Possibilty of Playing Every Numerical Combination to Guarantee Win?

Question

In the state of CT, there exists Powerball for which six numbers must be chosen out of a pool of 75 numbers. No two numbers can be chosen twice. Is it possible to play every numerical combination to gurantee a wim of more than 11 million dollars.

Answer 1

Answer...it is possible to play every numerical combination to gurantee a wim of more than 11 million dollars.

However, the cost of playing every numerical combination will far exceed the 11 million dollar payoff.

In other words, if I said that I would give you $50 if you correctly picked a number between 1 and 100 that I was thinking of, but it cost $1 a guess, you could just give me $100 and guess every number to win $50. But you'd have trouble convincing me to feel guilty about taking your money.

Answer 2

According to the lottery web site, the odds of picking all six numbers are 1 in 146,107,968. So it would cost more than 146 million dollars to win 11 million dollars. The winnings would actually be higher since a person can also win the combinations with less numbers.

Answer 3

I think others have already answered your question, but you might be interested to know that it has already been acted on in the Virginia state lottery. A few years ago there had been no jackpot for some time and the accumulated value of the jackpot actually exceeded the total cost of all the tickets required to cover all the possibilities. A consortium of Australian investors bought up all the tickets, or almost all, and won the jackpot. They were rather sneaky about this and bought the tickets without anyone noticing. Some people got really upset and thought the Australians should be denied the payoff, but they got their money.

Answer 4

The number of tickets you would have to buy is:
75*74*73*72*71*70 =
144,978,876,000.

So it would cost you 145 billion tickets to win 11 million dollars

Answer 5

Yeah, but you'd have to buy 145 million tickets.