Daily3 game: 3 numbers from 0 to 9 are drawn w/ repetitions allowed...

[mrnerd]

Related to another thread: Here is another problem that is similar that I know the answers to and perhaps it will shed some light on the issue.

In the Daily3 game, three numbers between 0 and 9 will be drawn in succession (repetitions allowed). The player marks three numbers on a game card and has a choice of how to play, straight (the player's numbers will match the three numbers drawn in exact order) or box (the player's numbers will match the three drawn in any order).
a. What is the probability of a box if 3 distinct numbers are drawn? Answer: 0.006
I was initially thinking that "drawn" means the sample space was changing. For instance, instead of 10x10x10 for total possible combinations, it was 10x9x8. However now I notice that if I take it to mean that I "pick" 3 distinct numbers I get the right answer. 3P3 = 6 total permutations of 3 distinct numbers, 6/1000 = 0.006.
I'm not sure if I'm just accidentally arriving at the correct answer in the wrong way though.

b. what is the probability of a box if 2 of the numbers drawn are the same? Answer: 0.003
again if I use the same interpretation of the problem then I can arrive at the correct answer.
If I pick XXY as my 3 numbers, there are 3 distinct ways to order XXY
XXY
XYX
YXX
3/1000 = 0.003.

I'm not really sure how to do that ordering on part b more mathematically than just trial and error like that though, which is the problem I keep running into on tests.
I think the confusing part of this question was the use of the word "drawn", however I'm still not sure I'm interpreting it correctly.

Thanks for you help

 

[Dr.Peterson]

unfortunately my TA didn't show up for his office hours.
However, here is another problem that is similar that I know the answers to and perhaps it will shed some light on the issue.

In the Daily3 game, three numbers between 0 and 9 will be drawn in succession (repetitions allowed). The player marks three numbers on a game card and has a choice of how to play, straight (the player's numbers will match the three numbers drawn in exact order) or box (the player's numbers will match the three drawn in any order).
a. What is the probability of a box if 3 distinct numbers are drawn? Answer: 0.006
I was initially thinking that "drawn" means the sample space was changing. For instance, instead of 10x10x10 for total possible combinations, it was 10x9x8. However now I notice that if I take it to mean that I "pick" 3 distinct numbers I get the right answer. 3P3 = 6 total permutations of 3 distinct numbers, 6/1000 = 0.006.
I'm not sure if I'm just accidentally arriving at the correct answer in the wrong way though.

b. what is the probability of a box if 2 of the numbers drawn are the same? Answer: 0.003
again if I use the same interpretation of the problem then I can arrive at the correct answer.
If I pick XXY as my 3 numbers, there are 3 distinct ways to order XXY
XXY
XYX
YXX
3/1000 = 0.003.

I'm not really sure how to do that ordering on part b more mathematically than just trial and error like that though, which is the problem I keep running into on tests.
I think the confusing part of this question was the use of the word "drawn", however I'm still not sure I'm interpreting it correctly.

Thanks for you help

This really should have stayed in the original thread so we could compare better.

But my image now is that your card has three columns, each containing 0 through 9, and you "mark" any one in each column. The sample space is 1000 3-digit numbers (since repetition is allowed). The "draw" amounts to "them" marking a card in the same way. Of the 1000 things you could have done, IF we know that "they" picked two numbers the same, then we know that 3 possible outcomes are wins for you. There is a way to count this 3 formally, but just counting by hand is easy enough. I don't have time to tell you more now; I'll get back later.

 

[mrnerd]

I'm sorry, I thought I was replying in the other thread. I'm not sure what happened. I clicked the reply button same as I did just now.

Thanks for your help

 

[Dr.Peterson]

I'm sorry, I thought I was replying in the other thread. I'm not sure what happened. I clicked the reply button same as I did just now.

Thanks for your help

It's possible someone moved it, thinking it belonged elsewhere; that's why I phrased my comment as "should have stayed". It's also why I responded without having time to say much, just to get the context on the record.

Here's the original problem:

In the Daily4 game, four numbers between 0 and 9 will be drawn in succession (repetitions allowed). The player marks four numbers on a game card and has a choice of how to play, straight (the player's numbers will match the four numbers drawn in exact order) or box (the player's numbers will match the four drawn in any order).
a. What is the size of the sample space? 10^4 = 10,000
b. what is the probability of a straight? 1/10,000 = 0.0001

I'm not understanding what I'm doing wrong here
c. What is the probability of a box if four distinct numbers are drawn?
4 distinct numbers = 10*9*8*7 = 5040
4 permutations of a "box" = 4P4 = 4
4/5040 = 1/1260 <-- I know this is wrong
Another way of doing the same problem:
each number is distinct: 10C4 = 210
1 combination of winning "box" = 1/210 <-- I'm also sure this is wrong

d. What is the probability of a box if two of the numbers drawn are the same?
pick 3 numbers, the remaining number has to be 1 of the 3 already picked.
10*9*8*3 = 2160
4 permutations of winning "box" = 4/2160 = 1/540

Now that I've seen the example problem, and reoriented my mental model of the drawing, what you're doing doesn't fit what they evidently mean. The idea is that you have filled in a card, allowing repetitions according to the rules, so there are not 10P4 = 5040 ways you could have done so, but 10^4 = 10000. But we are told (in c) that their numbers are distinct; if we were to list all possible cards that would be winners, there are 4P4 = 4! = 24 of them, so the probability that your card is one of those winners is 24/10000 (not 24/5040, which would have given your answer of 1/210).

For (d), you know that two of their 4 numbers are the same (like XYZZ). So the numerator has to be the number of ways to arrange such a set of numbers. Have you learned a way to count rearrangements of a word with repeated letters? That, or something equivalent, is what you need. Let me know what counting methods you have learned, and I can show how to use one of them here.

 

[mrnerd]

thank you! I figured it out. I'm not going to repost my answer in case anyone else wants to try and figure it out for future reference.
I used the counting technique where you count "zz" as 1 letter, so then you would have 3!*2 right? and that gives you all the ways to reorder "xyzz".
I figured it out when you mentioned rearranging letters in a word. I don't know why but I didn't make the connection that this was the same technique.

thanks again

 

[Dr.Peterson]

thank you! I figured it out. I'm not going to repost my answer in case anyone else wants to try and figure it out for future reference.
I used the counting technique where you count "zz" as 1 letter, so then you would have 3!*2 right? and that gives you all the ways to reorder "xyzz".
I figured it out when you mentioned rearranging letters in a word. I don't know why but I didn't make the connection that this was the same technique.

thanks again

Yes, that's one of several ways to think about the letter problem. Good work.

And a large part of the answer to your initial question, "I can never really figure out how I'm supposed to apply formulas" is to make connections, as you mentioned, with techniques for other problems. To do that, you need to think about other ways to represent the problem (to reframe it, or use a different model). Here, you yourself used a notation like XYZZ, which suggested that we equate this part of the problem to arranging letters. Other problems can be modeled by rolling dice, picking balls from an urn, and so on, each of which has a different behavior. And one problem may require several different models, so you need to keep your mind free to consider alternate perspectives.