It is quite difficult to calculate the probability of losing even for one day, since it is unknown how many bets you will make on this day. As I understand it, this is a random value. But I can calculate the probability of getting 5 losses in an infinite game: 2(5+1)-2 = 62 bets. This is provided that we have averaged your odds to 2.
So, if you get a loss (5 in a row) every 62 bets, then the minimum value of the total loss will be 62x6=372 bets. But we need to take into account that you can "return" to the previous day.
The minimal value for the lowest model I got is lower, as you need to cut down the timing between them to the average also so it boils down to n=6/(0.5)
(5)+5, each chain is not waiting for the average to finish tossing, the worst scenario assumes evey single one of those happens in the middle so half of the bets are voided as they happen after the chain is restarted, what you have is the middle average.
If we take into account the margin, then 1.9(5+1)-2 = 45 bets to get day lost (5 lost bets in a row).
To lose completely you need 63x45 = 2865 bets.
Yup, I got the same on this model, this is the average with average spreads between chain of 5 losses.
I need to rest

You need to rest a ton because
- When a day has only 4 bets left, there is no betting as no cain would happen
- only two horses starting at 1.10 and losing mean 1+10+100+200+400...
and..
Horse racing is not coin tossing,

, if two horses from the same yard, same sire are going to a claimer race and they are both at 2.2 odds for their races, the first flops , the second will be 3-4x. If a stable has three winners, the odds for their 4th racing horse will go down, if a yard is losing 4 times, the odds will double, a bit like the hand is getting more accurate at tossing over the day.
Anyhow, Sunday was a nearly empty day, Monday and Tuesday as I type, there were no losing chains, I will update this somewhere Thursday or Friday.