[QUOTE=MentalGuy]
The first question seems correct to me, though I would have probably done it slightly differently.
On the second question, is it supposed to be “How many numbers between 100 and 999 have 3 distinct digits?”?
If so, then you have only 9 choices for the first number, but also 9 for the second, and then 8 possibilities for the third. Multiplying those you get 648. There is no need to divide, unless numbers like 478 and 784 are to be considered the same.
[/QUOTE]
My bad guy, reviewing the thread I saw your post for the first time, managing to miss it the first time, and you answered both questions properly the first time.
Malacandra
Yes, I did notice the numbers were the same as a previous posting but I was just saying my approach to the solution was different. Whereas everyone else was talking about ‘n’ ways of ordering ‘m’ objects, I went at the problem one card at a time. Different approach but the same answer.
[QUOTE=wolf_meister] Malacandra
Yes, I did notice the numbers were the same as a previous posting but I was just saying my approach to the solution was different. Whereas everyone else was talking about ‘n’ ways of ordering ‘m’ objects, I went at the problem one card at a time. Different approach but the same answer.
[/QUOTE]
That’s the way I was thinking it through, too. Probably 'cos I was talking about the birthday puzzle not too long ago - as in, how many randomly-selected people do you need before it is odds-on that two have the same birthday (day and month) - and that’s the approach to take with that one.
[QUOTE=CJJ*]
There are 52! ways to order a deck of cards.
There are 26! ways to order the black cards, and the same number of ways to order the red cards. Thus, there are 26!26! ways to draw first 26 red cards, then 26 black cards. Double this to get all the ways to order one color first, then the other = 226!*26!
The probability then is (2*26!*26!)/52! = 4.033 x 10^-15, or about one in every 250 trillion attempts. That’s the easiest way i know to solve this…
[/QUOTE]
Wait, I’m not a math whiz, but shouldn’t calculating the odds stop after the 26th card? After, the next 26 cards will, by definition, be of the opposite suit? Right? No?
[QUOTE=magellan01]
Wait, I’m not a math whiz, but shouldn’t calculating the odds stop after the 26th card? After, the next 26 cards will, by definition, be of the opposite suit? Right? No?
[/QUOTE]
I take it you’re looking at the numerator, seeing 2(26!)(26!), and wondering why the second 26! is there?
Focus on the numerator and keep in mind that the numerator alone is not calculating the odds–it’s counting the number of different ways that the black and red cards can be segregated.
The denominator, 52!, counts the number of different ways to arrange all 52 cards, without any restrictions.
In that case, the numerator should count ALL the different ways to segregate the cards (then divide by 52! to get the probability we want).
If we just have one factor of 26! in the numerator, we’ve counted all the different ways to arrange, say, all the black cards amongst themselves, but we’ve completely ignored all the different ways to arrange the red cards amongst themselves. Even assuming that all the red cards are grouped together, we still need to count the number of DIFFERENT ways this grouping can happen, and that’s why there’s a second factor of 26! in the numerator (along with the 2, indicating the cards can be arrange red first, then black, or black first, then red.
[QUOTE=Cabbage]
I take it you’re looking at the numerator, seeing 2(26!)(26!), and wondering why the second 26! is there?
Focus on the numerator and keep in mind that the numerator alone is not calculating the odds–it’s counting the number of different ways that the black and red cards can be segregated.
The denominator, 52!, counts the number of different ways to arrange all 52 cards, without any restrictions.
In that case, the numerator should count ALL the different ways to segregate the cards (then divide by 52! to get the probability we want).
If we just have one factor of 26! in the numerator, we’ve counted all the different ways to arrange, say, all the black cards amongst themselves, but we’ve completely ignored all the different ways to arrange the red cards amongst themselves. Even assuming that all the red cards are grouped together, we still need to count the number of DIFFERENT ways this grouping can happen, and that’s why there’s a second factor of 26! in the numerator (along with the 2, indicating the cards can be arrange red first, then black, or black first, then red.
Does that make sense?
[/QUOTE]
Kinda. May I ask it this way (the way I would have attempted to solvce it): You pick any first card (red), then the odds of choosing a red second card has to be 25 of 51, then 24 of 50, then 23 of 49…2 of 28, 1 of 27. And you stop right there, as the odds of all the next cards being the opposite suit are 100%. I don’t know how to express it mathematically, but shoudn’t the idea be
26!
[QUOTE=magellan01]
Kinda. May I ask it this way (the way I would have attempted to solvce it): You pick any first card (red), then the odds of choosing a red second card has to be 25 of 51, then 24 of 50, then 23 of 49…2 of 28, 1 of 27. And you stop right there, as the odds of all the next cards being the opposite suit are 100%. I don’t know how to express it mathematically, but shoudn’t the idea be
26!
52! (stopping at 27)
?
[/QUOTE]
That’s almost perfect, only realize that the first card might be black rather than red. So your answer should be exactly double what you have:
2(26!)
52! (stopping at 27)
Now compare this with the given answer:
(2*26!*26!)/52! = 2(26!) [26!/52!]
I grouped it this way so we could focus on [26!/52!] which, written out, is
26 * 25 * 24 *… * 1
52 * 51 * 50 * … * 1
Notice that a lot of those factors cancel out, leaving:
1
52! (stopping at 27)
and so the given answer is
2(26!)
52! (stopping at 27)
which is exactly the same as your answer!
I think the difference between the two approaches is what might be confusing you:
Your method calculates the probability “directly”, more or less (calculating the probability of each individual card being the correct color). The other method treats it as a combinatorial (counting) problem: Count how many different ways there are to segregate the red and black cards, then count how many different ways there are to arrange all 52 cards, period. Now divide to get the probability.
[QUOTE=Cabbage]
That’s almost perfect, only realize that the first card might be black rather than red. So your answer should be exactly double what you have:
2(26!)
52! (stopping at 27)
Now compare this with the given answer:
(2*26!*26!)/52! = 2(26!) [26!/52!]
I grouped it this way so we could focus on [26!/52!] which, written out, is
26 * 25 * 24 *… * 1
52 * 51 * 50 * … * 1
Notice that a lot of those factors cancel out, leaving:
1
52! (stopping at 27)
and so the given answer is
2(26!)
52! (stopping at 27)
which is exactly the same as your answer!
I think the difference between the two approaches is what might be confusing you:
Your method calculates the probability “directly”, more or less (calculating the probability of each individual card being the correct color). The other method treats it as a combinatorial (counting) problem: Count how many different ways there are to segregate the red and black cards, then count how many different ways there are to arrange all 52 cards, period. Now divide to get the probability.
[/QUOTE]
Maybe I’m more of a math whiz than I thought. Thanks for the hand-holding. One question though—why do you need the 2 in the numerator. Shouldn’t it be there only if you cared if you started with red or black?
Maybe a better way to ask it is: what would that equation look like if you mandated that you start with red?
[QUOTE=magellan01]
Maybe I’m more of a math whiz than I thought. Thanks for the hand-holding. One question though—why do you need the 2 in the numerator. Shouldn’t it be there only if you cared if you started with red or black?
Maybe a better way to ask it is: what would that equation look like if you mandated that you start with red?
[/QUOTE]
If you mandated that you start with red, the probability would be what you had in your previous post:
26!
52! (stopping at 27)
Of course, if you don’t care whether the first card is black or red, it’s a little different. In talking through the above solution, we would say, “there’s a 26 out of 52 chance that the first card is red; a 25 out of 51 chance for the second card,…”
However, if we don’t care if the first card is red or black, there’s no probability involved for the first card–that first card can be any one of the 52.
From there, we would say, “There’s a 25 out of 51 chance the second card MATCHES the first, a 24 out of 50 chance the third matches the first two,…” This gives the correct probability:
[QUOTE=magellan01]
Excellent. Except for the step above. I don’t see how they are equal. My apologies if I’m being dense.
[/QUOTE]
No problem. We can write 26! as 26(25!) (both are the product of all positive integers less than or equal to 26). Same thing for 52!, it’s equal to 52(51!).
So (2*26!26!)/52! = 226(25!)(26!) / [52(51!)] = 52(25!)(26!) / [52(51!)]
[QUOTE=Cabbage]
No problem. We can write 26! as 26(25!) (both are the product of all positive integers less than or equal to 26). Same thing for 52!, it’s equal to 52(51!).