It’s the festive season and that time of the year where new deals, business offers, discounts, buy one get two free, and (err..and even scams?) offers light up internet buyers around the world. And this time…Bitcoin’s on offer too.
Financial tech firm and wallet player Revolut announced its Bitcoin Black Friday offering earlier this week, with the multi-billion pound business giving out one free Bitcoin to those who take part in a menial competition
Cryptonauts, we’re giving away a Bitcoin to one lucky customer in the UK! Just set up and execute a Recurring Buy this week to enter our prize draw.
T&Cs apply. Remember to consider your personal circumstances before purchasing.
— Revolut (@RevolutApp) November 25, 2020
The offering, as per a release, calls out Revolut users to smash the ‘recurring buy’ feature for Bitcoin on…you guessed it…Revolut’s app! It does come with a caution, however, “Remember to consider your personal circumstances before purchasing, as the price of crypto can be extremely volatile.”
Cautions aside, the recurring buy feature allows users to automate buying crypto worth a fixed amount of fiat currency on a fixed schedule. For example, buying €200 worth of Bitcoin every week.
Dollar-cost averaging as a Bitcoin strategy
In popular culture (such as graduate classes and the /biz section on 4chan), this is also referred to as ‘Dollar Cost Averaging’ and is popular with a lot of users as it averages out the daily volatility. “Most of our current users opt for a weekly recurring buy,” claims Revolut.
To be eligible, users must purchase €10 worth of Bitcoin on the set-it-and-forget-it recurring buy feature. Then, there’s gonna be a magic draw (one that doesn’t use Chainlink’s verifiable randomness to ensure the pick is actually random) to pick a winner and award them the free Bitcoin.
Meanwhile, Bitcoin, on its part, is seeing somewhat of a holiday fall. The asset fell 11% from over $19,300 to nearly $16,000 today, and is trading around $16,700 as of press time.
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