[Precision1C]

For a reasonably large number(say 10+) of samples your result will look like normal binomial distribution*$1000 shifted positive $50*N where N is the number bets. To determine risk of ruin you use standard deviation to determine on how widely you can tolerate the short term swings, 1 std dev is around 68%, 2 std dev is 90+%, and 3 std dev is 99.9%. By my calculations with a $1,000,000 bankroll your odds of ruin are miniscule, way beyond 5 std deviations.

F(N)=$50*N - 3(sigmaN)*1000

The above equation represents the maximum loss for N bets if you are one in a thousand kind of unlucky.

sigmaN is the standard deviation for a binomial distribution for coin flips

sigmaN=SQRT(N*.5*.5)=.5SQRT(N)

F(N)=$50*N- 1500SQRT(N)

To find the minimum value of F(N) take the derivative with respect to N and set to 0

F'(N)=50-750/SQRT(N) skipping some basic algebra steps

when F'(N)=0 N=[750/50]^2=225

plugging in N=225 for F(N)

F(225)= $50*225 - $1500*SQRT(225)=$-11,250

Notice that this sum is vastly smaller than $1,000,000 and even if you use 5 standard deviations instead of 3 for the equations. If you somehow lose $1,000,000 watch out for asteroids and falling airplane debris because they are bigger dangers. This is the reason that your competent sports book isn't going broke. However this test is affected by assumptions of event independence, for example if you take in 100*$1,000 on the same bet that isn't the same as setting N=100 with a $1,000 bet but is equivalent to setting N=1 with a $100,000 bet.