Lottery Question

Hello everyone!
I was part of a conversation in my office about lottery in general. We are talking about a specific contest that occurs every new year's eve in Brazil and has huge prizes. I tried to determine the odds of this game and got confusing results, with huge +EV.
Would you guys help me find the mistakes?

There are 60 numbers, 1 to 60.
You have to pick 6 different numbers among those 60, so let's say: 23, 25, 29, 50, 51, 54.
The bet value is R$2 (roughly U$1)

The prizes are:

R$240,000,000.00 if you get the 6 numbers right,
R$27,400.00 if you get 5 out of 6 numbers right,
R$470.00 if you get 4 out of 6 numbers right.

As usual I followed Wiz's method and determined the probabilities, which were:

Get 6/6 - 0,0000000199745
Get 5/6 - 0,0000064717379
Get 4/6 - 0,0004288164666

So, multiplying the probability with the prize we would get:

0,0000000199745 X 240,000,000.00 = 4.79
0,0000064717379 X 27,400.00 = 0.18
0,0004288164666 X 470.00 = 0.20

Total = 5.09

I know that probably my mistake is that I don't know how to consider the hypothesis of two or more people getting 6/6 and splitting the prizes. But how to do so?

I get a return of 5.17271640660548. I think you just added your 3 numbers wrong. It is +EV. It's because 5% of that years lotto sales go to that jackpot.
From http://www.lottery.net/mega-sena-info.asp

Quote:

The overall value of all prizes in Megasena calculates to 46% of overall income from Megasena ticket sales. From this total the prize values are broken down in the following manner:

35% is dedicated to the match 6 prize (Sena)
19% is allocated for the match 5 prize (Quina)
19% goes to the match 4 prize (Quadra)
22% is saved to a prize fund for a special MegaSena draw held every 5 draws (called the 'Final 0 or 5' draw).
5% is saved to a prize fund for a the last Final 0 or 5 draw of each year ('Mega da virada' draw, or 'Mega Turn').

Insufficient information, maybe?
How is the number of participants incorporated into the results?
If the EV is positive, it supposedly attracts lots of rational players. The more they are, the more the lottery organiser collects money.
A. If they pay the same fixed amount (to be divided among winners), then their EV becomes positive -- hence that of the players negative. This is what you seem to say about splitting the prizes.
B. On the other hand, if the prizes are fixed per individual winner, then the organisers are crazy. I doubt it. So it must be A.

So the EV depends on the number of players, and the lottery hopes that they reach the breakeven number that makes it negative (positive for them).