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Author Topic: [LUCKY HEX] - Bet 0.1 for a 1/8 chance to win 770%!!!  (Read 2464 times)
PandaMiner
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July 14, 2011, 06:48:05 AM
 #21

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

Just curious, how is this a 1/8 odds?
2 * 1/16 = 1/8

You have two chances to match your hex.

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



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Konzul
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July 14, 2011, 08:55:51 AM
 #22

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

I feel slightly guilty here Wink. 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?
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July 14, 2011, 08:46:04 PM
 #23

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

I feel slightly guilty here Wink. 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! Wink

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PandaMiner
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July 14, 2011, 09:07:55 PM
 #24

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.

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bitlotto
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July 14, 2011, 09:35:09 PM
 #25

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?

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PandaMiner
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July 14, 2011, 11:35:15 PM
 #26

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

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bitlotto
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July 14, 2011, 11:59:37 PM
 #27

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.

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PandaMiner
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July 15, 2011, 12:07:14 AM
 #28

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.

My Litecoin address: Lf9WKM61AhmXchG2Ph2cftHKSQhxHutcdk
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bitlotto
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July 16, 2011, 12:59:51 AM
 #29

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

*Next Draw Feb 1*  BitLotto: monthly raffle (0.25 BTC per ticket) Completely transparent and impossible to manipulate who wins. TOR
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Donations to: 1JQdiQsjhV2uJ4Y8HFtdqteJsZhv835a8J are appreciated.
Konzul
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July 16, 2011, 08:07:32 AM
 #30

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

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