Help with math? payouts? this is so wrong

Ok I dont really know what section to put this under as I need help to prove to someone that if I bet on a dice from 0.00 to 99.99 that 10.89 or over would not be 1.01x payout even with it having to be same odds as 55x2

My reasoning: 100 rolls of 1M @ 1.01x payout chance of winning = 89%, so chances are you'd win 89x = 890k (1,000,000 x 1.01 = 10,000 per win)
chances of losing = 11% so 11 rolls = 11m

pretty sure thats at least 90% odds in whoever owns the dice favour.
Apparently i'll make a fool out of myself for trying to help him fix his odds

(I am in no way affiliated with BBB, just tried to help him out so odds are correct)
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http://stattrek.com/online-calculator/binomial.aspx for a decent binomial calculator
https://primedice.com/play for checking roughly "fair" odds, i.e., "am I off by a mile?"

Assuming you're phrasing the question correctly (double-check your numbers, you might have flubbed a decimal or something)... it's the same as rolling a 10,000-sided die, no? Each time you roll, you have an 89.11% chance of winning each time you roll as the host, regardless of how many rolls you make. (99.99-10.89+0.01, since 0.00 is a potential roll). 55x2 gives you a 54% chance of winning as the host.

In 10.89x1.01:
-If you do 10,000 rolls, you will win on average 8911 of them, profiting 8911m on a 1m bet each time. You will lose on average 1089 of them, losing 1089*(1.01) = 1099.89m. Net profit: 7811.11m.

In 55x2:
-If you do 10,000 rolls, you will win on average 5400 of them, profiting 5400m on a 1m bet each time. You will lose on average 4600 of them, losing 4600m. Net profit: 800m.

So yeah, the other game is like 9.75 times more profitable for the host.

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Edit: apparently it's the bettor with the 89.11% chance of winning, so we're ok here.

 

I believe you switched the winning chances. If I have not misread the first post, host has a 89% lose chance.