Confused about Lottery Probability!?

Question

Ok, logically, I know that any ball is as likely to come up as any other. BUT, what is the probability of 6 evens coming out? Is it less than a mixture of odd and even? Or is it exactly the same? I have taken notice of the lottery result recently, and it is NEVER all odd or all even. I also looked at a list of winning numbers, and noticed that in nearly every winning combination, the first number (when put in ascending order) was below 10, and the last above 40. How can this be? Of course there are exceptions, but it seems that most of the time, the numbers are evenly spread. Back to the original query, I am certain that my parents are less likely to win the lottery by having all their numbers even. Is there any mathematical explanation for this, or am I wrong?

Answer 1

probability of 6 evens will be lower than a mixture of odd and even appearing. However, this may seem contradictory, but your parents have the same chance as everyone else if they choose all even, or all odd for that matter, or even, or even: 1,2,3,4,5,6 because the balls are randomly chosen.

Answer 2

If there are 48 balls (sorry, but entering the Lottery is just a voluntary tax paid by stupid people, so I don't know the exact rules because I've never bought a ticket) then the chances of getting 6 even numbers is



24.23.22.21.20.19 / 48.47.46.45.44.43 = 0.0109. So you would only expect it to happen 1 time in 100, and it would be no big surprise if it didn't happen for one time in several hundred.



(Explanation: there are 24 choices for the first even number, then 23 for the second one, etc, compared with 48 choices for any first number, 47 for the second one, etc).



But your parents are just as likely to win with ONE PARTICULAR set of all-even numbers as with any other set of numbers. If they were betting on ANY set of even numbers coming up, compared with ANY mixed set, that would be a very bad bet, but for a fixed choice of all 6 numbers it makes no difference. They are just as likely to win choosing 1 2 3 4 5 6 every week!



About the smallest and largest numbers, the point is the smallest number can be any one of the 6 balls so it is naturally biased to being a small number.



Take a simpler example of only 5 balls and you draw two. The 20 possible options are

12 13 14 15

21 23 24 25

31 32 34 35

41 42 43 45

51 52 53 54



There are 8 out of 20 with the smallest number 1

6 out of 20 with the smallest number 2

4 out of 20 with the smallest number 3

and only 2 out of 20 with the smallest number 4.

and none with the smallest number 5, of course



For 3 balls out of 5 there are 60 possible draws and 36 of them (more than half) have the smallest number 1.

123 124 125

132 134 135

142 143 145

152 153 154

213 214 215

231 234 235

241 243 245

251 253 254

312 314 315

321 324 325

341 342 345

351 352 354

412 413 415

421 423 425

431 432 435

451 452 453

512 513 514

521 523 524

531 532 534

541 542 543



The same idea applies to the smallest of 6 balls from 48, but counting the possibilities is a bit harder to do.



The "even distribution" thing is also easy to explain. Suppose the first 3 balls were 5, 40, and 10 say. There is a large range of 28 numbers between the 10 and the 40, compared with the other 17 split into 3 small ranges. So the most likely place for the 4th number to be is somewhere in the largest gap. If the 4th number doesn't fall in the gap, there is the same high chance that the 5th or 6th number wll. So on average, all the gaps between the numbers will usually finish up about the same size.

Answer 3

this is a permutation problem

ok lets say that there are 60 balls total and 6 show up in the lottery the odds of the first ball being even is 1/2

the odds of the second ball being even is 29/59

the third ball 28/ 58

the fourth 27 / 57

the fifth 26 /56

the sixth 25 / 55

to find the probability that they are all even we find the product of all probabilities



(1/2)*(29/59) * (28/58) * (27/57) * (26/56) * (25/55) = 35.88% roughly



so it not likely but its also certainly not a long shot either.

Answer 4

The probability of 6 even numbers is less than 1/2^6 (less because you use up an even number after each ball is selected.)

However the probability even-odd-even-odd-even-odd is also less that 1/2^6 (but very slightly higher).

So it is strange, but any choices of numbers for lottery is equally likely to occur even if you pick the numbers to be consecutive numbers.

Answer 5

It is a Bernoulli Trials problem. Assume an experiment has probability for success p, and probability for failure 1-p. Then the probability of exactly k successes out of n trials is given by C(n,k) * (p)^k * (1-p)^(n-k) where C(n,k) = n! / ( (n-k)! * k! ) is the binomial coefficient. (Recall here that 0! = 1) The only way that you will not win at least one prize, is if you lose all 5 times. The probability of losing is 1 - 0.11 = 0.89. The probability of losing 5 out of 5 times is C(5,0) * (0.11)^0 * (0.89)^5, where C(5,5) is the binomial coefficient. This turns out to be 0.56. Thus the probability of winning at least once is 1 - 0.56 = 0.44.