You and 2 friends each pay 1 dollar and get 3 lottery ticket. You agree to split the jackpot of 100 million dollars which would mean you get 33.33 million.
Is it to your advantage to pay an extra dollar so that there are 4 tickets thereby making your split 50% of the jackpot?
In other words, does the expected value of your winnings change enough to justify the initial outlay of another dollar?
kinda difficult to say if you don't know the expected value and/or odds of a ticket.
If the expected value of a 1.00 ticket is 1.00, then it doesn't matter, any split would be totally break-even. That is regardless of how many people are involved in the winnings sharing or what your portion is.
Originally posted by forkedknight kinda difficult to say if you don't know the expected value and/or odds of a ticket.
If the expected value of a 1.00 ticket is 1.00, then it doesn't matter, any split would be totally break-even. That is regardless of how many people are involved in the winnings sharing or what your portion is.
I don't think the odds matter. You should buy the extra ticket.
The problem is that your friends should buy extra tickets as well, so you would end up buying dozens of tickets.
Originally posted by Thomaster I don't think the odds matter. You should buy the extra ticket.
The problem is that your friends should buy extra tickets as well, so you would end up buying dozens of tickets.
Of course the odds matter! If the expected value on a ticket is 0.70, then you're losing .30 with every additional ticket you buy.
On the other hand, if the expected value is 1.30, then you should continue buying tickets, and so should your friends.
Originally posted by forkedknight Of course the odds matter! If the expected value on a ticket is 0.70, then you're losing .30 with every additional ticket you buy.
On the other hand, if the expected value is 1.30, then you should continue buying tickets, and so should your friends.
you can even just think in the most extreme cases: if your chance of winning is 100% (or even something close to 100), wouldn't you want as large a portion of the share as was fiscally advantageous? (i.e. buy tickets until the $1 you spend on the ticket gives you less money in return than its cost) and on the opposite end of the spectrum, if you are certain you will all lose (0% chance to win) then you should buy as few tickets as you are allowed, because you will get no return on your investment.
at some point in between 0% and 100% chance to win there is a "switch-over" as to whether it is advantageous to you to buy more tickets, and as such the odds of each ticket to win certainly will matter in deciding what to do.
Originally posted by forkedknight Of course the odds matter! If the expected value on a ticket is 0.70, then you're losing .30 with every additional ticket you buy.
On the other hand, if the expected value is 1.30, then you should continue buying tickets, and so should your friends.
Originally posted by Thomaster hmmm ... yes. I wonder what I did
And like I said before, if the expected value of a ticket is 1.00, then it's even money regardless. Would you rather win 2.00 half the time or 10.00 a tenth of the time?
Originally posted by forkedknight And like I said before, if the expected value of a ticket is 1.00, then it's even money regardless. Would you rather win 2.00 half the time or 10.00 a tenth of the time?
I can give you a game where your expected value is infinite and yet you'd be pressed to find someone willing to pay even 20 dollars to be able to play it.
Originally posted by Aetherael you can even just think in the most extreme cases: if your chance of winning is 100% (or even something close to 100), wouldn't you want as large a portion of the share as was fiscally advantageous? (i.e. buy tickets until the $1 you spend on the ticket gives you less money in return than its cost) and on the opposite end of the spectrum, if you are certa ...[text shortened]... ets, and as such the odds of each ticket to win certainly will matter in deciding what to do.
After some calculations I got that point at a chance of 1,444*10^-13
So if there are less than ~6,925 billion possibilities, you should buy the other ticket.
Originally posted by Thomaster After some calculations I got that point at a chance of 1,444*10^-13
So if there are less than ~6,925 billion possibilities, you should buy the other ticket.
there are only about 14 million combinations in a choose 6 out of 49 lottery
More 6/49 lotto fun
20 Feb '09 22:11
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