First let me explain what I mean by "expected value". In games of chance such as lotto, expected value is the amount of winnings (or loss) multiplied by the probability of winning. Suppose the game of chance is tossing a fair coin and you get P100 if heads come up and you get zilch if tails come up, so you have a 50/50 chance of winning and getting zilch. In this case, the expected value from the coin-toss game is
EV = (P100)(0.5) + (P0)(0.5) = P50
So basically the expected value is the amount you can expect to gain from playing a game before you actually play it (e.g., when you're thinking of buying the ticket). Of course, it "makes sense" to play if the cost of the ticket is less than or at least equal to how much you can expect to win. In this case you probably won't (or shouldn't) play the game if you have to gamble more than P50 to have a 50% of winning P100, but it'll make sense to play if your bet is only, say, P25.
Now let's look at Lotto 6/49. The game involves a player picking 6 numbers from 1 to 49 (hence the name). A lotto ticket, on the other hand, costs P20, which the player has to pay if he wants to have a chance of winning. PCSO, which manages the game, then (presumably) randomly selects 6 numbers and publishes it. If the player gets 6 out of 6 numbers he gets the pot, which could run to the tens to hundreds of millions of pesos. Getting 5 out of 6 numbers gets you P20,000; 4 out of 6 gets you P500; and 3 out of 6 numbers results in balik taya, or you get your P20 back.
So, and this is where it gets dicey, what are your chances of winning? If you buy one ticket, then you will win the pot if your exact 6 numbers are selected out of all the possible combinations of 6 numbers out of 49, which are 13,983,816 possible combinations (mathematically, this is 49C6). Therefore, the probability of your 6 numbers winning the pot is 1/13,983,816 = 0.00000007151.
But then winning the pot isn't the only way you can win-- you will also win something if you get 5/6, 4/6, or 3/6 numbers correct. Choosing 6 numbers is like getting 6 chances of getting 5 numbers correct since there are 6 possible combinations of 5 numbers from the 6 numbers you chose. Eh? Suppose you very creatively chose the following 6 numbers:
[1, 2, 3, 4, 5, 6]
Then the possible combinations of 5 numbers from the 6 numbers you chose are:
[1, 2, 3, 4, 5], [1, 2, 3, 4, 6], [1, 2, 3, 5, 6], [1, 2, 4, 5, 6] [ 1, 3, 4, 5, 6], and [2, 3, 4, 5, 6]
Similarly, you get 15 chances of getting 4 numbers correct and 20 chances of getting 3 numbers correct from the set of 6 numbers you chose. Using combinatorial mathematics, the expected value from the lotto game can be computed as:
EV = (1/49C6)(X) + (1/49C5)(6C5)^2(P20,000) + (1/49C4)(6C4)^2(P500) + (1/49C3)(6C3)^2(P20)
where X = the winning pot, which can change. The squared terms are there to take into account that both the player and PCSO choose 6 numbers.
Since the cost of a lotto ticket is P20, how much should the pot be so that the expected value from playing equals the cost of a ticket? Setting EV = P20, we can calculate X as... P261 million.
Yes, the pot will have to be at least P261 million before it actually makes sense to play Lotto 6/49. Any pot lower than that and you're basically throwing P20 into the crapper, just like paying P75 to play the coin-toss game above.
But then if buying that P20 ticket gives you the right to hope and to dream about what you will do in the off-chance you will win the pot, then by all means buy the ticket. After all, that's what PCSO (and all casinos, for that matter) are hoping you will think.
