[Redwing41]

Say you went to a casino and hypothetically they gave you +101.5 odds and they removed the greens from the roulette wheel giving you 50% odds. +101.5 at 50%. Over time you would win consistently.

Using math to explain the situation, what are my odds of losing 100 percent of my bankroll if I had bet 1 unit every time with a bankroll of:

5
10
20
40
50
100
200
300
400
500
1000
10,000.

[Redwing41]

For example if I had 1 unit and total of 1 unit bankroll my odds are 50% I will go bankrupt.

2 units bankroll I would be 25% chance of going bankrupt?? So I multiply by the power of 2?

[Alfa1234]

Yes correct, power of 2 for every extra spin of the wheel. E.g. 5 spins ending all on the same colour in a row would give a 1 in 32chance of happening.

However, make no mistake as there there is a flaw in that theory. After the first spin, the chance of a 2nd spin ending on the same colour as the first spin is not "25%" but it would be 50% again. This is a mistake a lot of gamblers make.

[Redwing41]

Yes correct, power of 2 for every extra spin of the wheel. E.g. 5 spins ending all on the same colour in a row would give a 1 in 32chance of happening.

However, make no mistake as there there is a flaw in that theory. After the first spin, the chance of a 2nd spin ending on the same colour as the first spin is not "25%" but it would be 50% again. This is a mistake a lot of gamblers make.

I am asking about bankroll size. How much bankroll is necessary to be able to withstand the amount of bets for the streaks to even out.

[Alfa1234]

The odds of losing 10 times in a row with a 50% chance of something happening are about 0.00098%. Logically, if the payout is +101.5 your chances of going bankrupt with a bankroll of 50 units are so close to 0 the number is almost negligable. The math is in your other thread. This is taking variance into account (betting a fixed stake with 50% odds and a positive payout using 2% or even 4% of your bankroll will reduce the risk of bankruptcy to something as close to 0 as possible).

Reading your other thread however, this is not the case and your theory is flawed...you cannot by adjusting the stake turn a bet with juice into profit over the long term. Any game in a casino has juice and gives the casino an edge (sometimes it's a small edge, yes, but there's always an edge), even baccarat. This is a mathematical certainty.

[benjy]

Looks like you want to calculate "risk of ruin". The first google result gives a handy calculator:

http://www.automated-trading-system....culation-tool/

For this you'll need to add number of trials. As stated the question posed seems to be lose "ever". Using infinity as the number of trials the answer to ALL of the bankrolls is 100%. Infinity is an unusual thing.

[Redwing41]

The odds of losing 10 times in a row with a 50% chance of something happening are about 0.00098%. Logically, if the payout is +101.5 your chances of going bankrupt with a bankroll of 50 units are so close to 0 the number is almost negligable. The math is in your other thread. This is taking variance into account (betting a fixed stake with 50% odds and a positive payout using 2% or even 4% of your bankroll will reduce the risk of bankruptcy to something as close to 0 as possible).

Reading your other thread however, this is not the case and your theory is flawed...you cannot by adjusting the stake turn a bet with juice into profit over the long term. Any game in a casino has juice and gives the casino an edge (sometimes it's a small edge, yes, but there's always an edge), even baccarat. This is a mathematical certainty.

Can you explain the math "Logically, if the payout is +101.5 your chances of going bankrupt with a bankroll of 50 units are so close to 0 the number is almost negligable"" It would look very likely at +101.5 odds w/ 50 units at bj table you will strike out? If you hit a big streak of 10 losses then win/lose similar amount, then a loss of 4, and then a continual bad break. In about 100 hands it could become 70 lose rate for that short period of time.

[Alfa1234]

As mentioned above, the chances of hitting a 10 loss streak is 0.00098%. Twice in a row is extremely unlikely. Another loss of 4...and you'd still have 50% of your bankroll left if you were working with 2% stake.

Black jack does not give you a 50% chance of winning every hand. You said something with a 50% chance of winning and a positive payout.

[Redwing41]

Lets say you have a bank roll of 50 units at 1 unit a bet. Out 80 total games in that particular session you win 30 out of 80 games = -10 = 50 -10 = 40 units. Lets say from there you get unlucky and hit a 10 loss streak. -30 units. then go up some and down some, then you lose again and your down to your last 20 units. You go back up to 35, then down, then up until you lose.
You have some safety but 50% win rate has some large swings.

If you had 10,000 units you would 99.999999999999% safe. what is the amount of units where you are getting 99.9%, 50% safety.

Yes, I said 50% chance of winning at positive payout as a reference.. It is easier to calculate with those values, and I was trying to prove a thought, even at +105 at 50/50 you can still lose your bankroll if you start off too small a bankroll or too many units.

[Alfa1234]

50 units with a 50% chance of winning and a 1 unit bet size has such a small risk of bankruptcy as the risk being negligent. It takes some very difficult math to calculate what you're asking but I believe the chance is lower than 1 in 2 to the power of 50.

[Redwing41]

50 units with a 50% chance of winning and a 1 unit bet size has such a small risk of bankruptcy as the risk being negligent. It takes some very difficult math to calculate what you're asking but I believe the chance is lower than 1 in 2 to the power of 50.

I dont think the math works out like that. There is definately a chance to lose, if you meet unusually large losses and small wins. But it does seem pretty safe, at 10 units, I would say it is very risk, at 20 units it still is risky, 40-50 starting to get safer, at 100-150 units is very safe, unlikely you will lose, 500 units is 99.999999.



That is what im trying to figure out, how to calculate this, even a very crude way to find a rough estimate by maths.
because at 10 units, 15 units and 25 units is very different in terms if you will go bankrupt even 50/50 coin flip and +100 odds. With 50 total bets you may get as low as -35 units in a down swing. It is not unlikely to see 6 losses, 5 losses, 4 losses in a row and they will add up, given enough units you will even out at even odds + win%.

[Alfa1234]

I believe you are trying to figure out the standard deviation. This depends on the number of bets you are going to make. But obviously there is some deviation to the standard method as you would be expecting to make a profit after x number of bets because of you +EV.

[Redwing41]

If I find the standard deviation of X how do i then connect that to the amount of bets necessary to break even on average? and the starting size of br to accommodate the lowest losing streaks and making your bankroll 99.99 percent safe or to whatever percent you feel safe 90, 95.