Probability question

[George.W89]

This might be quite straightforward but I'm not 100% sure how to calculate this so help will be much appreciated.

If I have an 8 sided die, and want to hit a 1,2,3 combination in any sequence the probability would be (3/8)(2/8)(1/8) given that I had 3 attempts, correct?

My question is, how would the probability change/increase if I have 5 attempts?

Thanks!

 

[Steven G]

There are 8^3 possible outcomes when you toss this die three times. How many different outcomes are favorable to you (ie have 1, 2 and 3)? That answer is 6 (123, 132, 213, 231, 312, 321). So the probability of tossing this die three times and getting a combination of 1, 2 and 3 is 6/8^3.

So your answer must be wrong as you say that you get three attempts at this. 6/8^3 is what you get if you have exactly one attempt. Having more attempts, like 3, will increase the probability. Do you see that? If you buy 2 lottery tickets you have a better chance of winning compared to just buying one ticket.

So please go back and rethink the original question before doing the 2nd problem.

 

[JeffM]

Your answer is correct. Jomo just got there a different way.

He went

[MATH]\text {number of ways to permute three distinct objects} = 3! = 3 * 2 * 1 = 6.[/MATH]
Then he went

[MATH]\text {number of possible outcomes of rolling an 8-sided die three times} = 8^3 = 512.[/MATH]
And arrives at [MATH]\dfrac{6}{512} = \dfrac{3}{256} \approx 1.2\%.[/MATH]
You used the fact that what you get is independent on each roll.

Probability of getting one of the three desired numbers on first roll = 3/8.

Probability of getting one of the other two desired numbers on second roll = 2/8.

Probability of getting the remaining desired number on third roll = 1/8.

[MATH]\dfrac{3}{8} * \dfrac{2}{8} *\dfrac{1}{8} = \dfrac{6}{512} \approx 1.2\%.[/MATH]
So your thinking was fine, but it does not extend as easily to your new problem as jomo's does.

I presume your new problem is what is the probability of rolling 1, 2, and 3 plus two numbers that are not 1, 2, or 3 in five rolls. Or is it something different?

 

[Steven G]

Your answer is correct. Jomo just got there a different way.

He went

[MATH]\text {number of ways to permute three distinct objects} = 3! = 3 * 2 * 1 = 6.[/MATH]
Then he went

[MATH]\text {number of possible outcomes of rolling an 8-sided die three times} = 8^3 = 512.[/MATH]
And arrives at [MATH]\dfrac{6}{512} = \dfrac{3}{256} \approx 1.2\%.[/MATH]
You used the fact that what you get is independent on each roll.

Probability of getting one of the three desired numbers on first roll = 3/8.

Probability of getting one of the other two desired numbers on second roll = 2/8.

Probability of getting the remaining desired number on third roll = 1/8.

[MATH]\dfrac{3}{8} * \dfrac{2}{8} *\dfrac{1}{8} = \dfrac{6}{512} \approx 1.2\%.[/MATH]
So your thinking was fine, but it does not extend as easily to your new problem as jomo's does.

I presume your new problem is what is the probability of rolling 1, 2, and 3 plus two numbers that are not 1, 2, or 3 in five rolls. Or is it something different?

There you go presuming again!

 

[JeffM]

There you go presuming again!

I must admit that I am presumptuous

 

[Steven G]

Your answer is correct. Jomo just got there a different way.

He went

[MATH]\text {number of ways to permute three distinct objects} = 3! = 3 * 2 * 1 = 6.[/MATH]
Then he went

[MATH]\text {number of possible outcomes of rolling an 8-sided die three times} = 8^3 = 512.[/MATH]
And arrives at [MATH]\dfrac{6}{512} = \dfrac{3}{256} \approx 1.2\%.[/MATH]
You used the fact that what you get is independent on each roll.

Probability of getting one of the three desired numbers on first roll = 3/8.

Probability of getting one of the other two desired numbers on second roll = 2/8.

Probability of getting the remaining desired number on third roll = 1/8.

[MATH]\dfrac{3}{8} * \dfrac{2}{8} *\dfrac{1}{8} = \dfrac{6}{512} \approx 1.2\%.[/MATH]
So your thinking was fine, but it does not extend as easily to your new problem as jomo's does.

I presume your new problem is what is the probability of rolling 1, 2, and 3 plus two numbers that are not 1, 2, or 3 in five rolls. Or is it something different?

Not that it is your fault but why can't one roll 1,2,3,2 again! and then an 8? The problem is not worded very well. This was the point to the OP I was trying to make (and hopefully did) in my initial post.

 

[Steven G]

the probability would be (3/8)(2/8)(1/8) given that I had 3 attempts, correct?

My question is, how would the probability change/increase if I have 5 attempts?

Thanks!

What exactly do you mean my attempts?
I assume that you get three tries at getting the 1,2, 3 combination. That is, you roll a die three times (because you want three numbers), then you roll the die three times again for your 2nd attempt and then you roll the die three times for your 3rd attempt. Now you want the probability of getting the 1,2,3 combination at least once? Is this the way you want us to think about this problem. For the 2nd one, you do as just outlined but then do it another two times for a total of 5 attempts?

 

[George.W89]

Sorry about phrasing the question so poorly and thanks for your replies. The attempts referred to rolls of the die.

If I roll the die 3 times its (3/8)(2/8)(1/8) and that is correct. Although your method of 6 possible preferred outcomes divided by 8^3 outcomes is also perfect.

Now, if I roll the die 5 times, what is my increased probability? I know the possible outcomes are 8^5 now, but is there a faster way to determine the possible preferred outcomes rather than (12311,12312,12313......)

 

[Romsek]

I interpret the problem as rolling an 8 sided die 3 or 5 times and a success is there being at least a 1, 2, and 3 in the roll string.
OP got the correct answer for rolling 3 times.

Rolling 5 times is actually much more complicated.
We have to consider 5 cases of successful rolls
i) there are 3 of 1, 2, or 3
ii) there are 2 each of 2 of 1, 2, or 3
iii) there are 2 of 1 of 1, 2, or 3, and 1 of 4-8
iv) there are 1 each of 1, 2, and 3, and there are 2 of 1 of 4-8
v) 5 different numbers with 1,2, and 3 included

I'll state how many of each occur. I leave it to you to work through these and digest them.

i) \(\displaystyle \dbinom{3}{1}\dbinom{5}{3}2!\)

ii) \(\displaystyle \dbinom{3}{2}\dbinom{5}{2}\dbinom{3}{2}\)

iii) \(\displaystyle \dbinom{3}{1}\dbinom{5}{1}\dbinom{5}{2}3!\)

iv) \(\displaystyle \dbinom{5}{1}\dbinom{5}{2}3!\)

v) \(\displaystyle \dbinom{5}{2}5!\)

Summing these yields an overall probability of getting a roll string containing 1, 2, and 3 as

\(\displaystyle p = \dfrac{1275}{16384}\)

 

[George.W89]

Thanks Romsek! Through practical experience that is what I estimated the probability to be. But, I would love to know the method you used as well. Also, I'm not familiar with the notation you used for the numbers in brackets?

 

[JeffM]

Thanks Romsek! Through practical experience that is what I estimated the probability to be. But, I would love to know the method you used as well. Also, I'm not familiar with the notation you used for the numbers in brackets?

[MATH]\dbinom{n}{k} = \dfrac{n!}{k! * (n - k)!}, \text {where }[/MATH]
[MATH]n \ge k \ge 0,\ j = 0 \implies j! = 1, \text { and } j > 0 \implies j! = j * (j - 1)!.[/MATH]
[MATH]\dbinom{n}{k}[/MATH] is called the binomial coefficient and tells you how

many ways you can select k distinct items out of n distinct items without regard to the order of the k items.

How many ways can you select three distinct letters from the letters A, B, C, D, and E.

[MATH]\dbinom{5}{3} = \dfrac{5!}{3! * (5 - 3)!} = \dfrac{5 * 4 * 3!}{3! * 2!} = \dfrac{5 * 4}{2} = 10.[/MATH]
Let's verify that there are 10 combinations (remember that order does not count, meaning ABC, ACB, BAC, BCA, CAB, and CBA count as one combination).

ABC
ABD
ABE
ACD
ACE
ADE
BCD
BCE
BED
CDE

 

[Steven G]

Below is how I would work your problem. Let's just cover something 1st.
In how many ways can you arrange the letters ABCDE. The answer is 5! . How about AAABB? The answer is \(\displaystyle \frac{5!}{3!2!}\). How about AABBC? The answer is \(\displaystyle \frac{5!}{2!2!}\)

Now let's look at cases Romsek's listed as this is the way to do it.

i) there are 3 of 1, 2, or 3 (and 1 of each for the other two numbers)
An example would be 22213---This can be arranged in \(\displaystyle \frac{5!}{3!}\) ways. But we can choose 3-1's, 3-2's or 3-3's (3 ways)
The total numbers of ways is \(\displaystyle {3 \choose 1} \frac{5!}{3!}\)

ii) there are 2 each of 2 of 1, 2, or 3
An example would be 11223---This can be arranged in \(\displaystyle \frac{5!}{2!2!}\) ways. Again we can choose one number from the three to be the single = # of ways to pick two from the three to be double.
The total numbers of ways is \(\displaystyle {3 \choose 1} \frac{5!}{2!2!}\).


iii) there are 2 of 1 of 1, 2, or 3, and 1 of 4-8
An example would be 11236---This can be arranged in \(\displaystyle \frac{5!}{2!}\) ways
We can choose one number from the three to be the double and then pick 1 from 5 (4, 5,6,7,8)
The total numbers of ways is \(\displaystyle {3 \choose 1}{5 \choose 1} \frac{5!}{2!}\)


iv) there are 1 each of 1, 2, and 3, and there are 2 of 1 of 4-8
12377---This can be arranged in \(\displaystyle \frac{5!}{2!}\) ways. We can pick the 1 double from 5
The total numbers of ways is \(\displaystyle {5 \choose 1} \frac{5!}{2!}\)


v) 5 different numbers with 1,2, and 3 included
An example is 12357---This can be arranged in \(\displaystyle 5!\) ways
From {4,5,6,7,8} we must pick 2 which can happen in \(\displaystyle {5 \choose 2}ways\)
The total numbers of ways is \(\displaystyle {5 \choose 2}5!\)

Amazingly I got the same sum as Romsek.

 

[George.W89]

Excellent, thank you all! Got everything I needed. Keep up the good work, much appreciated.

 

[Romsek]

Amazingly I got the same sum as Romsek.

Why is it amazing? The correct answer is the correct answer. Or are you just amazed that I was able to get the correct answer?

 

[Steven G]

Why is it amazing? The correct answer is the correct answer. Or are you just amazed that I was able to get the correct answer?

There was so many little mistakes that I could have made.
I'm just joking that It is amazing that little me got the same answer as the the giant Romsek.
Take it as a complement.

 

[Romsek]

meh I'm no giant. I'm just a dumb engineer. You all are the math whizzes.

 

[Steven G]

meh I'm no giant. I'm just a dumb engineer. You all are the math whizzes.

Not from what I see. Are you really an engineer and not a mathematician? I usually get that one correct. Some people on this forum, for example, are excellent in math but other helpers you can see that they are just different. They speak math, they live math, they are math! As Denis would have said--them are the mathematicians. I thought that you fit right into that.

 

[Romsek]

I interpret the problem as rolling an 8 sided die 3 or 5 times and a success is there being at least a 1, 2, and 3 in the roll string.
OP got the correct answer for rolling 3 times.

Rolling 5 times is actually much more complicated.
We have to consider 5 cases of successful rolls
i) there are 3 of 1, 2, or 3
ii) there are 2 each of 2 of 1, 2, or 3
iii) there are 2 of 1 of 1, 2, or 3, and 1 of 4-8
iv) there are 1 each of 1, 2, and 3, and there are 2 of 1 of 4-8
v) 5 different numbers with 1,2, and 3 included

I'll state how many of each occur. I leave it to you to work through these and digest them.

i) \(\displaystyle \dbinom{3}{1}\dbinom{5}{3}2!\)

ii) \(\displaystyle \dbinom{3}{2}\dbinom{5}{2}\dbinom{3}{2}\)

iii) \(\displaystyle \dbinom{3}{1}\dbinom{5}{1}\dbinom{5}{2}3!\)

iv) \(\displaystyle \dbinom{5}{1}\dbinom{5}{2}3!\)

v) \(\displaystyle \dbinom{5}{2}5!\)

Summing these yields an overall probability of getting a roll string containing 1, 2, and 3 as

\(\displaystyle p = \dfrac{1275}{16384}\)

I guess I can explain a bit of this reasoning.

i)
pick 1,2, or 3 as the triple number. That's the \(\displaystyle \dbinom{3}{1}\)
now place those 3 numbers. That's the \(\displaystyle \dbinom{5}{3}\)
The remaining two numbers have 2! permutations.

ii)
pick the 2 digits of 1,2,3 that will be doubled. \(\displaystyle \dbinom{3}{2}\)
place the first 2 doubled digits. \(\displaystyle \dbinom{5}{2}\)
place the next 2. \(\displaystyle \dbinom{3}{2}\)
the final digit takes the remaining slot

iii)
pick the 1 digit of 1,2,3 that will be doubled. \(\displaystyle \dbinom{3}{1}\)
pick the 1 digit of 4-8. \(\displaystyle \dbinom{5}{1}\)
place the doubled digits. \(\displaystyle \dbinom{5}{2}\)
the remaining 3 numbers have 3! permutations

iv)
pick the digit from 4-8 that will be doubled. \(\displaystyle \dbinom{5}{1}\)
place the doubled digits. \(\displaystyle \dbinom{5}{2}\)
the remaining digits, 1,2,3, have 3! permutations

v)
pick the 2 digits from 4-8. \(\displaystyle \dbinom{5}{2}\)
the 5 unique digits have 5! permutations

As you see I look at things a bit differently from Jomo but both are equivalent.
Actually I kinda like the way he does it.

 

[Romsek]

Not from what I see. Are you really an engineer and not a mathematician? I usually get that one correct. Some people on this forum, for example, are excellent in math but other helpers you can see that they are just different. They speak math, they live math, they are math! As Denis would have said--them are the mathematicians. I thought that you fit right into that.

MSEE, though I did also get a BS Math. I have just enough math to know that I'd make a lousy mathematician.
I was never able to get through graduate level analysis, ala Royden. Just too much rigor!

 

[Steven G]

MSEE, though I did also get a BS Math. I have just enough math to know that I'd make a lousy mathematician.
I was never able to get through graduate level analysis, ala Royden. Just too much rigor!

Royden was so much fun. Some people are just not cut out for this stuff (not necessarily you). My wife has a PhD in Biology and refuses to listen to me when I try to talk math with her. Math is something that you just have to love. My mathematician friend defines a mathematician as someone who has a love for the subject. I am much less generous and think that a mathematician is someone who does mathematical research.