Let's say I'm playing 8/5 bonus or 9/6 jacks.
Assuming a short term play with no royal, how would one calculate the
average number of hands from a given stake.
For example, 0.25 8/5 bonus no royal, put in a $20 - whats the average
number of hands you can play on that $20 before its gone assuming
perfect play?
In Dan Paymar's Book Precision play he gives risk of ruin tables which have how long a given number of bets will last.
Royal Flusher <royalflusher@gmail.com> wrote: Let's say I'm playing 8/5 bonus or 9/6 jacks.
Assuming a short term play with no royal, how would one calculate the
average number of hands from a given stake.
For example, 0.25 8/5 bonus no royal, put in a $20 - whats the average
number of hands you can play on that $20 before its gone assuming
perfect play?
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--- In vpFREE@yahoogroups.com, "Royal Flusher" <royalflusher@...>
wrote:
Let's say I'm playing 8/5 bonus or 9/6 jacks.
Assuming a short term play with no royal, how would one calculate
the
average number of hands from a given stake.
For example, 0.25 8/5 bonus no royal, put in a $20 - whats the
average
number of hands you can play on that $20 before its gone assuming
perfect play?
For 9/6 JOB, the answer is 655 hands or 134 hands, depending on what
you mean by "average".
Do you want to know the mean number of hands you will play? (That's
the expected value of the number of hands you will play.) Or, do
you want to know the median number of hands you will get to play?
The "mean" number of hands is gotten directly from the ev and the
bankroll. 9/6 JOB returns 99.54%, so your expected loss is 0.46% of
what you bet. If you want to rule out the royal, then you miss out
on another 1.98% of ev. Your total loss rate is 2.44%
If you are playing a 25c 5-coin machine, you are betting 1.25 a
hand. You are losing "on average" 2.44% * 1.25 = 3.05 cents a pull.
You can "expect" to get $20.00/0.0305 = 655.7 hands. In other
words, if you started with $20 a million times, and you added up all
the hands you played before you went broke and divided by 1 million,
your answer would be pretty close to 655.7.
Even though the "mean" number of hands you will get to play from
your $20 is 655.7, there is an 82% chance you will go broke before
655 hands!* The median is much less than the mean, in this case.
The "median" result is the number of hands that you will exceed
exactly 50% of the time. To get that figure, I can use my program,
Dunbar's Risk Analyzer for Video Poker. I simply look for the
number of hands that gives me a 50% RoR. That number of hands is
134.
You can do the same thing for the other game you were interested in.
--Dunbar
* I got the 82% figure from DRA-VP also.
--- In vpFREE@yahoogroups.com, "Royal Flusher" <royalflusher@...>
wrote:Let's say I'm playing 8/5 bonus or 9/6 jacks.
Assuming a short term play with no royal, how would one calculate
the
average number of hands from a given stake.
For example, 0.25 8/5 bonus no royal, put in a $20 - whats the
average
number of hands you can play on that $20 before its gone assuming
perfect play?For 9/6 JOB, the answer is 655 hands or 134 hands, depending on what
you mean by "average".Do you want to know the mean number of hands you will play? (That's
the expected value of the number of hands you will play.) Or, do
you want to know the median number of hands you will get to play?The "mean" number of hands is gotten directly from the ev and the
bankroll. 9/6 JOB returns 99.54%, so your expected loss is 0.46% of
what you bet. If you want to rule out the royal, then you miss out
on another 1.98% of ev. Your total loss rate is 2.44%If you are playing a 25c 5-coin machine, you are betting 1.25 a
hand. You are losing "on average" 2.44% * 1.25 = 3.05 cents a pull.
You can "expect" to get $20.00/0.0305 = 655.7 hands. In other
words, if you started with $20 a million times, and you added up all
the hands you played before you went broke and divided by 1 million,
your answer would be pretty close to 655.7.Even though the "mean" number of hands you will get to play from
your $20 is 655.7, there is an 82% chance you will go broke before
655 hands!* The median is much less than the mean, in this case.The "median" result is the number of hands that you will exceed
exactly 50% of the time. To get that figure, I can use my program,
Dunbar's Risk Analyzer for Video Poker. I simply look for the
number of hands that gives me a 50% RoR. That number of hands is
134.You can do the same thing for the other game you were interested in.
--Dunbar
* I got the 82% figure from DRA-VP also.
I don't think that answered Mr. Flusher's question. I think he was
asking for the mean number of hands it takes to first get to the point
of being behind by $20, not the mean number of hands each $20 loss
takes, which your figure of 655 hands indicates. Losing $20 before
655 hands are played 82% of the time suggests that the mean number of
hands it takes to first lose $20 is much less than 655. 655 would be
the mean only if, after losing $20, more play could be done which will
often result in winning some back and later returning to the $20 net
loss or often losing another $20 or more and returning to the original
$20 net loss. I don't know what the answer is, but I'd guess it would
be close, but not exactly to, the 134 hands that is the median. Only
assuming, as I do, the results aren't distributed evenly around 134
hands would it be different.
--- In vpFREE@yahoogroups.com, Tom Robertson <thomasrrobertson@...>
wrote:
>--- In vpFREE@yahoogroups.com, "Royal Flusher" <royalflusher@>
>wrote:
>>
>> Let's say I'm playing 8/5 bonus or 9/6 jacks.
>>
>> Assuming a short term play with no royal, how would one
calculate
>the
>> average number of hands from a given stake.
>>
>> For example, 0.25 8/5 bonus no royal, put in a $20 - whats the
>average
>> number of hands you can play on that $20 before its gone
assuming
>> perfect play?
>
>For 9/6 JOB, the answer is 655 hands or 134 hands, depending on
what
>you mean by "average".
>
>Do you want to know the mean number of hands you will play?
(That's
>the expected value of the number of hands you will play.) Or, do
>you want to know the median number of hands you will get to play?
>
>The "mean" number of hands is gotten directly from the ev and the
>bankroll. 9/6 JOB returns 99.54%, so your expected loss is 0.46%
of
>what you bet. If you want to rule out the royal, then you miss
out
>on another 1.98% of ev. Your total loss rate is 2.44%
>
>If you are playing a 25c 5-coin machine, you are betting 1.25 a
>hand. You are losing "on average" 2.44% * 1.25 = 3.05 cents a
pull.
>You can "expect" to get $20.00/0.0305 = 655.7 hands. In other
>words, if you started with $20 a million times, and you added up
all
>the hands you played before you went broke and divided by 1
million,
>your answer would be pretty close to 655.7.
>
>Even though the "mean" number of hands you will get to play from
>your $20 is 655.7, there is an 82% chance you will go broke
before
>655 hands!* The median is much less than the mean, in this case.
>
>The "median" result is the number of hands that you will exceed
>exactly 50% of the time. To get that figure, I can use my
program,
>Dunbar's Risk Analyzer for Video Poker. I simply look for the
>number of hands that gives me a 50% RoR. That number of hands is
>134.
>
>You can do the same thing for the other game you were interested
in.
>
>--Dunbar
>
>* I got the 82% figure from DRA-VP also.I don't think that answered Mr. Flusher's question. I think he was
asking for the mean number of hands it takes to first get to the
point
of being behind by $20, not the mean number of hands each $20 loss
takes, which your figure of 655 hands indicates. Losing $20 before
655 hands are played 82% of the time suggests that the mean number
of
hands it takes to first lose $20 is much less than 655. 655 would
be
the mean only if, after losing $20, more play could be done which
will
often result in winning some back and later returning to the $20
net
loss or often losing another $20 or more and returning to the
original
$20 net loss. I don't know what the answer is, but I'd guess it
would
be close, but not exactly to, the 134 hands that is the median.
Only
assuming, as I do, the results aren't distributed evenly around 134
hands would it be different.
Dang, but you're absolutely correct, Tom. I'll run a sim that will
keep track of how many hands were played until $20 ruin (for,
say, 100,000 trials), and get the mean value that way.
Thanks for catching my error!
--Dunbar
I don't think I was wrong after all, Tom. The reason the mean is as
high as 655 hands is simply because occasionally a HUGE number of
hands will be played. Sometimes the player will get to play over
50,000 hands before losing $20. That happened 3 times in a 40,000-
trial sim I just ran. Those kind of "lucky" streaks have the effect
of pulling the mean way above the median.
The fact that 82% of the time you will go broke in 655 hands doesn't
tell us anything about the mean number of hands.
Here's another sim I just ran. I let 1,000,000 trials go until $20
was lost, and I counted the total hands played. The mean number of
hands played for those 1,000,000 trials is 659.7 I'm pretty sure
the difference between 659.7 and the theoretical 655.7 is just noise.
--Dunbar
--- In vpFREE@yahoogroups.com, Tom Robertson <thomasrrobertson@...>
wrote:
>--- In vpFREE@yahoogroups.com, "Royal Flusher" <royalflusher@>
>wrote:
>>
>> Let's say I'm playing 8/5 bonus or 9/6 jacks.
>>
>> Assuming a short term play with no royal, how would one
calculate
>the
>> average number of hands from a given stake.
>>
>> For example, 0.25 8/5 bonus no royal, put in a $20 - whats the
>average
>> number of hands you can play on that $20 before its gone
assuming
>> perfect play?
>
>For 9/6 JOB, the answer is 655 hands or 134 hands, depending on
what
>you mean by "average".
>
>Do you want to know the mean number of hands you will play?
(That's
>the expected value of the number of hands you will play.) Or, do
>you want to know the median number of hands you will get to play?
>
>The "mean" number of hands is gotten directly from the ev and the
>bankroll. 9/6 JOB returns 99.54%, so your expected loss is 0.46%
of
>what you bet. If you want to rule out the royal, then you miss
out
>on another 1.98% of ev. Your total loss rate is 2.44%
>
>If you are playing a 25c 5-coin machine, you are betting 1.25 a
>hand. You are losing "on average" 2.44% * 1.25 = 3.05 cents a
pull.
>You can "expect" to get $20.00/0.0305 = 655.7 hands. In other
>words, if you started with $20 a million times, and you added up
all
>the hands you played before you went broke and divided by 1
million,
>your answer would be pretty close to 655.7.
>
>Even though the "mean" number of hands you will get to play from
>your $20 is 655.7, there is an 82% chance you will go broke
before
>655 hands!* The median is much less than the mean, in this case.
>
>The "median" result is the number of hands that you will exceed
>exactly 50% of the time. To get that figure, I can use my
program,
>Dunbar's Risk Analyzer for Video Poker. I simply look for the
>number of hands that gives me a 50% RoR. That number of hands is
>134.
>
>You can do the same thing for the other game you were interested
in.
>
>--Dunbar
>
>* I got the 82% figure from DRA-VP also.I don't think that answered Mr. Flusher's question. I think he was
asking for the mean number of hands it takes to first get to the
point
of being behind by $20, not the mean number of hands each $20 loss
takes, which your figure of 655 hands indicates. Losing $20 before
655 hands are played 82% of the time suggests that the mean number
of
hands it takes to first lose $20 is much less than 655. 655 would
be
the mean only if, after losing $20, more play could be done which
will
often result in winning some back and later returning to the $20
net
loss or often losing another $20 or more and returning to the
original
$20 net loss. I don't know what the answer is, but I'd guess it
would
be close, but not exactly to, the 134 hands that is the median.
Only
assuming, as I do, the results aren't distributed evenly around 134
hands would it be different.