One for the Mathematics boffins!

UnsungHeroGuitars

UnsungHeroGuitars

Fractal Fanatic
One for the Mathematicians!

Apologies for the rather left-of-field nature to my question, but I figure that there'd be some brains around here that could clarify this for me :encouragement:

I'm trying to wrap my head around this and whilst it may appear pretty simplistic, I feel there might be more to it than first meets the eye...

  • Imagine lottery balls number 1 - 49.
  • They are randomly drawn and their face values added up.
  • Is it possible to define two (or three) ranges of totals that would each stand an equal chance of occurring..?

By that I mean, given that the combined face value of all the 49 balls adds up to 1225, it could be considered that the 'average' value of each ball would therefore be 25 (1225 / 49 = 25) agreed..?
Assuming that six balls are drawn, is it then correct to state that the 'mid-way point' is 150, given that 6 x 25 = 150..?

Is it then correct to assume that a collective total below 150 has a perfectly equal chance of occurring to one above 150, or does 'probabilty' start to enter the fray..?

Expanding this one step further, is it possible to determine 3 bands of totals that also stand an identically equal chance of occurring in a random draw of 49 numbers..?

I look forward to your thoughts on this! :D
 
Last edited:
Hi Clive
I will try to answer to the best of my understanding.
The first question I would ask is the ball you pull first put back into the lottery, because that will determine the probability of the next pull. Think about a lottery of 5 balls, if you pull a 5 on the first pull, you have only 1-4 left to go instead of 1-5. Fewer combinations.
Second, you need to think of all the combinations of 6 balls (assuming no repeats). To get a total of 21, you must pull 1, 2, 3, 4, 5, and 6, To get 279, you need 49. 48, 47, 46, 45, and 44. These are the only combinations that give these totals. There are many more combinations to give the "average" total of 150, so in the end you will come out with a bell curve. The "average" is much more probable than either extreme.
It is equally probable that the total will be above or below average, but the "middle" band of 3 "bands" will be much more probable than either of the other 2.
Does this help?
 
Thanks very much guys! :D

As I suspected, once you delve further into this, it does indeed get pretty complex, so trying to explain it in simple terms to others will likely leave everyone going
blink.gif


I think I have an easier way to achieve what I'm after, so I think I'll head down that route instead :)
 
Let me try again, Clive, after rereading your OP. To simplify, I am thinking of the classic 2-dice bell curve, the possible combinations are:
dice total
1,1 2 2,1 3 3,1 4 4,1 5 5,1 6 6,1 7
1,2 3 2,2 4 3,2 5 4,2 6 5,2 7 6,2 8
1,3 4 2,3 5 3,3 6 4,3 7 5,3 8 6,3 9
1,4 5 2,4 6 3,4 7 4,4 8 5,4 9 6,4 10
1,5 6 2,5 7 3,5 8 4,5 9 5,5 10 6,5 11
1,6 7 2,6 8 3,6 9 4,6 10 5,6 11 6,6 12
This gives the number of ways to make the totals:
Total: 2 3 4 5 6 7 8 9 10 11 12
Combinations: 1 2 3 4 5 6 5 4 3 2 1
whew. So, to answer what i think is your question, can you divide the combinations into three parts that each have an equal probability of occurring? In the dice table there are 36 possible combinations, and so you would want 12 in each band. The middle band would include the 7 total with 6 combinations plus part of the 6 total and part of the 8 total (actually 3 combinations from each). The lower band would be all the totals below the 6 and the upper would be all the totals above the 8, with the remaining fractions of 6 and 8 added back respectively. Expand this concept to your 6-ball question and you should have your answer. How to do this mathematically and not by hand I have no friggin idea. :)
 
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