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I have six boxes each with fithteen randomly placed numbers from 1-90.a number is nominated ( say no 30) which happens to be in box 4 ( say ) what are the odds that no 30 will be the last no to be called ( fithteenth ) in box no 4 before any of the other boxes are completed.
If I interpret your question correctly, you are asking what are the odds that a particular number will be the last one pulled from the first box to be totally emptied, right? This is really pretty simple - that particular number could be any one of the 90, so the odds are 1 in 90.
This is the wrong answer.
So what do you think the answer is?
If you know it's the wrong answer then please tell us how you know. You didn't indicate how the numbers are selected, and so I made an assumption that your phrase "what are the odds that no 30 will be the last no to be called" indicated that the numbers from 1-90 were being randomly selected - is that assumption incorrect?Quote:
Originally Posted by duggieboy
Please read the question again.
duggieboy: Please read my response again.
to recap.
to satisfy question- no 4 box must be filled first silmultaneously with no 30 being called. Example-should after 84 calls each box requires 1 number to complete,unless no 4 requires no 30 the exercise is abandoned. Similarly if no 30 is called before 14 calls no 4 box cannot meet the requirements. All numbers randomly called and placed.
duggieboy - I know English is not your first language but your description of the problem is still confused. Please tell me if this is the process that you are trying to describe:
1. The numbers 1 -90 are written on slips of paper which are randomly distributed into 6 boxes, with 15 slips of paper per box.
2. A number from 1-90 is randomly generated, and the slip of paper with that number is removed from its box.
3. A second number is randomly selected from the remaining 89 numbers and its slip of paper is removed from its box.
4. This process continues until the number 30 happens to be drawn.
Your question is: what are the odds that the slip with number 30 is in box 4, that it is the last slip of paper to be removed from box 4, and that this happens before any of the other 5 boxes are empty - do I have that right? If so, then the odds can be calculated as follows:
The probability that number 30 is places in box 4 is 1/6.
The probability that box 4 is the first t be emptied is 1/6.
The probability that number 30 is the 15th number to be pulled from its box is 1/15.
Consequently the odds that number 30 is the 15th number to be pulled from box 4, and that this occurs before any of the other boxes are emptied, are (1/6)(1/6)(1/15) = 1/540.
Your understanding of the original question is correct. Your answer is not. Because no4 box has a handicap the other five boxes have an increased chance of being first.also the odds are entirely dependent on the number of calls made-- your answer does not specify how many. The right answer is on several websites. p.s english speaker 80 years.
Oh, so this wasn't a real question you had but rather a test to see if we get the right answer? OK, I'll bite - please provide a link to one of these web sites you mention that has the right answer.
I don't understand what you mean by "no 4 box has a handicap" - what handicap? And the answer is not dependent on the number of calls made, unless the process I described is not what you have in mind. The number of calls will vary from trial to trial, but what I calculated are the overall odds before you start the game, and hence before you know how many calls are going to be made.
The question is genuine. It relates to the poor cash prize offered by a promoter.--- I would have thought that as box 4 can be eliminated at any call it would be classed as a restriction-penalty-handicap. Please refer to lotto/bingo odds sites. They don't answer my question but their figures are higher than yours even without a " handicap ".
But ANY box can be eliminated the same way, right? So box 4 is no different than any of the others, it's just that it's the box that you are rooting for.Quote:
Originally Posted by duggieboy
The game you have described is not any Lotto or Bingo game that I am aware of. If you could give me a specific site to look at it would be helpful.Quote:
please refer to lotto/bingo odds sites. They don't answer my question but their figures are higher than yours even without a " handicap ".
Try wizardofodds.com/bingo see tables. Are they relative to my question ?
I don't see any games described in that web site that are equivalent to what you have been asking about. In Bingo the goal is to get numbers called to complete a column, or a row, or a diagonal on your card. There is no requirement that a particular number (30 in your example) be called, or that all 15 numbers of a column be called before any any other columns are finished, or that the specific number (30) be the last number called of the 15. Also in most Bingo games the numbers are not randomly distributed across columns as you described - if number 30 is present on your card it is always in the second column.Quote:
Originally Posted by duggieboy
Perhaps there's another version that you are thinking about - there are a lot of pages on that web site so if you can give me the specific page URL that has a probability table for the game you are thinking about that would be helpful.
**EDIT** What I described above is US Bingo, but I see that there is a British version which sounds more similar to what you described, but still isn't a very good match. In British Bingo each player is given 6 tickets, with 15 numbers per ticket arranged in 3 rows (5 per row), such that all numbers 1-90 are represented across the 6 tickets. Numbers are called out randomly, and you win if you get a row of 5 complete, or two rows, or all 3 rows of a ticket (called a "full house"). So it sounds like you're thinking of the odds of getting a full house. However, what you asked about is the odds of (a) number 30 appearing on the 4th of the 6 tickets, and (b) number 30 being the last number called on the 4th ticket, and (c) the 4th ticket being filled before any of the other 5 that you have are filled. But I still don't see how this set of circumstances relates to British Bingo. Also - the odds that they list for 90-number Bingo are regarding being able to cover all 15 numbers on a card given N numbers drawn, which is an entirely different question - see: http://wizardofodds.com/games/bingo/...-number-bingo/
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