True Odds on Hop/Any 7 Bets

I understand that the house odds on a single Any 7 bets is 16+%.
My question is this: Is there a way to mitigate that based on a 5 or 6 count method? Here is why I ask -

The cumulative percent that someone should have already crapped out after 5 rolls is 42.3%.
The cumulative percent that someone should crap out between 6 and 10 rolls is 29.9%
Leaving the cumulative percent that someone should crap out after 10 rolls at 27.8% (eventually it will happen).

Thus, if you bet the five Any 7 rolls between the 6th and 10th rolls, your odds of winning versus loosing are 29.9%/27.8% - Roughly 7.6%

Doesn't that mitigate the Any 7 roll standard probability of 16+% house edge when given 4:1 odds? Or are the odds simply not true odds and am I missing something here? (5 bets with 4:1 odds, one of which wins, making it true odds?)

So let's look at it slightly differently:
Crap out 2-4 = 33.1% (1/3rd of the time)
Crap out 5-9 = 34.8%
Crap out 10+= 32.1% (just less than 1/3rd of all random shooters make it here)

So 34.8/32.1 = 108.4%, or a 8.4% player advantage.
8.4% of the time the shooter will crap out more in the 5 to 9 range than after it (10+).

The bet: Hop 7s pays 15:1 per 7 combination, or 5:1.
A $30 Hop 7s pays $150 + the winning $10 bet remains (-$20 on two losing bets).

The worst paying instance is when the bet hits on the 9th number rolled. This means you have bet $30, $30, $30, $30, $30. The return is $160, for $150 bet.
The 5 paying results are +$130, $100, $70, $40, $10 profits. The 10+ (41.6% of the time) is -$150.

And this is where I start to get a little lost in the math of it all...
(1.495*chart percent) (that is 100% of 34.8+32.1 percents that make up the remaining rolls)
13.6% of the time +160
11.81% of the time +130
10.16% of the time +100
8.82% of the time +70
7.62% of the time +40
(52.01... off by .01% due to rounding, but close enough for me)
47.98% of the time -150

So I guess I need help finishing the math of the losses versus gains....

Let's take a number of rolls, say 1000.
136 = +160 21760
118 = +130 15340
102 = +100 10200
88 = +70 6160
76 = +40 3040
+56500
480 = -150 -72000

Or is it 136 * 130, 118 * 100, etc? For a total of +40900. Either way, the house suddenly has way better odds than 16%, namely 21% or 76%!!

Which leaves only the "gambler's" way to win at a casino; notice if a table if hot or cold and gamble that it will continue to be... I guess I worked out my question, but feel free to comment if you see glaring errors or what have you...

>I understand that the house odds on a single Any 7 bets is 16+%.
>My question is this: Is there a way to mitigate that based on a 5 or 6 count method?
Mitigate: to change the math from what is to what I want it to be.

> Here is why I ask - https://wizardofodds.com/games/craps/number-of-rolls/

>The cumulative percent that someone should have already crapped sevened out after 5 rolls is 42.3%.
The cumulative percent that someone should crap out seven out between 6 and 10 rolls is 29.9%
Leaving the cumulative percent that someone should crap seven out after 10 rolls at 27.8%.

> (eventually it will happen). YES!! Including those history making three hour rolls. Eventually it will happen.
YOU will be broke before that three hour roll ends.
YOU will be broke before that two hour roll ends.
YOU will be broke before that one hour roll ends.

>Thus, if you bet the five Any 7 rolls between the 6th and 10th rolls,
>your odds of winning versus loosing are 29.9%/27.8% - Roughly 7.6%

>Doesn't that mitigate the Any 7 roll standard probability of 16+% house edge when given 4:1 odds?
> Or are the odds simply not true odds and am I missing something here?
> (5 bets with 4:1 odds, one of which wins, making it true odds?)

Here's your sign!