How much Carbon Dioxide would be required to displace enough oxygen to prevent ignition? In air, how much $\ce{CO2}$ would be required to prevent a match to light or a butane lighter to work? 

I have a basement office that is unusable because of an apparent lack of oxygen, currently it is my guess that the oxygen is being displaced by $\ce{CO2}$.  

A bit research suggests that it would take a reduction to around 14% for a safety match to fizzle out without any flame. But my naive estimate suggests 7% $\ce{CO2}$, and at that level isn't $\ce{CO2}$ deadly? 
And this isn't even accounting for the displacement of nitrogen, surely that would mean the levels of $\ce{CO2}$ would be even higher. 

Would a 6% increase in $\ce{CO2}$ displace oxygen by only small percentage say 2% as it's only a smaller ratio of the atmosphere? Or would the presence of $\ce{CO2}$ contribute to the lack of ignition compounding the lack of oxygen?

See my related question on the DIY stack exchange

UPDATE

An insepector from the local council has installed a data-logger, whilst he was in the basement he took a reading that I've attached below. When this reading was taken the air felt poor, but by no means the worst it's been, a candle wouldn't stay alight. An breathing during conversation seemed laboured. It looks like the assumption of displacement doesn't look correct, there's more to it. 
Here's the readings: 

Gas - Reading (Normal)

$\ce{CO2}$ 3.3% (0.04%) 
$\ce{O2}$ 17.5% (20.8%)
Other 79.2% (unchanged?)
The parameter you are asking about is called the Minimum Oxygen Concentration (MOC), and is a measurement of the minimum amount of oxygen present in air to propagate a flame. The amount depends on the fuel being burned, and varies from about 4% for hydrogen to about 18% for some solid combustibles.

Let us consider your example of butane, and round some numbers to make our calculations simple. Butane has a MOC of 12%. Normal oxygen content in air is 20%, with nitrogen and sundries making up the remaining 80%. Assume uniform mixing and displacement of gases is equal. In order to reduce the oxygen content to 12%, you need to introduce enough CO2 to make up 40% of the air content:

$O2 + N2 = 100$  =>20 + 80 = 100%

$O2 + N2 + CO2 = 100$    => 12 + (12*80/20) + CO2 =100%

However, let us consider a couple of other very significant points. 


An oxygen level of 12% is unsafe. Anything below 19.5% is unsafe. At levels of 12%, a person will have impaired judgement and coordination, and experience increased pulse rates. Prolonged exposure can have fatal consequences.
A CO2 level of 40% is fatal. Full stop. Levels above 7% can cause loss of consciousness and just 10% can be fatal.
CO2 is actually denser than air, so mixing of the main components is unlikely to be uniform. This means that higher CO2 concentrations will be found lower down in the room.


I hope you stay well clear of your basement until you have resolved this problem and have appropriate active ventilation installed, as well as possibly an oxygen monitor. 

In the very least, get yourself a canary.