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Author Topic: tradehill BTC withdrawal bug?  (Read 902 times)
pennytrader
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July 08, 2011, 03:25:02 AM
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There are several different types of Bitcoin clients. The most secure are full nodes like Bitcoin-Qt, which will follow the rules of the network no matter what miners do. Even if every miner decided to create 1000 bitcoins per block, full nodes would stick to the rules and reject those blocks.
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DamienBlack
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July 08, 2011, 03:26:59 AM
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Hmm... perhaps they are attaching a fee. Try 3.995.

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July 08, 2011, 03:27:41 AM
 #3

that's weird.

try 3.9999999.
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July 08, 2011, 04:41:52 AM
 #4

yeah probly because they want to charge a few bitcents transaction fee.

elggawf
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July 08, 2011, 04:47:55 AM
 #5

I don't think they charge fees for BTC withdrawals - I've withdrawn exactly what's in my account before.

However if you have slightly more than 3.99995 BTC, their shitty software will round up to 4 and then give you that message. It's infuriating - Bitcoin is divisible to 8 decimal places, yet they only show four. If you only want to work in 4 decimal places, that's cool - don't let our transactions go lower than 0.0001 BTC and there won't be an issue.

MtGox and CampBX do this too. I whined about it in CampBX's thread, and apparently they're going to be fixing it and showing the full number. I think I complained to TradeHill about it, but it was around the time of the MtGox hack so it probably got lost in the mess. Sad

^_^
pennytrader
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July 08, 2011, 05:55:18 AM
 #6

So should I draw 3.99995 or should I wait for them to fix the issue?

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pennytrader
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July 08, 2011, 06:08:42 AM
 #7

I figured it out by keep changing sell orders.

So the balance is 3.9999737917
typing 3.9999737918 will reject my order.

Amazing, they have 10 decimal places Smiley

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July 08, 2011, 06:13:16 AM
 #8

has anyone successfully withdrawn dollars to their bank straight from TradeHill?

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