You are assuming that the load is resistive, in most cases with an active semiconductor load the current will not decrease with the voltage, in fact in many cases the current will increase with the voltage decreasing as the active regulators will increase the current to generate the constant voltages the load is requiring.
But even if the load was resistive, lets say we have 12v supplying a resistive load that needs 300w of power, that will need 300/12 A of current, So 25A of current. So lets look as the load and contact resistance as a 0.48R resistor, Initially the contact in the plug may have (lets say) 0.05 ohm, so there is 0.48 - 0.05 = 0.43 ohm in the actual load. So the voltage across the plug is 25 * 0.05 = 1.25v, and the load 25 * 0.43 = 10.75v. So that's 25 * 1.25 = 31.25W dissipated in the contact and 267.W in the load. The load will have active cooling and will be fine with that. Now lets say the contact resistance goes to 0.2 ohm. Now we have 0.43 + 0.2 ohm in total, so 0.63 ohm in total. The 12v supply will then pass 12 / 0.63 = 19A in total. And that will be 19 * 0.43 = 8.17v across the load and 3.8v across the contact. So power is now 19 * 8.17 = 155.23W across the load and 19 * 3.8 = 72.2W across the contact.
The increased resistance of the contact has not reduced the current enough to mean that more power is not dissipated across the contact so contact temperature increase and the resistance increases and it gets worst. Thats with a purely resistive load, which as I said is not the real world case here.