How would I solve a problem that asks me to calculate the probability of drawing the following cards from a deck, if I were to draw the cards one at a time:
{Jack, Queen, King, Ace}?
What are the steps that I have to take to begin solving this? Would it be the same process for solving the same problem except with a different sequence:
{Jack,Jack,Jack, Jack}?
Thanks
Ok, sorry for being unclear. Let me clear things up.
I have a deck of 52 cards. From it I draw one card and I get a Jack. I draw again and get a Queen. On the third draw I get a King. On the fourth draw I get an Ace. What is the probability of this happening?
Same scenario for the other one:
First draw = Jack, second draw = Jack, third draw = Jack, fourth draw = Jack. What is the probability of this happening?
You still have not said exactly what you are talking.
Are you asking about a 'hand' of four cards drawn in any order?
OR, are you asking about a 'hand' of four cards drawn in particular order?
And you did not address Bob's question: "is the drawn card returned to the deck after each draw?
If you are asking "what is the probability of getting "{Jack, Queen, King, Ace}" in that order without replacement, then there are 4 Jacks in 52 cards so the probablity of getting a Jack on the first draw is 4/52= 1/13. There are now 4 Queens in 51 cards so the probability of getting a Queen on the second draw is 4/51. Similarly, the probability of gettig a King on the third draw is 4/50= 2/25 and the probability of getting an Ace on the last draw is 4/49. The probability of getting "{Jack, Queen, King, Ace}", in that order, without replacing the card in the deck after drawing it (and reshuffling) is (1/13)(4/51)(2/25)(4/49)= 32/812175 which is about 0.0000394.
IF you mean "what is the probability of getting "{Jack, Queen, King, Ace}" in any order then one can show that the probability of those cards in any single order is the same, 32/812175, so we just need to multiply by the number of such orders which is 4!= 24. The probability is (24)(32)/812175= 768/812175 which is about 0.0009456.
IF you mean "in that order" but you replace the card (and reshuffle) after each pick then you are always picking from 52 cards so the denominators in those fractions is always 52 and the denominators are always 4 so the probability is (4/52)^4= (1/13)^4= 1/28561 which is about 0.0000350.
IF you mean "in any order" and you replace the card (and reshuffle) after each pick then you need to multiply that by 4!= 24. The probability is 24/28561 which is about 0.0008403.
Suppose that instead of asking for J,Q,K,A , you ask for red card, red card, red card, red card.
Is the probability 26/52 + 25/51 + 24/50 + 23/49 = 242351/124950 = 1.94 approximately,
or is it 26/52 . 25/51 . 24/50 . 23/49 = 46/833 approx 0.055 ?