What are the odds of this numbers game?

Stan Loh · #1 09-20-2007, 04:30 PM

Basically a numbers game. You pick 4 digits, pay $1.

Payout table is for each $1:

exact 4 numbers: $4000
exact last 3 numbers: $500
exact last 2 numbers: $50

5% back on every $1 spent.

Just want to calculate the odds. It's a lottery, yes. But what's the insane house edge? No, I am not going to play this, but I would just like to know.

Thanks.

KurtSF · #2 09-20-2007, 04:43 PM

Without doing the math, looks like about a 50% payout. Unless its for charity it should be criminal.

jba · #3 09-20-2007, 05:43 PM

it's profitable. you sure the numbers are right? $.30 profit for each play, without calculating the 5% back. I almost feel like I'm missing something.

KurtSF · #4 09-20-2007, 06:51 PM

Boy was my seat-of-the-pants wrong. I saw a 10000:1 shot paying out at 4000:1 and assumed -EV. Since jbs disagrees I did the math. Assuming the wins stack (that is, if you get 4 right you win the $4k but you also got 3 right so you win the $500 and the $50 too), and the $0.05 back is frictionless (cost of $1 bet is only $0.95), here's my payout table:

odds pay weighted
1 $4549.05 $4549.05
9 $549.45 $4941.45
90 $49.50 $4414.50
9900 -$0.95 -$9405

Summing the weighted payoffs we get +$4500 EV on $10,000 in bets. This pays out 45%.

Note, that if the wins don't stack it pays out 35% like jbs stated. I'm confused, but this looks right. How messed up am I?

Stan Loh · #5 09-20-2007, 10:31 PM

Wait a sec how can this be profitable? You only can win one of those for each $1. They don't stack. That means if you win on exact last 3 digits, exact last 2 digits don't pay. Only one 4 digit number result each draw.

So each $.95c bet has 3 chances, a 1:10000 of winning $4k, a 1:1000 chance of winning $500, and a 1:100 chance of winning $50. I think, something like that. I might have heard the numbers wrong. Is the 1:100 chance to win $50 causing the EV to go up?

tyler9768 · #6 09-20-2007, 11:04 PM

I think the easiest way to look at it is if you play this game 10,000 times, you'll spend $9500.
One time you'll win $4000.
9 times you'll win $500, for $4500 (I'm assuming the 10th time is included in the match 4, I could be wrong, but if I am it's only more profitable)
90 time's you'll win $50 for $4500.

So that puts you ahead $3500. Which seems crazy, am I screwing something up here?

Edited: I just realized Kurt did almost the same thing 4 hours ago, but atleast it gives me more confidence in my numbers.

Stan Loh · #7 09-20-2007, 11:25 PM

I think you guys are right. But damn these dodgy neighborhood numbers game anyway. Probably won't pay up if hit first prize.

jba · #8 09-20-2007, 11:26 PM

[ QUOTE ]
I think you guys are right. But damn these dodgy neighborhood numbers game anyway. Probably won't pay up if hit first prize.

[/ QUOTE ]

how do they "draw" the number?

Stan Loh · #9 09-20-2007, 11:29 PM

I think they follow a certain local lottery corporation's results.

KurtSF · #10 09-21-2007, 01:14 PM

How about this play: enter 100 times:

0000
0001
0002
...
0099

The first two digits, of course, could be anything.

1 in 100 you win $4k
9 in 100 you win $500
all other times you win $50

So take $50 off of all the prizes and call the cost of the play $45. +$40 EV from trifecta ($450 * 9%) and +$40 EV from quads, total EV of this play is -$45 + 40.5 + 39.5 = +$35.

You could do this every week!

How does this work for the house? Obviously you run the game until someone hits quads then take the money and run. I think your instincts might be right on Stan.