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Adding a pinch of gambling. :P

 

Miq and myself will be throwing dice on May 23rd, fighting each other for a certain creature. Whoever throws the dice and gets odd values on most throws, wins the creature. This is where you come in. What you need to do is guess the outcome of the throws: will it be an even value, or will it be an odd value? Here are the details:

 

- Number of dice rounds = number of gamblers.

- Minimum number of gamblers is 3.

- Gamble is between 5 silver coins and 3 gold coins per player.

(you may put in up to 3 gold if you wish, but this gives you no better chances; it only makes the pot larger)

- Sing-ups are required, to determine the number of rounds Miq and I will be playing.

- Signup here and send the gambling fee to dst until May 15th, 23:59 ST.

- Post proof of the fee transfer here.

- Number of people who post proof determines the number of guesses (rounds) you each need to make.

- Send these guesses via forum or in-game PM to me, anytime before May 22nd, 23:59 ST.

- Form of guess is any of the following: odd-odd, even-even, odd-even or even-odd. (Miq-myself)

- Game between Miq and myself will commence on May 23rd.

- May 24th, logs will be posted, guesses will be made public and results will be announced.

- Prize distribution will occur afterwards.

 

Score is % based. How many dice you get right out of how many will be thrown.

 

Prizes are:

1st place: 50% of pot

2nd place: 30% of pot

3rd place: 20% of pot

(values are rounded up, and additional coin requirements would be sponsored by anniversary rewards)

 

Now... this is a gamble. Because of this, there's a possibility that 2 or more people might get 1st place. Should this happen, prizes get split accordingly. (for instances, 2 players getting 1st place will win 25% of the pot, each)

 

[log='Example']3 players: A, B, C = 3 rounds. Pot = 15sc
Player A says: odd-odd, odd-even, even-odd
Player B says: even-odd, odd-even, odd-even
Player C says: odd-odd, even-even, odd-odd

Rounds, Miq-Asthir:
1. 7-10
2. 6-12
3. 3-7

Results:
Player A: yes-no, no-yes, no-yes = 3/6 = 50%
Player B: no-no, no-yes, yes-no = 2/6 = 33.33%
Player C: yes-no, yes-yes, yes-yes = 5/6 = 83.33%

Rewards:
1st = Player C = 7.5sc rounded up = 8sc
2nd = Player A = 4.9995sc = 5sc
3rd = Player B = 3sc (no roundup required)[/log]

Edited by Myth
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  • 3 weeks later...

Actually, I don't really see why not, considering the randomness of the event :p

 

Which reminds me; we'll throw:

 

12-sided Dice
A beautiful 12 sided dice

 

Found in MB Capitol. Make sure you have one when the time comes. :p

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Prizes are:

1st place: 50% of pot

2nd place: 30% of pot

3rd place: 20% of pot

(values are rounded up, and additional coin requirements would be sponsored by anniversary rewards)

 

Now... this is a gamble. Because of this, there's a possibility that 2 or more people might get 1st place. Should this happen, prizes get split accordingly. (for instances, 2 players getting 1st place will win 25% of the pot, each)

 

I probably won't participate in this nice game (although I will if I'm around), but may I suggest changing the prize rules to if two players get 1st place both, they divide 1st and 2nd place rewards (50%+30%)/2 rather than 50%/2 where the 2nd place gets higher reward while actually being third?

If more than 2 players tie on the 1st, just split the whole pot on them equally, and that's it.

 

 

...or I just read it too fast and superficially and you got it covered? :P

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  • 2 weeks later...
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