Location-based services are currently in a race to lock up the biggest brands and retailers they can. Earlier this month Facebook ran a major promotion with the Gap; Chipotle is also using the platform to offer buy-one-get-one-free deals. This morning, Fast Company reported that Foursquare has forged a deal with Pepsico. And today SCVNGR is announcing that it’s scored a partnership with arguably the biggest brand of all: Coca-Cola.
SCVNGR CEO Seth Priebatsch says that the company has been gunning to land Coke for some time (and that its competitors have as well). The campaign will take the form of custom, coke-themed SCVNGR challenges that will appear in 10 major malls across the country through the end of the year. Players who complete these challenge will be eligible for rewards like Coke bottle openers (meh) and $100,000 worth of American Express gift cards (more tempting). Players will be able to redeem these challenges at each malls’ information booth. And on Black Friday, Coke will actually have teams of people in the malls urging customers to try out the challenges.
Obviously this is a big win for SCVNGR, especially if it can maintain a relationship with Coke to run more campaigns in the future. Since its rewards platform launched over the summer, Priebatsch says that SCVNGR players have redeemed some $1.5 million worth of free goods.
The timing of today’s soft-drink related announcements seems like more than a coincidence, but it’s worth noting that Foursquare’s deal is quite different from SCVNGR’s. As reported in Fast Company today, Foursquare has actually partnered with Safeway/Vons to integrate with the grocery chain’s existing rewards program — this helps lower the barrier to get users playing the game, because it happens automatically at check-out. Users are rewarded with different Pepsi products based on their Foursquare history.
In contrast, SCVNGR probably won’t be going the auto-checkin route any time soon, because it’s focused primarily on offering engaging challenges rather than the check-in itself.