Check my understanding
April 22nd, 2023 at 11:25:16 AM
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I have a deck of cards. there are 2 copies of each card, numbered 1 - 16 for a total of 32 cards.
I want to figure out the odds of drawing 2 pairs from the top 3 (i.e. cards numbered 14 15 and 16).
First card = 6/32 (could be 1 of the 6 top cards of the set).
Second card = 5/31 (could be any 1 of the 5 remaining top 6 cards).
The third card is where i get stuck. If the 2nd card is the pair to the first then the probability is 4/30 (2/15 simplified), as the third card could be any one of the remaining two pairs.
If the 2nd card is not the pair to the first, then there are only 2 cards out of 30 (1/15 simplified) which could complete the hand.
the last card is 1/30 as there is only 1 remaining card in the deck that could complete the condition.
How do I handle this sort of situation? Is this the right way to think about this probability situation or would a better approximation be to treat each card as an individual event?
I want to figure out the odds of drawing 2 pairs from the top 3 (i.e. cards numbered 14 15 and 16).
First card = 6/32 (could be 1 of the 6 top cards of the set).
Second card = 5/31 (could be any 1 of the 5 remaining top 6 cards).
The third card is where i get stuck. If the 2nd card is the pair to the first then the probability is 4/30 (2/15 simplified), as the third card could be any one of the remaining two pairs.
If the 2nd card is not the pair to the first, then there are only 2 cards out of 30 (1/15 simplified) which could complete the hand.
the last card is 1/30 as there is only 1 remaining card in the deck that could complete the condition.
How do I handle this sort of situation? Is this the right way to think about this probability situation or would a better approximation be to treat each card as an individual event?
April 22nd, 2023 at 1:00:43 PM
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I will assume you are trying to draw two pairs numbered 14-16 from the top four cards of the deck.
Since the order of the cards doesn't matter, I would count the total number of sets of 4 cards, and then how many of those have two pairs numbered 14-16.
The total number of sets of four cards is C(32,4) - i.e. the number of "combinations" of 4 items taken from a set of 32 - which is 35,960.
For each set of two high pairs, there are three pairs of ranks - {14,15}, {14,16}, and {15,16} - and for each pair, there is only one way to draw the pair as there are only 2 cards of that number in the deck, so the total number of 4-card sets that have two high pairs is 3.
The probability is 3 / 35,960.
Doing it your way, I would separate the second card draws into ones that match the first card, and ones that did not; calculate the final probabilities in those two cases separately, and then add them.
Since the order of the cards doesn't matter, I would count the total number of sets of 4 cards, and then how many of those have two pairs numbered 14-16.
The total number of sets of four cards is C(32,4) - i.e. the number of "combinations" of 4 items taken from a set of 32 - which is 35,960.
For each set of two high pairs, there are three pairs of ranks - {14,15}, {14,16}, and {15,16} - and for each pair, there is only one way to draw the pair as there are only 2 cards of that number in the deck, so the total number of 4-card sets that have two high pairs is 3.
The probability is 3 / 35,960.
Doing it your way, I would separate the second card draws into ones that match the first card, and ones that did not; calculate the final probabilities in those two cases separately, and then add them.
