If I win it I'll host a $100k freeroll for all the forum members for sure. I would definately play some WPT and EPT events. At least for the experience if nothing else. Maybe visit Vegas during the WSOP time and play an event or two there.
I can tell you one thing though. I would quit my job so damn quick it would be a Guiness world record.
Good luck to everyone that bought a ticket. I have mine.
Play poker and golf at the World's best Casino and Golf Locations:
Tunica, Las Vegas, Costa Rica etc. etc.
Enter WSOP main event for the pure chance I amy be able to make Phil Hellmuth freak out
This statement indicates that the statistical distribution of the balls that fall out of the ball dropping doohickey depends on the number of Lotto 6/49 tickets sold. It doesn't get any more rigged than that.
This is actually pretty cool if you think about it. I'm now tempted to go buy a Lotto 6/49 ticket just to feel the raw and awesome power of actually being able to alter the odds of winning the Lotto 6/49 for every single Lotto 6/49 ticket holder.
Technically it's +EV to play (provided the jackpot isn't split, which it almost certainly will be).
It depends on the number of tickets sold.
For example, if exactly 1 ticket is sold, tonight's draw is obviously a +EV game due to the expected share of the jackpot (in the 1 ticket case, $40 million) being greater than $28 million.1
To generalize, if the expected share of the jackpot is greater than $28 million, then Lotto 6/49 is +EV.
For today's draw, let's assume that at least as many tickets will be sold as were for the last $30 million draw. There were about 27 million tickets sold for that draw. To make the math a little easier, assume that 28 million tickets will be sold for tonight's draw. That means that the expected number of jackpot winners is approximately 2. (Using the binomial distribution, with n = 28 million, and p = 1/14 million, the expected number of successful outcomes is n*p = 2.)
Therefore, the expected share of the jackpot is $20 million. (In fact, my guess is that a lot more that 28 million tickets will be sold for tonight's draw.) This is short of the $28 million required to make the game +EV.
A specific Lotto 6/49 draw being +EV is incredibly unlikely, and I doubt that there has ever been a single Lotto 6/49 draw which has been +EV for the player.
To make a bit of a looser (but still correct in principle, IMO) statement, Lotto 6/49 is not, and will never be, a +EV game for the player.
The house edge for Lotto 6/49 is generally around 50%, and usually goes up (meaning, of course, that the draw is worse for the players) for "large jackpot" draws. After we find out how many tickets were sold for tonight's draw, we can then compute the expected share of the jackpot, and hence the house edge. Roughly, if more than 42 million tickets are sold for tonight's draw, the house edge will be greater than 50%.
ScottyZ
1In terms of analysing the house edge of Lotto 6/49 and similar lotteries, the prizes other than the jackpot are negligible.
dude.....I don't think 40 Million would last very long!
Although you could do the Dr. Evil thing......."I raise One Million Dollars....."
And then Tyson would come over the top....."I'm all-in...for about 40 Mill or so. Give or take.."
The house edge for Lotto 6/49 is generally around 50%, and usually goes up (meaning, of course, that the draw is worse for the players) for "large jackpot" draws. After we find out how many tickets were sold for tonight's draw, we can then compute the expected share of the jackpot, and hence the house edge. Roughly, if more than 42 million tickets are sold for tonight's draw, the house edge will be greater than 50%.
This is apparently incorrect.
On closer inspection of the OLGC Lotto 6/49 rules page
"47% of LOTTO 6/49 draw sales is dedicated to the Prize Fund."
and
"Any amount not won in the 6/6, 5/6 + Bonus, or 5/6 prize categories is added to the 6/6 jackpot prize for the next draw."
So, the long run house edge of Lotto 6/49 is 53%. However, the house edge is actually reduced during draws when the previous draw's jackpot prize1 is not won.
Theoretically, after a long string of no jackpots won, Lotto 6/49 could conceivably be a +EV game. That would be the case if the carryover amount was greater than 53% of the current draw's ticket sales, essentially "filling in" the house edge portion of the prize pool.
Is this scenario something to shoot for then? Definitely not. Such a large carryover would indicate that the ex post house edge of the last few draws had been nearly 100%, since no jackpot was handed out. (Note that ex post house edge is a weird animal, and is not a good tool for analysing the a priori expectations of lottery payoffs.)
there was one winning ticket, paying out approximately $54 million. Incidentally, the expected number of jackpot winners from ~50 million tickets sold was about 3.6.
ScottyZ
1Or, I guess, the previous 5/6 + Bonus, or 5/6 prizes.
"The unidentified man bought the ticket at a Mac's convenience store in Camrose. Looking dazed, he retuned to the store late last night to tell the clerk about his good fortune."
Brainy.
Reports indicate that the following discussion ensued:
Store Clerk: May I see?
Unidentified Man: Ho, ho ho! See with your eyes, not with your hands!
Store Clerk: Please! We are all amigos here!
Random Guy in Store: Unidentified Man, I think we can trust the Store Clerk.
Unidentified Man: (hands over the ticket)..... Now give it back!
Store Clerk: Give what back?
There are a total of 13,983,816 different groups of six numbers which could be drawn from the set {1, 2, ... , 49}. To see this we observe that there are 49 possibilities for the first number drawn, following which there are 48 possibilities for the second number, 47 for the third, 46 for the fourth, 45 for the fifth, and 44 for the sixth. If we multiply the numbers 49 x 48 x 47 x 46 x 45 x 44 we get 10,068,347,520. However, each possible group of six numbers (combination) can be drawn in different ways depending on which number in the group was drawn first, which was drawn second, and so on. There are 6 choices for the first, 5 for the second, 4 for the third, 3 for the fourth, 2 for the fifth, and 1 for the sixth. Multiply these numbers out to arrive at 6 x 5 x 4 x 3 x 2 x 1 = 720. We then need to divide 10,068,347,520 by 720 to arrive at the figure 13,983,816 as the number of different groups of six numbers (different picks). Since all numbers are assumed to be equally likely and since the probability of some number being drawn must be one, it follows that each pick of six numbers has a probability of 1/13,983,816 = 0.00000007151. This is roughly the same probability as obtaining 24 heads in succession when flipping a fair coin!
Supposed to be various employees at a gas plant. Some
secretaries, technicians, plant workers, etc. 17 of them
each getting a little over 3 million dollars. (hate to be
one of the ones who didn't buy in) They all have to keep
working now. lmao
PS, some math dude was quoted as saying winning the draw
was equal to flipping a coin 24 times and having them all
come up heads.
A Massachusetts biweekly lottery that was not popular had +EV draws last year because jackpot fever never took hold as the winless streak continued for over a year. Popular lotteries such as 6/49, Super 7 and Powerball have never had an +EV draw.
By the way, lottery retailers usually get up to 1% bonus of prizes won. This is not the case in Alberta so the Mac's owner in Camrose that sold the winning ticket lost out on more than $540,000!
A specific Lotto 6/49 draw being +EV is incredibly unlikely, and I doubt that there has ever been a single Lotto 6/49 draw which has been +EV for the player.
Comments
I can tell you one thing though. I would quit my job so damn quick it would be a Guiness world record.
Good luck to everyone that bought a ticket. I have mine.
Red
Tunica, Las Vegas, Costa Rica etc. etc.
Enter WSOP main event for the pure chance I amy be able to make Phil Hellmuth freak out
Of the two, it actually seems more likely that the Tamiflu dose will be the one that's worth $40 million in a couple of weeks.
ScottyZ
http://toronto.cbc.ca/regional/servlet/View?filename=to_lottery20051026
This statement indicates that the statistical distribution of the balls that fall out of the ball dropping doohickey depends on the number of Lotto 6/49 tickets sold. It doesn't get any more rigged than that.
This is actually pretty cool if you think about it. I'm now tempted to go buy a Lotto 6/49 ticket just to feel the raw and awesome power of actually being able to alter the odds of winning the Lotto 6/49 for every single Lotto 6/49 ticket holder.
ScottyZ
It depends on the number of tickets sold.
For example, if exactly 1 ticket is sold, tonight's draw is obviously a +EV game due to the expected share of the jackpot (in the 1 ticket case, $40 million) being greater than $28 million.1
To generalize, if the expected share of the jackpot is greater than $28 million, then Lotto 6/49 is +EV.
For today's draw, let's assume that at least as many tickets will be sold as were for the last $30 million draw. There were about 27 million tickets sold for that draw. To make the math a little easier, assume that 28 million tickets will be sold for tonight's draw. That means that the expected number of jackpot winners is approximately 2. (Using the binomial distribution, with n = 28 million, and p = 1/14 million, the expected number of successful outcomes is n*p = 2.)
Therefore, the expected share of the jackpot is $20 million. (In fact, my guess is that a lot more that 28 million tickets will be sold for tonight's draw.) This is short of the $28 million required to make the game +EV.
A specific Lotto 6/49 draw being +EV is incredibly unlikely, and I doubt that there has ever been a single Lotto 6/49 draw which has been +EV for the player.
To make a bit of a looser (but still correct in principle, IMO) statement, Lotto 6/49 is not, and will never be, a +EV game for the player.
The house edge for Lotto 6/49 is generally around 50%, and usually goes up (meaning, of course, that the draw is worse for the players) for "large jackpot" draws. After we find out how many tickets were sold for tonight's draw, we can then compute the expected share of the jackpot, and hence the house edge. Roughly, if more than 42 million tickets are sold for tonight's draw, the house edge will be greater than 50%.
ScottyZ
1In terms of analysing the house edge of Lotto 6/49 and similar lotteries, the prizes other than the jackpot are negligible.
Who wrote that gem?? Nice journalism!!! Donks!!!
Anyway... all you people playing the 6/49 and similar are +EV for me... keeps my taxes lower!
P.S. Thanks, Scotty for killing my dream.
Although you could do the Dr. Evil thing......."I raise One Million Dollars....."
And then Tyson would come over the top....."I'm all-in...for about 40 Mill or so. Give or take.."
JohnnieH
Cheers
Lou
This is apparently incorrect.
On closer inspection of the OLGC Lotto 6/49 rules page
http://lotteries.olgc.ca/howtoplay.do?game=649&flag=null
we see that
"47% of LOTTO 6/49 draw sales is dedicated to the Prize Fund."
and
"Any amount not won in the 6/6, 5/6 + Bonus, or 5/6 prize categories is added to the 6/6 jackpot prize for the next draw."
So, the long run house edge of Lotto 6/49 is 53%. However, the house edge is actually reduced during draws when the previous draw's jackpot prize1 is not won.
Theoretically, after a long string of no jackpots won, Lotto 6/49 could conceivably be a +EV game. That would be the case if the carryover amount was greater than 53% of the current draw's ticket sales, essentially "filling in" the house edge portion of the prize pool.
Is this scenario something to shoot for then? Definitely not. Such a large carryover would indicate that the ex post house edge of the last few draws had been nearly 100%, since no jackpot was handed out. (Note that ex post house edge is a weird animal, and is not a good tool for analysing the a priori expectations of lottery payoffs.)
According to the Loto Quebec site
http://diffusion.loto-quebec.com/sw3/res/asp/index.asp?l=1&pRequest=2&cProduit=4
there was one winning ticket, paying out approximately $54 million. Incidentally, the expected number of jackpot winners from ~50 million tickets sold was about 3.6.
ScottyZ
1Or, I guess, the previous 5/6 + Bonus, or 5/6 prizes.
You are either going to win...or you're going to lose...50/50
Hobbes
That's hot!
Let the lawsuits fly.
"The unidentified man bought the ticket at a Mac's convenience store in Camrose. Looking dazed, he retuned to the store late last night to tell the clerk about his good fortune."
Brainy.
Reports indicate that the following discussion ensued:
Store Clerk: May I see?
Unidentified Man: Ho, ho ho! See with your eyes, not with your hands!
Store Clerk: Please! We are all amigos here!
Random Guy in Store: Unidentified Man, I think we can trust the Store Clerk.
Unidentified Man: (hands over the ticket)..... Now give it back!
Store Clerk: Give what back?
ScottyZ
isnt it that the chance of hitting the first number is 1/49.... the chance of hitting the 2nd number is 1/48.... 3rd is 1/47.... and so on...
so you have to multiply all those numbers to get your odds-
so your chances are - 1/(49*48*47*46*45*44)
1/ 10,068,347,520 ??? am i way off?
Your chances of hitting the first number drawn are 6/49, not 1/49.
The probability of hitting all 6 numbers is 6*5*4*3*2*1/(49*48*47*46*45*44), which equals about 1 in 14 million.
If you are familiar with it, you might also recognize the fraction above as the reciprocal of 49 choose 6.
ScottyZ
secretaries, technicians, plant workers, etc. 17 of them
each getting a little over 3 million dollars. (hate to be
one of the ones who didn't buy in) They all have to keep
working now. lmao
PS, some math dude was quoted as saying winning the draw
was equal to flipping a coin 24 times and having them all
come up heads.
By the way, lottery retailers usually get up to 1% bonus of prizes won. This is not the case in Alberta so the Mac's owner in Camrose that sold the winning ticket lost out on more than $540,000!