ROR/Expected Win Question

[LovinItAll]

If one were calculating the risk of ruin for a particular game, would the following be true?

Parameters:

- Game has no statistical player/house advantage (a coin flip, for example)

If the risk of ruin of losing 40 units over 300 samples is 5%, would the chance of winning 40 units over the same number of samples also be 5%? To me, it seems obvious that it would be, but I have a friend that begs to differ.

Thanks in advance ~ LIA

 

[MangoJ]

If one were calculating the risk of ruin for a particular game, would the following be true?

Parameters:

- Game has no statistical player/house advantage (a coin flip, for example)

If the risk of ruin of losing 40 units over 300 samples is 5%, would the chance of winning 40 units over the same number of samples also be 5%? To me, it seems obvious that it would be, but I have a friend that begs to differ.

Risk of ruin depends not only on the game, but also on the strategy. If you would bet your entire bankroll, risk of ruin is 50%.

I assume you are flat betting a single unit each time on a fair coin flip. In this case you have a one-dimensional lattice random walk (see http://en.wikipedia.org/wiki/Random_walk).
The expected number of bets to either hit -40 or +40 units is 40^2 = 1600, so 300 is a rather short game for your target. Hence I will neglect stop win/ stop loss events and just calculate your chance being below -40 after exactly 300 bets.

EV on each bet is zero, and stdev of each bet is 1.0. 300 rounds is large enough to justify a Gaussian approximation (instead of an exact binomial distribution). The stdev of 300 bets is sqrt(300) * 1.0 = 17.3, so your loss limit is 2.3 stdev.

Your chance of losing 40 or more units in 300 bets is (1-erf(40/sqrt(300)/sqrt(2)))/2 = 1.04%. Same is winning 40 or more units.

 

[LovinItAll]

50%, not 5% - of course, sorry about that. We were looking at a BJ table where the house has an x% edge, and I said, "Well, if they didn't have an edge, the ROR and the Win Rate would be the same." He said, "No, it wouldn't."

Thank you for the math behind that. It was a silly discussion anyway, as he was describing his 'betting system' that reduces the house advantage to exactly 0%, yet he contends somehow gives the player an advantage of 2% (despite the house's edge of 0%, which makes the whole thing ridiculous and him impossibly stupid).

Please.....don't ask. I still have a headache from trying to explain that if a game's EV=0, it's the same for both sides.

Thanks again ~ LIA

Edit: Flat betting isn't a part of the 'strategy', obv., but neither is counting cards.

 

[k_c]

Risk of ruin depends not only on the game, but also on the strategy. If you would bet your entire bankroll, risk of ruin is 50%.

I assume you are flat betting a single unit each time on a fair coin flip. In this case you have a one-dimensional lattice random walk (see http://en.wikipedia.org/wiki/Random_walk).
The expected number of bets to either hit -40 or +40 units is 40^2 = 1600, so 300 is a rather short game for your target. Hence I will neglect stop win/ stop loss events and just calculate your chance being below -40 after exactly 300 bets.

EV on each bet is zero, and stdev of each bet is 1.0. 300 rounds is large enough to justify a Gaussian approximation (instead of an exact binomial distribution). The stdev of 300 bets is sqrt(300) * 1.0 = 17.3, so your loss limit is 2.3 stdev.

Your chance of losing 40 or more units in 300 bets is (1-erf(40/sqrt(300)/sqrt(2)))/2 = 1.04%. Same is winning 40 or more units.

I have a risk of ruin calculator on my website. It doesn't answer what you are asking because it doesn't limit number of trials (300 in your question,) but instead computes probability of going broke when risking x units in an attempt to profit by y units (at which point you quit) for however many trials it takes to reach either of the 2 outcomes - lose x units or win y units.

1 unit per trial is bet, payout is even money, and win rate/EV is a fixed amount that is input.

Sample input would be for a pass line bet in craps (EV=-1.41%.) Risking 10 units to win 10 units results in failure 57.004% of the time and success 1-57.004 = 42.996% of the time.

You may or may not find it interesting. I'm not that well grounded in statistics but this is a purely mathematical calculation. You need javascript enabled in your browser for the program to work properly.

Link: Fixed EV Ror calculator

 

[MangoJ]

LovinItAll, ROR is certainly not 50%. ROR means risk of RUIN. This means the probability that you lose your entire bankroll during the game.
It doesn't mean the probability that to come out lose something.

Thanks for the link kc. Wikipedia also tells that for a coin flip losing x bets before winning y bets is y / (x+y). I peeked in your code (^^), and you do this same for even games.
For non-even games you seem to use the line

ror = (Math.pow(p, bank) - Math.pow(p, goal)) / (1 - Math.pow(p, goal));

Is there any kind of mathematical literature to that, just for curiosity ?

Personally I'm not using ROR calculations, as I'm from the Kelly fraction fraction. Thats the benefit of not being dependent on winnings for life costs.
I'm resizing bets with wins and losses. Increasing Kelly fraction with low bankroll and decreasing Kelly fraction with higher bankroll. Hence never overbetting, ROR=0 all times.