this by calculating the probability of drawing one card.
The probability of seeing an ace in your first five draws is 34.12%
If there were only two aces in your deck of 52 cards the probability would be 18.48%
18.48 + 18.48 is 36.96 not 34.12 so addition is certainly not the correct way to do this.
Click to expand...
If only two cards were drawn, you would have two disjoint events, namely "an ace and a 2" and "a king and a queen". But since 5 cards are drawn, both events can happen together (A, 2, K, Q, and one other card). So, as @blamocur says, you need the more general formula for a union.

Can you explain, though, what you mean by "the hypergeometric calculator", and how you are doing your calculations?
 
The reddit thread is wrong. One of the fundamental laws of probability is:

[math]\text {P(A or B) = P(A) + P(B) - P(A and B).}[/math]
So, the desired probability is

[math]\text {P(one ace and one 2) + P(one king and one queen) } \\ - \text { P(one ace, one king, one queen, and one 2)}[/math]
 
I believe I've solved it, but I did not find it easy at all :( In particular, finding the probability of having one of each rank (i.e., "P(one ace, one king, one queen, and one 2)" from @JeffM's post) took some time. My steps were:
  1. Consider 4 events [imath]X_i[/imath] where [imath]1\leq i \leq 4[/imath], which correspond to having [imath]i[/imath]-th rank (out of four, i.e. ace, 2, king and queen) in the hand of 5.
  2. Compute the probabilities of unions of those events. Because of the symmetry we need only 4 numbers, i.e., unions of one (being simply the probability of the event), two, three and four events;
  3. Using the general formulae for probabilities of unions of two, three and four events compute the probabilities of intersections of two, three and four events.
  4. Now apply the formula supplied by @JeffM.
 
JeffM said:
The reddit thread is wrong. One of the fundamental laws of probability is:

[math]\text {P(A or B) = P(A) + P(B) - P(A and B).}[/math]
So, the desired probability is

[math]\text {P(one ace and one 2) + P(one king and one queen) } \\ - \text { P(one ace, one king, one queen, and one 2)}[/math]
Click to expand...
I think it would be slightly cleaner to say "at least one..." instead of "one" since the latter might be interpreted as "exactly one". But it would make the text almost unreadable :)
 
I am not sure why you find the computation of the probability of picking one ace, one king, one queen, one 2, and one from jack through 3 conceptually difficult:

[math]\left \{ \left ( \dbinom{4}{1} \right)^4* \dbinom{52 - 16}{1} \right \} \div \dbinom{52}{5}[/math]
Do you see where that comes from?

Obviously, we cannot check your answer if you do not provide it. Nor can we help identify any computational errors (if there are any) in your work if you provide no computations.

EDIT: I realize now that I have assumed that the problem requires exactly one of the designated ranks. We would need to see the problem‘s own words to be sure that my interpretation of the problem is warranted.
 
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