Hello, The lottery is currently at $586 Million. I heard today the odds of winning are 1 in 259,000,000. (It used to be lower, but since increasing the numbers you can select from to 75 now, it has increased dramatically). With the number so high, and expectations it could hit $1 Billion eek if unclaimed, couldn't a billionaire theoretically go in and purchase 259 million lottery tickets (one of each number possibility) to guarantee the win? Based on the odds and historical data, it's highly likely there will only be one winner. Therefore, they could gain a quick $25M-$225M assuming you're the only winner. At $586 million, the expected lump sum is $416 million pre-tax, and $282 million post-tax. At $1 billion, the expected lump sum is $710 million pre-tax, and $483 million post-tax. You'd make $224 million overnight. Literally. Yes?
If you think 259 million lottery tix + 1 in 259mill odds = guaranteed win, you know nothing about probabilities
Several problems with this: 1. Labor involved in buying 259 million tickets in less than a week 2. Splitting the prize could turn it into a huge loss 3. If you have the money to do this, there are much safer investments without the massive effort this would take
You're right. I stayed up all night and haven't slept much. Essentially, you're increasingly the pool of purchased tickets, and while the combinations of winning are 1 in 259 million (and you could guarantee you'd win), you've diluted the pool so dramatically that you're guaranteeing a second winner. Nevermind!
There's a math formula. It is based on a multiplication of available numbers. It's not what you said. Pick five from 1-75, then one from 1-15: 75 (available in the first number) * 74 (available in the second, since one was already taken) * 73 (so on) * 72 * 71 * 15 (pick one, can be repeated, since it's a red ball) = 31,066,902,000 possible combinations You are forgetting that within those incorrect numbers for the jackpot prize, you will also win the lower 5 numbers, 4 numbers, 3 numbers, and RED BALL ONLY. Those don't count towards the jackpot probability, but you're sure to be guaranteed to make some of that money back. I've hit 2 a few times. Also, I've gotten the numbers correct for two a few times. Numbers hit and payoffs How to play Mega Millions. GOOD LUCK!
P.S. Answer to my question is theoretically yes you could guarantee you win, but possibly at tens or hundreds of millions of dollars in losses. ;-) Carry on.
This is very, very interesting though that I never thought about... You'd also guarantee every win for every lower category as well...
"The chance someone will match one of the 259 million possible number combinations that could land a jackpot..." http://www.chicagotribune.com/news/chi-mega-millions-jackpot-20131217,0,4732450.story So there are only 259 million combinations... so you'd win all of them. Every one though...
No they aren't. The odds are 1 out of total number of potential combinations. There are just under 259 million possible winning number combinations. That's for the grand prize. The odds are not based on anyone else buying tickets. The odds are not factoring in anything like auto picks, etc. and it is only accounting for a single drawing, so no probability issues to deal with about certain ones being more probably because of past drawings or anything like that. Your pick represents one part of the odds. 1 The total number of possible winnings numbers represents the other part: 258,890.850. If every person in America bought a ticket, it would not change the probability of the numbers you pick being the winning combination. If every person in American but you refused to play, again, it would not impact your odds of having the winning combination. Now of course, the probability of being a sole winner is vastly different.
That's incorrect. gatsby shows you what the possible winning combinations for the grand prize there are.
You aren't doing the math right. I'm not sure how to write an equation on the BBS, but you can find the equation on various lotto websites. Here's how you do it in excel. =COMBIN(75,5) this will tell you the possible number of 5 random combinations can be selected from a field of 75 total numbers. The 6th number is the mega number. There are only 15 ways it can be drawn obviously, so you take the number above and multiply it by 15. The format is 5/75 and 1/15
No, he is incorrect. You aren't doing the math right. But it's cool, it's a complicated system. I won't lie and say I knew it off the top of my head. I had this conversation with someone else and worked on it.
