Highest floor of a building from which to drop an egg without breaking it
Here are some comments on your question:


Bad problem description – I can't help but to make a method which find the input parameter, seems kind of silly. It could simply do return n-1 and you're done. It would have been better if the problem description gave a predefined method which you could call at most two times before it gave some error. But I digress...
Choose good names for your variables – Two store the lower and higher test floors in a list t is bad. What is t[0] or t[1]. Say low, highis way better, and even they can be improved upon
Assign multiple values at once – If you want to use one line to assign multiple values you can do: low, high = 1, 11
Don't use magic numbers – If you have magic numbers, like the number of floors, do either declare them as constants at the top or as default parameters to the method in question
Give proper text output – It took me quite some time to understand what your output actually was meant to describe. It would have been much clearer if you had said "First egg broke on floor x" and "Second egg broke on floor y"
Switch to new style print formatting and function – It is recommend to switch to the new string.format() style for print output. I.e. print ('First egg broke on {}'.format(high))
Error handling is good – As the problem was limited to a 100 floor building, one could/should also let the code reflect that. One way can be to assert legal floor levels, and another could be to make sure that when returning you don't return an illegal level.


Finally you can reduce the number of attempts by a little if you start out attempting with a little higher delta value/range and let it shrink as you get closer to the top1. It can be shown that the optimal delta offset is the result of  \$\frac{\sqrt{1 + 8\cdot {floors}} - 1}{2}\$. With these suggestions, and a simple loop to test for all available floors we get:

import math

MAX_FLOOR = 100

def safest_egg_drop_floor(n):
    """Return the safest floor to drop an egg from."""

    assert 1 &lt;= n &lt;= MAX_FLOOR, 'Building has floors from 1 through {}'.format(MAX_FLOOR)

    # Calculate the optimal delta to start with, to reduce attempt numbers
    delta = int(math.ceil((math.sqrt(1 + 8*MAX_FLOOR) - 1)/ 2))

    low, high = 1, delta

    while low &lt; MAX_FLOOR:
        if high &gt;= n:
            print('  First egg broke on floor {}'.format(high))

            for i in range(low, high+1):
                if i &gt;= n:
                    print('  Second egg broke on floor {}'.format(i))
                    if i == 1:
                        raise ValueError("Eggs will always break")
                    else:
                        return i - 1
        else:
            delta -= 1
            low, high = high, high + delta
            if high &gt; MAX_FLOOR:
               high = MAX_FLOOR

    return MAX_FLOOR


def main():

    for i in range(0, MAX_FLOOR+2):
        print('\nFind egg drop when it breaks at {}'.format(i))
        try:
            max_floor = safest_egg_drop_floor(i)
            print('    Safest drop from floor {}'.format(max_floor))
        except (ValueError, AssertionError) as error:
            print(error)


Which outputs:

Find egg drop when it breaks at 0
Building has floors from 1 through 100

Find egg drop when it breaks at 1
  First egg broke on floor 14
  Second egg broke on floor 1
Eggs will always break

Find egg drop when it breaks at 2
  First egg broke on floor 14
  Second egg broke on floor 2
    Safest drop from floor 1

...

Find egg drop when it breaks at 100
  First egg broke on floor 100
  Second egg broke on floor 100
    Safest drop from floor 99

Find egg drop when it breaks at 101
Building has floors from 1 through 100


1 See http://datagenetics.com/blog/july22012/index.html for a better explanation.