Could somebody please tell me what the odds are of a 25 bet swing on 50/50 bets, and also what bankroll would be necessary to have a 0.001% risk of ruin?

Thank you,
Scott

If you have a finite bankroll and flip coins at even money forever, you're a lock to go broke. You need to have a positive expectation to avoid going broke.

You're also a lock to have a 25 bet downswing at some point.

If you have an advantage in the game, then you can compute your risk of ruin for playing indefinitely. That is, if you have 25 bets ( or buy-ins) , then the probability of busting with 25 bets is

(q/p)^25.

if p=51%, then the probability of busting is (.49/.51)^25 ~ 36.78%

If we know the win-rate and standard deviation of the coin flipping game, then the risk of ruin is

r = e^(-B*2*wr/s.d^2)

The s.d is approximately 1 for p close to 50%.

r= e^(-25*2*0.02/1^2)

r ~ 36.78% ( same as before )

If you wish to determine B with a ror of 0.001% for some p ,say p=51%

Then we use the same equation written in a different form

B = -s.d^2/(2*wr)*lnr

B = -1^2/(2*0.02)*ln(0.00001)
B ~ 287.82

The prob of a 25 bet swing is 1 (100%), with an infinite number of flips. As stated above. To answer your question in a meaningful way you must state how many total flips you will be doing. For instance losing 25 out of 25 is almost zero.

Okay, let's say I'm going to do 1,000,000 flips. What are my odds of seeing a 25 bet swing, and what bankroll would I need to survive that many flips?

This has been discussed here before. The probability that you drop 25 buy-ins at some point in the 1 million flip interval is exactly twice the probability that you drop at least 25 buy-ins at the end of the 1 million flip interval.

Refer to this post by BruceZ.

http://archives1.twoplustwo.com/show...313&page=&vc=1

The standard deviation per flip is 1 and the standard deviation over 1 million flips is sqrt(1 million) = 1000

EV = 0.

Plug in the numbers in the equation quoted in the link and you get

49% + 1*49% = 98%

There is a 98% chance that at some point you will drop 25 buy-ins after 1 million flips have been made. Eventually this number approaches 100% as the number of hands approaches infinity.

jay_shark is correct. To answer your other question, you need $3,291 to have less than a 0.001 probability of going broke in 1,000,000 flips. Each extra dollar you start with buys you another 608 flips while keeping your 0.001 risk of ruin (if you add or subtract a lot of dollars, the approximation breaks down).

Cool, thank you guys. There's a 50/50 game within another game that I play, and I've been betting on it for fun, but don't want to go broke in the mean time. I was sizing my bets to have 30 buy-ins, but quickly saw a 27 bet swing (25 when I posted) and figured out that my guess on bankroll size was way off.

Since I'm playing this game upwards of 1,000 times a day, it looks like I need to massively increase my bankroll, or else just stop playing.

Scott

I have played poker for many years and have an observation for you. The players who go broke are the ones who make prop bets. This is true even if they are clever about their prop bets so they generally have an edge.

I've thought about this a lot. One possibility is people who like prop bets do other things that make them go broke. But I think it's something else. I think even the small additional volatility added by prop bets is the difference between recovering after a bad run and going broke.

It took me a long time to come to this conclusion. The statistician in me said a few thousand bets, never more than 1% of your bankroll, couldn't add substantially to total variance. It's trivial compared to your poker variance, hundreds of thousands of bets, some considerably more than 1% of your bankroll and not independent like the prop bets.

Everyone knows poker players who like casino games go broke, but they usually ascribe it to the negative edge in those games. I think it's mainly the variance.

My opinion is, if you want to be a serious poker player and not go broke, you have to give up gambling.