Chances of no hits on three lotto / lottery tickets

My wife buys a Megamillions Just Jackpot ticket a couple times a month for $3 each time. (The JJ ticket gives you two chances to win the jackpot, but no chance of winning any other prizes, so each chance of winning the jackpot is only $1.50, vs. the normal $2 ticket.) But because the Texas Lotto is now at a 12-year high ($50M), she's using that $3 to buy 3 Texas Lotto tickets instead. So for the same amount of money spent, she improved her odds of winning from 1 in 300M (MM/JJ), to 1 in 8.3M (1 in 50M, ÷ 3).

However, she didn't hit a single number on her three tickets! I wondered what were the odds of that? So let's see if I did this right. (EDIT: Actually wrong, as pointed out by ThatDonGuy, but leaving up for posterity.)

The probability of winning a game with multiple trials is 1 - pⁿ, where p is the probability of winning one trial, and n is the number of trials.

Using an interesting example, if the odds of hitting a slot jackpot are 1 in 216,000, and you play 216,000 spins, what are your odds of hitting the jackpot?

1 - (215,000 ÷ 216,000)^216,000 = 0.63, or 63%.

So, for the lotto problem, first I figure the chances of not hitting the first number. Texas Lotto uses balls numbered 1-54, so the chances of matching one number with one pick are 1/54. Six balls are drawn, so each ticket contains six picks. So with three tickets each containing six numbers, we have 18 trials to try to match the first ball drawn, so our chances of that are:

1 - (53/54)¹⁸ = 0.2857 (chances of matching 1 number with 18 tries, i.e. 3 tickets)

That's just for the first ball drawn and there are six, so we run through the formula again, to see the chances of getting a hit with six tries:

1 - 0.2857⁶ = 0.99456 (chances of matching any of 6 numbers with 3 tickets)

So the chances of *not* having a hit are 1 - 0.999456 = 0.000544, or 1 in 1838.

Dude.

Years ago the Wizard did the math on a popular slot called "It's Good To Be Bad", where you got a pay if you had something like 15 spins without a hit. I'm kinda thinking we deserve something for not a single hit on three tickets. Like, a pack of gum or something.

Not quite that high.
If I am understanding you correctly, none of the 18 numbers on your wife's ticket - and I will assume that all 18 numbers are different - were among the 6 numbers drawn.

There are C(54,6) ways to draw 6 balls from 54. Of these, there are C(36,6) ways to draw 6 balls from the 36 numbers not on the ticket.
The probability of doing this is C(36,6) / C(54,6), or about 1 in 13.26.

Thanks! In hindsight, combin() is clearly the way to go.

What I don't understand is why my method doesn't also work. I think it might have to do with the fact that I'm considering each ball drawn as a separate game, though in reality, any number on the ticket could match any ball.

Deleted

Quote: MichaelBluejay

Thanks! In hindsight, combin() is clearly the way to go.

What I don't understand is why my method doesn't also work. I think it might have to do with the fact that I'm considering each ball drawn as a separate game, though in reality, any number on the ticket could match any ball.
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I think the main problem is this:

Quote:

So, for the lotto problem, first I figure the chances of not hitting the first number. Texas Lotto uses balls numbered 1-54, so the chances of matching one number with one pick are 1/54. Six balls are drawn, so each ticket contains six picks. So with three tickets each containing six numbers, we have 18 trials to try to match the first ball drawn, so our chances of that are:

1 - (53/54)18 = 0.2857 (chances of matching 1 number with 18 tries, i.e. 3 tickets)

Yes - but you're trying to figure out the probability of not matching that number in 18 tries, which, using your method, is (53/54)^18 = 0.7143.
0.7143^6 = 0.1328, which is closer to the truth.
Another way you could look at it: there are 54 possible numbers that can be the first number, and 36 of them don't match any on the ticket, so the probabilty of not matching is 36/54, or 2/3; (2/3)^6 = 64/729, or about 1 / 11.39. However, that only really applies to the first number, as for the second number, there are now only 53 numbers remaining that can be drawn.