Don't.
So you can expect to win 47.5% of hands, or lose 52.5%.
Change that to 7 instead of ten, and use the exact .525 instead of .53.
0.525^7=0.010992972052001953125
So once every (1/0.01099) tries, or once every 90.96 tries.
So if a new 'round' starts after each win, as a loss would be still part of the last streak, you will lose 7 in a row once for every 90 wins.
90x$5 win= $450 right?
now your one loss goes like this:
$5($5 lost), $10($15 lost), $20($35), $40($75), $80($155), $160($315), $320($635).
So you can expect to win $450 from your many little wins, but when the big losing streak does happen, you'll lose $635, so you'll be down $185.
Ok so you say approximately every 90 tries you lose 7 times in a row. I have an idea why don't you start this way and once you will reach that losing streak you will end up at 185$ loss right? Ok here is my idea - when that happens, start with $50 instead of $5 and work your way up (by doubling). Now the probability of getting another streak of 7 times will be about 90 tries away so lets not be too optimistic and take 1/4 of 90 which will give us about 22 times (so probability of getting a streak of 7 times in a row a second time within the first 22 wins is very very slim). After the 22 wins of $50 each you will be up about 1000$. Once you reach this point start all over again with starting bet of 5 dollars until you get another streak of 7 in a row. I went to Atlantic City about 2 weeks ago and tried this strategy - but I started with 25$ bet instead of $5 - so finally I got a losing streak of 6 and I was down about $1,670. Once that happened I started with 50$ bet and it worked - I made my money plus another 1000 extra. Please let me know what you think of this system.