A VP probability question

Recently I was playing DDB on the Wizard of Odds free game site for fun. It is an excellent resource and I thank the Wizard for providing and maintaining the site. I am told the average number of hands to yield a quad is approximately 425. I was tracking the frequency of quads out of curiosity and ran into a stretch of 2600 hands without a quad. Can anyone tell me what the percentage chance is of playing that many hands without seeing a single quad? I'm trying to develop an understanding of how bad (and good) the game can be in its volatility

In 9/6 DDB the probability of any quad is 0.2393% or about 1 in 417.895, and the probability of hitting zero quads in 2600 perfectly-played future hands of 9/6 DDB is 0.197% or about 1 in 507.

Quote: JB

In 9/6 DDB the probability of any quad is 0.2393% or about 1 in 417.895, and the probability of hitting zero quads in 2600 perfectly-played future hands of 9/6 DDB is 0.197% or about 1 in 507.

The math is quite simple.

The formula for not hitting in N trials
(1-p)^N
P=1/417.895
N=2600
(1-0.002393)^2600

Let's say you want to know the median or the 50/50 mark

=LOG(1-Pct%) / LOG(1-P)
=LOG(1-0.5) / LOG(1-(1/417.895))
=289.32

just a bit more than Half of the Quads over time will hit by the 290th hand
290 = 50.0819%

(1-0.002393)^290

Enjoy!

Maria Ho
WSOP Main Event Champion 2012

Quote: JB

In 9/6 DDB the probability of any quad is 0.2393% or about 1 in 417.895, and the probability of hitting zero quads in 2600 perfectly-played future hands of 9/6 DDB is 0.197% or about 1 in 507.

So I have to say wow!! That looks like it may well be a once in a lifetime kind of drought. Thanks for the answer. Thanks also to Guido for posting the formula to achieve the result. I'm not sure I understand the median statement though. If the base probability is about 1 in 417.895, doesn't that represent an average where about half of quads appear before that point and about half appear after?

Quote: 4andaKicker

I'm not sure I understand the median statement though.
If the base probability is about 1 in 417.895, doesn't that represent an average where about half of quads appear before that point and about half appear after?

The 1 in 417.895 is the mean or average.
It is about 63.2654%.

It does NOT represent the median or the 50/50 mark.

The 1/N mean is close to 63.2% in most distributions like these.
A geometric distribution, for a binomial event wait time - how long we wait until we get a success,
most times has a median about 69% smaller than the mean.
here is it's relative freq dist
418 (mean) or even 290 (median) can not easily be seen
The mode happens to be 1. We can see that one.


Where a normal distribution, say the number of heads in 200 coin flips,
the mean, median and mode - the most common value, are most times the same. (100)
easy to see the mean, median and mode.
It is top and center

Thanks for all the work and details in your answer. I really appreciate it.