Secondhand Serenade
3 upcoming events ยท 3 venues ยท West Hollywood, Nashville, Philadelphia
Market Status
Moderate
Price Range
TBD
Events
3
Tickets Available
110
Demand Distribution
๐ฅ High Demand
๐ City Comparison
All Events
Venues
Secondhand Serenade โ Ticket Intelligence
Secondhand Serenade has 3 upcoming shows across 3 cities including West Hollywood, Nashville, Philadelphia. Performances are scheduled at Roxy Theatre - CA, City Winery - Nashville, The Foundry - Philadelphia.
Moderate
Current market analysis shows moderate demand with good availability for Secondhand Serenade shows. 33% of dates show high demand, 67% of dates show moderate.
Moderate
33% of dates are experiencing high demand, which may affect seat selection.
Unknown
Prices vary by date and venue. Checking multiple dates is recommended for the best value.
๐ฏ Buying Guidance
There is good availability across most dates. Comparing prices between dates can help find the best value.
Secondhand Serenade shows run from June 5 to July 11, 2026, a span of 36 days. The average gap between show dates is 18 days. Sunday is the most common performance day, with 2 of 3 shows (67%).
The market shows a balanced distribution, with 67% of dates at moderate demand and broader selection across price tiers. West Hollywood has the strongest demand signal (high demand), while Philadelphia offers broader availability (moderate).
Frequently Asked Questions
Where is Secondhand Serenade playing?โพ
Secondhand Serenade has upcoming shows in West Hollywood, Nashville, Philadelphia. West Hollywood has 1 shows, Nashville has 1 shows, Philadelphia has 1 shows.
When is the next Secondhand Serenade show?โพ
The next Secondhand Serenade show is on Saturday, June 6, 2026 at Roxy Theatre - CA.
Are Secondhand Serenade tickets still available?โพ
Yes. There are 110 tickets available across 3 upcoming dates. Current market status is moderate.
When is the best time to buy Secondhand Serenade tickets?โพ
There is good availability across most dates. Comparing prices between dates can help find the best value.