Bitcoin Forum

Over. Not really enough interest to make it worth the extra time. Maybe another day!

"Only" 5% fee was what you forgot to mention! ;)

It's a 2.5% fee.

Income 0.40
Outgoing 0.38

So far...(to help those who play figure out if they won)
1st ticket: 2 is your pick
next block is 134703 and last 2 digits are: 9, b
No win.
2nd ticket: 1 is your pick
next block is 134704 and last 2 digits are: 7,6
No win.
3rd ticket must wait till another transaction shows...
(if none in a few days I'll pay 0.01 to force a new block)

1JpqBHN...won .38 with tx b2fd5c76f9... because next unique block listed ended in ...cb

Going to put in an entry or two for fun.... but waiting for a block on the first one before sending the next.

I sent 0.01. The last 0.05 was a loser. 1/12 won...I'm sure one will win again soon.  ;)

Upped the price to .1 to enter. 770% return. People who send .05 will still get the .38 though. Good luck all!

Well I put in two more entries.... so we shall see.

Put in 3 too.  Hopefully i'll go 3/3 ? (: hahah, ridiculously lucky there

Just curious, how is this a 1/8 odds?

2 * 1/16 = 1/8

You have two chances to match your hex.

Why is it 1/16?  Sorry if I'm missing something obvious

Hex has 16 characters and it's pretty much random.

O duh. THanks for the clarification, don't know how I managed to overlook that fact

I'm guessing your bets were in Block 135727. They lost!  :'(
Next block was Block 135763 and it ended in 0,1
None of the inputs had a 0 or 1 as the first digit.

14HHEY... wins with tx b40bfae0df... because next block ended in 0, b

Oh nice!! I just won 0.77 BTC!!! Thanks bitlotto!

And now trying two more times, someone has to bankrupt bitlotto ;)

LOL. Done. I'm pretty much broke. My % for all my stuff is pretty much so low right now I just have enough to pay hosting/domain etc. If my games get big one day hopefully I have a tiny bit to spare.

19h6Y2... wins .77 with tx of efa19645de... because next block has 1, e at the end.

What happens if BOTH hexes match the tx (transaction) number? Is there a double payout? Hmm?  :P


I think your math is off, but it might be in error of the player, so I won't press the matter.

EDIT:

Your math is in error of the house... when both hex digits match! (because I assume there is no double payout)

The real formula is therefore:


   (n * ß) - (n - P)
----------------------
             ßⁿ


ß = base number system (# digits in system),
n = number of "places" player can win in,
P = number of paid matches. [say if all "places" match, how many times do you get paid?]

So, if we plug in the numbers, and then simplify, we get....


(2 * 16) - (2 - 1)                  31                     1
-------------------       =       -----       =        -----
          16²                          256                  8.26

I feel slightly guilty here ;). I had put in a couple of more bets and have won again!

Are you doing that all by hand or do you have it automated? Seems like a lot of work. Can't you integrate it into your bitlotto website?

Ya, I do it by hand. It's not THAT popular so it doesn't take me long. I just hover over all the blocks to see the hex then look at the previous transactions. Guess I have to earn my tiny cut! ;)

bitlotto, are you going to answer my previous post?  What happens, in the rare case that, both numbers match the TX first digit? There's only 1 in 256 chance of that happening.

Right now, its still .77 even if both "rolls" are what you need. I could change that I guess but the .77 would have to go down... Do you think it's worth changing?

No, just acknowledging it is fine.  The odds are still pretty good.. 1 in 8.26, just be exact. :p

Sorry, right now I have a headache so bad my vision is going blurry...so excuse the present dumbness but:
Wouldn't P be 2 since either place can win. Even if they are the same we still look at them and each one can win even though we aren't increasing the payout.

Nah. I hear ya, but that is why there is an "n" in the top right too.  "n" places to win on, but only "P" gets paid... so if all "n" places turns out to be the identical number, only "P" times will the jackpot be paid.  In this case, it's 2 minus 1.  Out of the 2 spots, only 1 gets paid at any given time, even if both spots have the same winning hex.

I spent over 35 minutes figuring out a formula that is scalable. Feel free to play with the numbers, you'll see it works for all base numbers and paid places!! (I'm pretty proud of that)

Try these numbers:
base 10 [0 - 9], 2 spots, 1 paid    = 19:100 (1 in 5.26)
base 8, 2 spots, 1 paid                = 15:64 (1 in 4.27)
base 16, 3 spots, 1 paid              = 46:4096 (1 in 89)
base 2 [binary], 3 spots, 1 paid   =  4:8 (1 in 2)
base 16, 2, 2                             = 32:256 (1 in 8 for single payout, 1 in 256 for double payout)

It works. The formula.  I do think it's easier to just pay out for any one instance.

1Az5sgYz... wins with 3a6cebd165...

Grats to 1Az5sgYz! I am delighted to say that it's not me this time... But I have already made a few more bets ;)