- 20 Feb '09 21:28 / 1 editIf anyone saw my post in the General Forum, the Ontario Lotto 6/49 jackpot is up to $48 million for this Saturday's draw! I've run the numbers, and the jackpot is actually so high that the expectation is positive this time, even with the expected splitting of the jackpot. Here are the Lotto 6/49 rules for anyone that's interested:

http://www.olg.ca/lotteries/games/howtoplay.do?game=lotto649

The prize structure is a little more complicated than I first thought, but it's an interesting exercise to find the expectation for these games.

**Q:**Using the rules for Lotto 6/49 described in the link provided, and assuming any values that you can't calculate explicitly (such as ticket sales, etc...), what is the minimum jackpot that makes this game a fair bet? (i.e. the expectation becomes positive)

EDIT: BTW, in case it doesn't mention the cost, each Lotto ticket costs $2 and provides you with one line of 6 numbers (1 play). - 24 Feb '09 16:50

$27.967.632,-*Originally posted by PBE6***If anyone saw my post in the General Forum, the Ontario Lotto 6/49 jackpot is up to $48 million for this Saturday's draw! I've run the numbers, and the jackpot is actually so high that the expectation is positive this time, even with the expected splitting of the jackpot. Here are the Lotto 6/49 rules for anyone that's interested:**

http://www.olg.ca/lotteri ...[text shortened]... the cost, each Lotto ticket costs $2 and provides you with one line of 6 numbers (1 play). - 24 Feb '09 19:07

There are 49 numbers you can choose. You have to choose 6 numbers.*Originally posted by PBE6***That's similar to my back of the envelope calculation, so I'm confident in your answer. Care to share the details of your calculation, and the assumptions that you made?**

My calculator told me there are 13983816 possibilities. (49nCr6)

13983816 * $2,- = $27.967.632,- - 25 Feb '09 15:21

Ah. There are other prizes you can win too, so I think the break-even point would be a touch lower. And remember, there are 2 different payout systems depending on the size of the last jackpot!*Originally posted by Thomaster***There are 49 numbers you can choose. You have to choose 6 numbers.**

My calculator told me there are 13983816 possibilities. (49nCr6)

13983816 * $2,- = $27.967.632,- - 25 Feb '09 16:05

ow I see, and if someone else has chosen the same numbers you only win half the money. I'm too lazy for this*Originally posted by PBE6***Ah. There are other prizes you can win too, so I think the break-even point would be a touch lower. And remember, there are 2 different payout systems depending on the size of the last jackpot!** - 26 Mar '09 01:55

Without knowing the ticket sales, I don't see how you can be so sure that the $48mil jackpot really represents a positive expectation. Seems to me you would need to know how many entries there are in the current draw, in order to assess how many players are likely to share the major prize(jackpot).*Originally posted by PBE6*

[b]If anyone saw my post in the General Forum, the Ontario Lotto 6/49 jackpot is up to $48 million for this Saturday's draw! I've run the numbers, and the jackpot is actually so high that the expectation is positive this time, even with the expected splitting of the jackpot. - 26 Mar '09 13:35

You can make an educated guess as to how many people the OLGC expects will play by looking at the difference between the current jackpot and the previous one.*Originally posted by luskin***Without knowing the ticket sales, I don't see how you can be so sure that the $48mil jackpot really represents a positive expectation. Seems to me you would need to know how many entries there are in the current draw, in order to assess how many players are likely to share the major prize(jackpot).** - 26 Mar '09 17:54 / 1 edit

Ah, the old St. Petersburg Paradox. I personally would be willing to pay something in the neighbourhood of $20, only because I expect to need to play several times in order to win some money back and I have limited "funny money", but of course the expectation of the game is infinite and therefore any price would theoretically be a fair price to pay. Interesting summary here:*Originally posted by Palynka***How much would you be willing to pay to play this game:**

A fair coin is tossed until a result of "tails" comes up. You get $1 if it's tails in the very first toss, $2 if only in the second, $4 if only in the third and so on, doubling the winnings every time heads is the result.

http://en.wikipedia.org/wiki/St._Petersburg_Paradox - 26 Mar '09 19:07 / 1 edit

Yep. The point being that positive expectation isn't necessarily all that great.*Originally posted by PBE6***Ah, the old St. Petersburg Paradox. I personally would be willing to pay something in the neighbourhood of $20, only because I expect to need to play several times in order to win some money back and I have limited "funny money", but of course the expectation of the game is infinite and therefore any price would theoretically be a fair price to pay. Interesting summary here:**

http://en.wikipedia.org/wiki/St._Petersburg_Paradox

Edit - But of course, it's still an interesting curiosity to consider.