Hey, I found online this problem about a deck of cards but I’m stuck on it,
We have three decks of 52 cards, 4 suits of 13 cards. We shuffle them separately and separate each deck into two piles of 26 cards. Now ask someone to take a card from 4 of the 6 available piles, one per pile, and we see that 4 kings have come out. So let's take the 2 piles from which no cards have been drawn, put the 4 kings inside, and shuffle again. You now have 56 shuffled cards. Divide into 2 piles and take the first 4 cards from one of the piles. What is the probability that these 4 cards are 4 kings with the same suits as those obtained at the beginning?
I first used the combinations formula, as we have 4 kings and 56 cards in the deck. Having this expression: 4!/56!. Afterwards I divide it for 52! as it its important not only the options in which those 4 kings are in the deck but also the other 52 cards which might be in a position or in another. Having as a final result 4!/((52!)(56!)). Nevertheless I suspect that this doesn’t considerate the possibility of having other kings (in the 52 cards deck before joining the 4 kings that we have taken outside) that are the same suit as the 4 kings we are adding to the deck; and those should also be taken into account. Having arrived here I’m also trying to approach the problem using the Bayes theorem P(A I B)=P( A intersection B)/( P(B), but as the A and B from the P( A intersection B) are not independent, I cannot use it.
If you would mind giving me some way to continue or recommendation, or anything it would be a lot of help,
Thanks!
We have three decks of 52 cards, 4 suits of 13 cards. We shuffle them separately and separate each deck into two piles of 26 cards. Now ask someone to take a card from 4 of the 6 available piles, one per pile, and we see that 4 kings have come out. So let's take the 2 piles from which no cards have been drawn, put the 4 kings inside, and shuffle again. You now have 56 shuffled cards. Divide into 2 piles and take the first 4 cards from one of the piles. What is the probability that these 4 cards are 4 kings with the same suits as those obtained at the beginning?
I first used the combinations formula, as we have 4 kings and 56 cards in the deck. Having this expression: 4!/56!. Afterwards I divide it for 52! as it its important not only the options in which those 4 kings are in the deck but also the other 52 cards which might be in a position or in another. Having as a final result 4!/((52!)(56!)). Nevertheless I suspect that this doesn’t considerate the possibility of having other kings (in the 52 cards deck before joining the 4 kings that we have taken outside) that are the same suit as the 4 kings we are adding to the deck; and those should also be taken into account. Having arrived here I’m also trying to approach the problem using the Bayes theorem P(A I B)=P( A intersection B)/( P(B), but as the A and B from the P( A intersection B) are not independent, I cannot use it.
If you would mind giving me some way to continue or recommendation, or anything it would be a lot of help,
Thanks!