Yes that is what I mean, I mean it would be nice to say, I want to have enough cash to cover 5 losses, if I roll 50 times, what are the odds of me winning/losing? Expected Profit, etc
To clarify,
"I want to have enough cash to cover 5 losses" would mean you have 31 units, or 1+2+4+8+16=31 . Or if you lose the first 5 times you play, then your bankroll will be exactly zero.
If you were playing
toss the fair coin, on average you would see 5 losses in a row with
62plays
If you were playing
Just Dice, on average you would see 5 losses in a row with
59.5 plays
If you were playing
European Roulette, on average you would see 5 losses in a row with
55.5 plays
If you were playing
American Roulette, on average you would see 5 losses in a row with
50 plays
The zero house edge probability is easily calculated as 2^(k+1)-2 where k=5 . It's easy to do that calculation and then subtract a little for different house edges.
I am ignoring the latter part of your question because you might be in the middle of a losing streak at exactly 50 rolls.
It is more exact to pick your game and designate what the average number of rolls before you see your designated losing streak.
If you are having trouble envisioning this, then consider a streak of only 2 losses. On average this will happen every 2^(2+1)-2=6 tosses of a fair coin.
I've simulated it in a spreadsheet if you are having trouble doing it yourself. Download the file so you can re-caculate the spreadsheet with different random numbers. You can add more than 100 runs. But the average before getting 2 losses in a row will be around 6.
https://docs.google.com/spreadsheet/ccc?key=0Aor7l039-pqHdDR1SXNONV9KQW1Id3F6cGQ0bUdPblE&usp=drive_web#gid=0
