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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
Quote:
Originally Posted by colerelm1
Sorry but you question is not very clear.
If you mean a set, $\displaystyle \{\text{Jack}, \text{Queen}, \text{King}, \text{Ace}\}$ that set occurs in $\displaystyle 4!(4^4)$ ways.
On the other hand, if you mean a $\displaystyle 4\text{-tuple}$, $\displaystyle (\text{Jack}, \text{Queen}, \text{King}, \text{Ace})$ that occurs in $\displaystyle (4^4)$ ways.
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?
I actually answered that question.Quote:
Originally Posted by colerelm1
Order 4-tuples are what you are talking about.
The 4-tuple $\displaystyle (4,2,3,1)$ is quite different from the set $\displaystyle \{1.2.3.4\}$.
In the first order makes a difference. In the second, order makes no difference.
I think the point that isn't clear is whether or not once a card has been drawn, is it returned to the deck, or is the second card drawn from a depleted deck, and so on for the other two cards ?
That never occurred to me to be a possible reading of the question.Quote:
Originally Posted by BobP
So I'm confused by what you mean here. Isn't a probability a percentage chance that it will happen? If you say that there are $\displaystyle 4!(4^4)$ ways the set {Jack,Queen,King,Ace} occurs, does that mean there exists a very big number n such thatQuote:
Originally Posted by Plato
$\displaystyle (4!(4^4)) / n$?
You still have not said exactly what you are talking.Quote:
Originally Posted by colerelm1
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.
I'd like to know the probability of drawing, one card at a time, without putting any cards back into the deck, a Jack, then a Queen, then a King, then an Ace.
Quote:
Originally Posted by colerelm1
$\displaystyle \frac{4}{52}\cdot\frac{4}{51}\cdot\frac{4}{50} \cdot \frac{4}{49}$
Why is it a product instead of a sum? I'm just trying to get a feel for the general approach to these types of problems and I feel like I could easily make the mistake of adding instead of multiplying for a similar problem.Quote:
Originally Posted by Plato
Quote:
Originally Posted by colerelm1
If you will kindly give an example where you think that a sum is used then maybe I can see your confusion.
Well like in the previous problem...Why is it:
instead of 4/52 + 4/51 + 4/50 + 4/49?Quote:
Originally Posted by Plato
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 ?
