Congtatulations! I know your the math guy and I'm sure your
estimate of 33 years is right on target. Could you share the math
process you used to get that number?
From: TomSkiLV <tomskilv@...>
To: vpFREE@yahoogroups.com
Sent: Sunday, March 23, 2008 2:31:09 AM
Subject: [vpFREE] Beating the odds....
Went to play $80 in free play this week on 2s wild. So 64 hands
on .25
machine and I hit four deuces twice and a wild royal. Now, odds of
hitting four deuces twice in 64 hands is about 1 in 12,000. I could
play $80 in freeplay everyday and this would only happen about once
every 33 years. TomSki
John Clark asked how you calculate that getting deuces 2x in 64 hands
is a 1 in 12000 occurance.
Here's a way to get this number.
Odds of getting deuces in FPDW is 1 in 4909 or in terms of
probability .000204. So the probability of getting deuces 2x
is .000204 x .000204 = 4.15E-08
Probability of not getting deuces is 1-.000204 or .999796. So the
probability of not getting deuces in the other 62 hands is .999796^62
= .987448
The probability of getting deuces 2x and 62 other non-deuces hands is
4.15E-08*.987448=4.09759E-08.
This would be the probability of doing this one way, but there are
combin(64,2) or 2016 ways to do this (for example you might get back
to back deuces and then 62 non-deuce hands, or one of the other 2015
ways).
Multiplying the probability of a way by the number of ways to do this
is 4.09759E-08*2016=8.2607E-07. You invert the probability to get
odds. So, 1/8.2607E-07=12105.4566 or about 1 in 12000.
The 33 years part is because TomSki said if he played just 64 hands
of freeplay every day and 12000/365=33.165 years. You could also
look at this as a once in every 968 hours of play occurance if you
play at 800 hands per hour or once a year for someone who plays a
fair amount of FPDW.
Bill
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--- In vpFREE@yahoogroups.com, John Clark <jaycee5353@...> wrote:
----- Original Message ----