Bingo permutations combinations

  • Thread starter Bassalisk
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  • #1
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Homework Statement


So 5 numbers are drawn from pile of 10 balls. After the draw, the ball is put back in the pile. So you always have 10 choices for the ball.After drawing 5 balls, another one is drawn for the sixth number.

What is the chance of getting the winning number-combination 848235.

Homework Equations


The Attempt at a Solution


So I have 105 possible solutions for the first part. Now I need to find the chance, to get a number that is a subset of (2,3,4,5,8,8).

I have these six numbers. I need to know how many 5-number combinations, can be made out of these 6 numbers.

I used the permutation formula and I got 360. I don't think that is right.

How can I know, how many 5-digit numbers can I make from these 6 numbers, given that one is repeating.

Example:

23458
23588

etc.

After I know how many combinations I have for those 5 digit numbers, I can make the probability of getting a combination of those 5 numbers in the first part.

For the second draw, I am still thinking. Maybe because, we again have 10 balls, and 5 different numbers to be drawn, chance is 1/2. So I would multiply the final result(probability for 5 digits) with 1/2.
 

Answers and Replies

  • #2
vela
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Why are you considering the sixth draw separately? How is it different from the other five?
 
  • #3
948
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Why are you considering the sixth draw separately? How is it different from the other five?
Hmmm that is the point. You have to do it that way. I mean you don't have to, but the exercise wants you to.
 
  • #4
948
2


But then again, you are right. Then the chance would be 1/1000000 ?
 
  • #5
1,033
1


http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition

This link will show you how to find a combination with repetition. If I understand your question properly, each ball is replaced after being picked, and the order in which they are picked doesn't matter, so I don't see why 10^5 is correct. This will lead to the result for the first question.

As for the second question, you want to know how many different ways you can arrange those 6 numbers into distinct numbers. If they were all different you would have 6 * 5 * 4 * 3 * 2 ways right? So how would you augment that formula to take into account that there are two 8's
 
  • #6
948
2


http://en.wikipedia.org/wiki/Combination#Number_of_combinations_with_repetition

This link will show you how to find a combination with repetition. If I understand your question properly, each ball is replaced after being picked, and the order in which they are picked doesn't matter, so I don't see why 10^5 is correct. This will lead to the result for the first question.

As for the second question, you want to know how many different ways you can arrange those 6 numbers into distinct numbers. If they were all different you would have 6 * 5 * 4 * 3 * 2 ways right? So how would you augment that formula to take into account that there are two 8's
@ First part: how is that not 10^5.

I am picking 5 balls.

I have 10 choices on the first draw. I pick the ball and I put it back. So on the second draw I would have again 10 choices to pick.

So thats 10*10.

Analogous for other 3

so that is 10*10*10*10*10.

@ Second part: We would divide by 2! ? because we have 2 8's repeating. That is from permutations with repetition I think.


Thing is, is that correct?

Will i have then numbers with shape like this:

2,3,4,5,8,8

23458
23488
23588
82345

etc. Is all that taken into the account?

But I don't want numbers like

22222
33333
44444

etc.

Are those taken into the account?
 
  • #7
1,033
1


@ First part: how is that not 10^5.

I am picking 5 balls.

I have 10 choices on the first draw. I pick the ball and I put it back. So on the second draw I would have again 10 choices to pick.

So thats 10*10.

Analogous for other 3

so that is 10*10*10*10*10.
First, you are picking 6 numbers are you not? 10 * 10 * 10 * 10 * 10 * 10 suggests that you need to pick numbers in a specific order. As far as i know, in bingo it doesn't matter what order your numbers get called out. On the first pick, you have the option of a 2 3 4 5 or 8. then on the second selection you have two scenarios. If you selected an 8 first, then you can once again pick a 2 3 4 5 or 8. if you selected a a non 8 (lets say a 2), then you can only pick a 3 4 5 or 8 on your second selection. And these "if" statements keep branching off until you have no more picks. The process that I started to define is compacted into the link that I showed you.
 
  • #8
1,033
1


@ Second part: We would divide by 2! ? because we have 2 8's repeating. That is from permutations with repetition I think.


Thing is, is that correct?

Will i have then numbers with shape like this:

2,3,4,5,8,8

23458
23488
23588
82345

etc. Is all that taken into the account?

But I don't want numbers like

22222
33333
44444

etc.

Are those taken into the account?
You've raised a key point. You would divide by 2! in all cases except the case that you only picked one 8, in which case you wouldn't need to divide at all.

And numbers like 22222 and 33333 suggest that you've done replacements after each number, which is not at all what you want to do. If you have 6 choices, it will never be 6 * 6 * 6 * 6 * 6 without replacement, it will be 6 * 5 * 4 * 3 * 2
 
  • #9
948
2


First, you are picking 6 numbers are you not? 10 * 10 * 10 * 10 * 10 * 10 suggests that you need to pick numbers in a specific order. As far as i know, in bingo it doesn't matter what order your numbers get called out. On the first pick, you have the option of a 2 3 4 5 or 8. then on the second selection you have two scenarios. If you selected an 8 first, then you can once again pick a 2 3 4 5 or 8. if you selected a a non 8 (lets say a 2), then you can only pick a 3 4 5 or 8 on your second selection. And these "if" statements keep branching off until you have no more picks. The process that I started to define is compacted into the link that I showed you.
Yes I just realised that, when I was fixing my broken pipe in bathroom. (By day I am student of electrical engineering, by night I am everything I have to be :D ).

Order doesn't matter.

Give me a few hours to work out this, I will come back with answers I hope.
 
  • #10
1,033
1


I have an answer that I am about 70% confident in so I look forward to discussing your answer later on.
 
  • #11
948
2


I have an answer that I am about 70% confident in so I look forward to discussing your answer later on.
So... I understood what you said, and what is your point.

I got the following.

I understood that the order doesn't matter. I misplaced that part. So I get way less combinations possible.

I got 2002. With 10 balls, for 5 picks.

Then I calculated, that out of those 6 numbers, that are the winning combination {2,3,4,5,8,8} I got 360 possible ways of arranging, to get 5 numbers.

So chance of getting one of those numbers is:

360/2002=0,1798=17,98 %.

Now, I am pulling out one more ball to get that winning combination.

Only numbers that are good for me are {2,3,4,5,8}. Only those numbers, can fill 5 digit number, to get that winning 6 digit number.

Taken in the account, that one 8 HAS to be in that 5 digit number, my chances are 5/10 to get that ball.

Multiply that by 0,1798 => 0,089 = 8,9%

Did I get it?

If I did, I can start on making simulations in matlab. That was the main point of this exercise, but I have to have a good analytical correct solution.
 
  • #12
1,033
1


So... I understood what you said, and what is your point.

I got the following.

I understood that the order doesn't matter. I misplaced that part. So I get way less combinations possible.

I got 2002. With 10 balls, for 5 picks.
Can you show me how you got 2002, and can you tell me why you only have 5 picks for a winning combination that requires 6?
 
  • #13
1,033
1


Then I calculated, that out of those 6 numbers, that are the winning combination {2,3,4,5,8,8} I got 360 possible ways of arranging, to get 5 numbers.
Can you show me how you got this number? If you have 6 numbers, and the order matters, then you have two possibilities to consider here. You have the case in which there are two 8s, and the case in which there is only 1.
 
  • #14
948
2


Can you show me how you got 2002, and can you tell me why you only have 5 picks for a winning combination that requires 6?
Im sorry for the delay I had a guest.

I went by this logic.

Find the chance of getting a number out of those 360 possible numbers.

I am picking first 5 numbers. Sixth will come later.

I have

[itex] \binom {10+5-1}{5}=2002[/itex]

But I am probably getting this wrong. Give me a sec.
 
  • #15
1,033
1


Ah yes, that looks correct. I did the calculation wrong initially.
 
  • #16
948
2


Ah yes, that looks correct. I did the calculation wrong initially.
But the question is:

Are we sure how do we get those chances to get those 5 digit number out of 6 possible.

For our case:

Getting

23458 and 85432

is the same.

So are we sure about that

6*5*4*3*2/2*1

We will get permutations like that.

And 23458 and 85432 are NOT the same numbers.


I am thinking of getting a result that tells you, How many possible numbers can you make from
2 3 4 5 8 8

when order DOESN'T matter.

After I get that number, I will divide that, by that number 2002.

Now I will have a chance, that tells us what proportion of 5 digit numbers, made from those 6 numbers, are appearing in those set of combinations of 2002.

After that, the sixth number is 5/10 chance to get. Multiply that by the result that I got before, and its done.

You follow me?
 
  • #17
1,086
2


Bassalisk, I think you should post the question in full, because from what you've posted it is not clear what the rules of the game are. At least to me it isn't, and the answer will fully depend on just that.
 
  • #18
1,033
1


Let me walk you through my logic about part B of this question. We have {2,3,4,5,8,8} and the question asks what are the chances of getting 848235 if we pick each number randomly.

So we need the first number to be an 8, so the chances of that are 2 in 6. The next number has to be a 4, so thats 1 in 5. The third one has to be an 8, so thats 1 in 4. The fourth one has to be a 2, so thats 1 in 3. The fifth one has to be a 3, so thats 1 in 2. And then there is only one left. So if you calculate out that though process you should get the probability of picking 848235 when you have {2,3,4,5,8,8}.

If you don't already have {2,3,4,5,8,8}, and you have just 10 balls which you need to pick 6 from with replacement, then you need to extend part A to part B of this question.
 
  • #19
1,033
1


You follow me?
I actually agree with Ryker that the bounds of this problem aren't entirely clear to me. If you could come up with a concise statement of what you hope to calculate that would help.
 
  • #20
948
2


Bassalisk, I think you should post the question in full, because from what you've posted it is not clear what the rules of the game are. At least to me it isn't, and the answer will fully depend on just that.
You are playing Bingo game. As in any Bingo game, you are winner if you get a certain combination.

You have 10 balls in total. You are drawing 5 balls. After each draw, you are putting the ball back into the pile. After that you are drawing the sixth ball.(like in the Bingo, Bingo+ or something)

Find the chance if the winning combination is the 848235.

My way of thinking is complicated. Lets just say we are drawing 6 balls, what is the chance of getting winning combination.
 
  • #21
948
2


I was trying to solve the problem in the way I presented because, It was totally my view of it.
 
  • #22
948
2


Are you guys familiar with MATLAB? Can I present to you my simulation, maybe it will help us get the result.
 
  • #23
1,033
1


Well because it is a combination, it doesn't matter what order you are picking it in does it?

So 848235 = 882345 = 234588 ...etc?

You will then have 15 choose 6, as opposed to the 14 choose 5 that we previously discussed. This gives 5005 possible ways to achieve that combination, when there are 10^6 possibilities.
 
  • #24
948
2


Well because it is a combination, it doesn't matter what order you are picking it in does it?

So 848235 = 882345 = 234588 ...etc?

You will then have 15 choose 6, as opposed to the 14 choose 5 that we previously discussed. This gives 5005 possible ways to achieve that combination, when there are 10^6 possibilities.
So we have 5005 possible combinations. (order doesn't matter, with repetition). Our number is unique.

So the chances are 1/5005?

1,998001998001998001998001998002 *10-4


Are you familiar with MATLAB?
 
  • #25
1,033
1


I am not familiar with MATLAB but I don't think it would be too hard for you to explain to me your algorithm.

And no, the chances are 5005 / 1000000. In a more meaningful discussion of the result, the answer is that there are 5005 ways to choose the combination out of 1000000 possibilities
 

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