You will win eventually

As I usually do with such problems, I'll start with simpler problems and look for a pattern. Maybe even a formula. Maybe a sequence in the OEIS.

Call the problem sought n=100 and I'll start with n=2: A 1/2 chance of winning and a 2/2 chance of winning if a second try is needed. E.V. = 1/2 * 1+1/2 * 2=3/2=1.5

n=3. 1st try: 1/3, 2nd try 2/3 * 2/3, 3rd try 2/3 * 1/3 * 3/3. E.V. = 1/3 * 1+4/9 * 2+4/9 * 3=17/9=.38888

n=4. 1st try: 1/4, 2nd try 3/4 * 2/4, 3rd try 3/4 * 2/4 * 3/4, 4th try 3/4 * 2/4 * 1/4 * 4/4, E.V.= 1/4 * 1+6/8 * 2 +18/64 * 3+6/64 * 4 = 71/32 = 142/64

n=5. Long story short E.V. = 1569/5^4

Ok so the denominators are n^(n-1) and the numerators, beginning the n=1 are 1,3,17,142,1569

A single hit in OEIS and the formula is there as a summation. The numbers are going to get huge but there's a link to the first 380 terms. I need the 100th.

$12209960630215980300253132956815012998412714159392662223854992521128809374822896606879062793565756480077431168694642794552798934366694577253573013735421843845957043306726309232640000000000000000000000 \approx 1.22099606302*10^{199}$

So the answer is $\large\frac{1.22099606302*10^{199}}{100^{99}} = \boxed{12.2099606302}$