Quote: Originally posted by Jake649 on May 02, 2004
Quote: Originally posted by prob987 on May 02, 2004
At the moment just before the first white ball is drawn you will have in the non-repeating set, a probability of 10 in 53 of having one number in either ticket. In the second set, sharing one number, you will have a probability of only 9 in 53 of having one number in either ticket. This means that you have an 18.9% of remaining in the game in the first case, and a 17.0% chance of remaining in the game in the second case. Clearly, before the first ball is picked, the non-repeating set is superior to the repeating set, by almost 2%, a small but significant number.
The repeating set has the advantage that if the first white ball matches the repeating number it becomes much better than than the non-repeating set. This advantage offsets the previously mentioned 2% advantage thereby making the two sets equivalent in value before any balls are drawn.
If you analyze this mathematically, you will see that if you are careful, and buy more than one ticket, it is impossible to improve your odds by the amount that represents the probability of winning on a single ticket divided by the number of tickets purchased.
I strongly disagree. Odds reduction can easily proven using a lottery with a small set of possible outcomes. This proof extrapolates to lotteries with large sets of possible outcomes.
You simply cannot improve your odds more than that, however. In order to be certain of winning the powerball, you need to purchase 120,526,770 tickets, being careful to select different numbers for each ticket. It turns out that if you buy 120,526,770 quick picks, however, you will only have a 73% chance of winning, since many numbers will be duplicates.
My calculations show that it is a 66% probability of winning.
Good luck,
Jake
In the first case, assuming only one trial, and that the other numbers in the set are all different (which is NOT the case in most wheeling systems as I understand them). The obvious extrapolation to the absurd case would be the case where two identical tickets were purchased. Your odds of winning the jackpot in that case with two identical tickets would be exactly equal to buying one ticket, and not the odds divided by 2. If one is playing eleven tickets and not two, one can choose numbers in such a way that one has reduced the odds from a combin(53,5) situation to a combin(52,4) situation with 100% certainty. One does this simply be choosing to play every number in the field at least once. In any case, you must go through the first trial (the first ball selected) in every draw, you cannot avoid it. You have stated a condition under which you have the first outcome known, but this is NOT what is obtained.
I would very much like to see this proof sketched out or a reference to it in a text, if we are indeed discussing the same thing. It is contrary to my own calculations, but if I've made an error, I would certainly like to know about it. If it were possible to reduce the odds by more than the integer equal to the number of tickets purchased, it would be theoretically possible to achieve a certainty of a win by playing less than the total number of tickets possible. That is counter-intuitive.
In case #3, I relied on my memory of the figure for the Poisson distribution for a number of tickets and for some reason had the number 27% "holes" in my head. The correct value is of course, 37% and I am wrong and you are right. The point was only that some strategies are better than others; so the basic idea remains correct.