Topic: DragonMint 16T MEGA Raffle [#3] (includes 5x miners + PSUs) (Read 246 times)
bump
maybe 32 entries and 0.007btc per entry,
for one miner and psu -kit
for one miner and psu -kit
I'll meet halfway and do the 32 entries at .01 .... .007 would eat me alive in shipping cost.... I'll go on and make those changes...thanks for the feedback you guys!
maybe 32 entries and 0.007btc per entry,
for one miner and psu -kit
for one miner and psu -kit
Number of entries is up to you based on what you think people will gamble. You can use 16 or 32 or a zillion. Its the prize value divided by amount of spots. I would expect my gamble would make sense to the marketplace. So current value of lot, plus a fraction premium for you troubles divided by the amount of spots. Otherwise i could go out and buy the same lot, offer spots at half price and still make an absolute killing. These raffles are done in the collectable section a lot and they usually follow that format.
yeah i saw that.... i figured that .1 per was too steep...
Maybe something like 0.01btc per entry,
and max. 20 entries.
and including worldwide shipping
note: this is my idea for one miner and psu -kit
and max. 20 entries.
and including worldwide shipping
note: this is my idea for one miner and psu -kit
hmm what about 32 entries @ .015 per each individual miner+psu (and yeah ww shipping incl)
Number of entries is up to you based on what you think people will gamble. You can use 16 or 32 or a zillion. Its the prize value divided by amount of spots. I would expect my gamble would make sense to the marketplace. So current value of lot, plus a fraction premium for you troubles divided by the amount of spots. Otherwise i could go out and buy the same lot, offer spots at half price and still make an absolute killing. These raffles are done in the collectable section a lot and they usually follow that format.
Maybe something like 0.01btc per entry,
and max. 20 entries.
and including worldwide shipping
note: this is my idea for one miner and psu -kit
and max. 20 entries.
and including worldwide shipping
note: this is my idea for one miner and psu -kit
Raffle is a nice idea.
But I don't like this, because he is trying to make a big profit with this raffle.
I don't think its something we should try to do here at the bitcointalk forum marketplace.
It's actually even worse than trying to get the noobs pay an overprice price at Ebay.com.
But I don't like this, because he is trying to make a big profit with this raffle.
I don't think its something we should try to do here at the bitcointalk forum marketplace.
It's actually even worse than trying to get the noobs pay an overprice price at Ebay.com.
okay I'll hear you out.... what size is fair iyo.... for number of entries and ticket price?
Raffle is a nice idea.
But I don't like this, because he is trying to make a big profit with this raffle.
I don't think its something we should try to do here at the bitcointalk forum marketplace.
It's actually even worse than trying to get the noobs pay an overprice price at Ebay.com.
But I don't like this, because he is trying to make a big profit with this raffle.
I don't think its something we should try to do here at the bitcointalk forum marketplace.
It's actually even worse than trying to get the noobs pay an overprice price at Ebay.com.
So you are valueing this lot at over 5 btc, ticket cost is prohibatively excessive no?
entries will be organized into blocks of 16. each block of 16 will have two digits assigned to them. the 2nd to last digit signifies which block of 16, the last digit is the winners raffle entry. these blocks of 16 assemble in ascending order... based on entry picked
So you would have (since there are only 128 tickets for the 16x16 table of options):One of 1-16 win if the second-last digit is 0, 1
One of 17-32 win if the second-last digit is 2, 3
etc?
that is correct
entries will be organized into blocks of 16. each block of 16 will have two digits assigned to them. the 2nd to last digit signifies which block of 16, the last digit is the winners raffle entry. these blocks of 16 assemble in ascending order... based on entry picked
So you would have (since there are only 128 tickets for the 16x16 table of options):One of 1-16 win if the second-last digit is 0, 1
One of 17-32 win if the second-last digit is 2, 3
etc?
How does the ending digit determine the winner when you have 128 different tickets? There are only 16 different characters in a block hash.
entries will be organized into blocks of 16. each block of 16 will have two digits assigned to them. the 2nd to last digit signifies which block of 16, the last digit is the winners raffle entry. these blocks of 16 assemble in ascending order... based on entry picked
How does the ending digit determine the winner when you have 128 different tickets? There are only 16 different characters in a block hash.
Edit: Closed :\ conversion to sale occurring momentarily.
Jump to: