Reducing voltage for analog input, or how to reduce number of wires needed?
There are a few ways to do this.


The Flex Gauge is a variable resistor, and already requires a voltage divider setup to work. All you would do is change the resistor value around. At 9V for a under 5V value, the bottom resistor should be about slightly greater than the lower resistance of the flex gauge. If the flex gauge is 30kohm unbent and 70kohm fully bent, then it should be around 35 kohm. A voltage divider calculator and multimeter can be used to figure it out. This removes the 5V wire.








simulate this circuit &ndash; Schematic created using CircuitLab


Modify the leds for 5V power. Two red diodes in series (2.2 + 2.2) and a resistor for 0.6V at 20mA is fine on 5V. The Blue diodes will need to be changed to parallel connections. One resistor for each blue diode. This removes the 9V wire.








simulate this circuit


Most complicated, place the flex gauges in parallel. Each with an offset resistor, this would result in a different value that could be interpreted by the ADC. In my calculations, one will have much less range than the others, and some overlap, but as simple on/off type sensors, that's not much of a problem. This removes 2 wires.








simulate this circuit


Combine 1 or 2 with 3. 3 wires down.


As a suggestion, use a USB power bank (one that doesn't turn off by itself or you may need to add a pulse setting). Much better than a 9V battery, rechargeable and convenient. 

Alternatively, go smd and bare bones. A bare microcontroller,  smd sot-23 transistors or mosfets, 1206 or 0805 are all small but hand solderable, keeping everything on the hand except for power, which could be a single 2 wire cable.

As a side note, I've done this to a UV led strip, modifying it for 5V power from a usb power pack for a costume as well. Worked great.