[vpFREE] Re: A Probability Question

 

This is easier than it may first seem.   Suppose 1M hands at $5 a spin.  2.5% of that is $125,000.


So now you will just want to calculate the probability of losing $125,000 or less.   With a flat game like 7/5 BP, you can pretty safely go with expected value after a million hands.   That game drops roughly 2%, so your expected value is about $25,000.   To lose would be to run just awful.   Assuming everything else besides royals runs about average (which it almost has to be on this game) you'd have to be 6.25 royals below average.  

I have forgotten the math, but there is a fairly accurate and simple formula that will tell you the odds of getting  fewer than 17 events  when the expected number of events is 23.      So, for example, if you were flipping a coin 46 times, what is the odds of only getting 16 heads (or less).  The answer to that question will be very close to the correct answer for this situation. 

If this is a real life game, you've got a decent return, but nothing special.  If you crank out 1000 hands an hour, you make roughly $25 an hour.  As for bankroll....   This game will be about as mellow as can be and I'm thinking 20K should be viable, though that assumes you have other money for things like eating.

Much easier that doing the math is to run a bankroll analysis on one of the many VP programs out there.
 That will, in effect, put your actual break-even point at a loss of $25,000 on the game.  It should tell you P(losing 25K or less), which will also be the answer to your question.  The VP program I have allows you to include cashback, which then allows you to simply look to see the answer with no need to do any further analysis to understand what the output means.   What you should see, if you have done it right, is an "average" result of +25,000 for a million hands and the frequency of results that deviate by more than $25,000 will be very low  (and only half of those will be losses)   If the software is good, it will also give you the probability curves for all the outcomes.


QZ

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Posted by: what7do7you7want@yahoo.com
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[vpFREE] Re: A Probability Question

 

Thanks for the responses, guys. 

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Posted by: mickeycrimm@yahoo.com
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Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

 

Or 1000 - (n<=1000).

Thx.


------ Original message------
Date: Mon, Feb 13, 2017 16:57
To: h_dunbar@hotmail.com [vpFREE];
Cc:
Subject:Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

Oh. Sorry. Now I get it. A 1000 - 12 long interval, as opposed to an interval of length 12.

Sent from my LG K10, an AT&T 4G LTE smartphone

------ Original message------
Date: Mon, Feb 13, 2017 16:55
To: h_dunbar@hotmail.com [vpFREE];
Cc:
Subject:Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

The 1000 as opposed to the 'n' aspect of it.

Thx.

Sent from my LG K10, an AT&T 4G LTE smartphone

------ Original message------
Date: Mon, Feb 13, 2017 16:44
To: h_dunbar@hotmail.com [vpFREE];
Cc:
Subject:Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

"BUT, those are the probabilities of hitting blanks on your first "n" hands or your next "n" hands.   If you're wondering about, say, losing 12 straight hands some time during a 1000-hand play, then the probabilities would be much higher because you've got 988 possible starting points for your bad streak.

--Dunbar"

Can I possibly get you to lay that out mathematically with possibly a little accompanying explanation? I just can't get a handle on it.

Thx.

------ Original message------
From: h_dunbar@hotmail.com [vpFREE]
Date: Sun, Feb 5, 2017 05:27
Cc:
Subject:[vpFREE] Re: what's % of losing hands. DDB 9/6

 

"What is unusual number of losses in row, same game [9/6DDB]?  Would 8, 10 or 12 losing games in row be unusual?"


The chance of getting "nothing" in the 9/6 DDB pay table is 55.28%.   

The chance of getting nothing twice in a row is 55.28% x 55.28% = 30.56%
The chance of getting nothing 3 times in a row is 55.28% x 55.28% x 55.28% = 16.89%
and so on...

Stick =55.28%^8 into any spreadsheet, and you'll find that the chance of hitting blanks on your next 8 hands is 0.87%, about one in 115.

Here are some values (that'll probably format miserably):

streak 1 2 3 4 5 6 7 8 9 10 11 12
chance 55.28% 30.56% 16.89% 9.34% 5.16% 2.85% 1.58% 0.87% 0.48% 0.27% 0.15% 0.08%
1 in … 1.8 3.3 6 11 19 35 63 115 207 375 679 1228

BUT, those are the probabilities of hitting blanks on your first "n" hands or your next "n" hands.   If you're wondering about, say, losing 12 straight hands some time during a 1000-hand play, then the probabilities would be much higher because you've got 988 possible starting points for your bad streak.

--Dunbar


---In vpFREE@yahoogroups.com, wrote :

Good afternoon folks.


Say we play 1000 hands, what is the theoretical loss %?   9/6 DDB. I'm thinking if average is ,say, 55% loss then  550 of 1000 would be losers.  Thats assuming I'm thinking right.


What is unusual number of losses in row, same game?  Would 8, 10 or 12 losing games in row be unusual? 


Hoping someone has the info at their finger tips?

 


Cheers.....Jeep

__._,_.___
Posted by: "APEppInk@yahoo.com" <APEppInk@yahoo.com>
Reply via web post • Reply to sender • Reply to group • Start a New Topic • Messages in this topic (8)
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Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

 

Oh. Sorry. Now I get it. A 1000 - 12 long interval, as opposed to an interval of length 12.

Sent from my LG K10, an AT&T 4G LTE smartphone

------ Original message------
Date: Mon, Feb 13, 2017 16:55
To: h_dunbar@hotmail.com [vpFREE];
Cc:
Subject:Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

The 1000 as opposed to the 'n' aspect of it.

Thx.

Sent from my LG K10, an AT&T 4G LTE smartphone

------ Original message------
Date: Mon, Feb 13, 2017 16:44
To: h_dunbar@hotmail.com [vpFREE];
Cc:
Subject:Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

"BUT, those are the probabilities of hitting blanks on your first "n" hands or your next "n" hands.   If you're wondering about, say, losing 12 straight hands some time during a 1000-hand play, then the probabilities would be much higher because you've got 988 possible starting points for your bad streak.

--Dunbar"

Can I possibly get you to lay that out mathematically with possibly a little accompanying explanation? I just can't get a handle on it.

Thx.

------ Original message------
From: h_dunbar@hotmail.com [vpFREE]
Date: Sun, Feb 5, 2017 05:27
Cc:
Subject:[vpFREE] Re: what's % of losing hands. DDB 9/6

 

"What is unusual number of losses in row, same game [9/6DDB]?  Would 8, 10 or 12 losing games in row be unusual?"


The chance of getting "nothing" in the 9/6 DDB pay table is 55.28%.   

The chance of getting nothing twice in a row is 55.28% x 55.28% = 30.56%
The chance of getting nothing 3 times in a row is 55.28% x 55.28% x 55.28% = 16.89%
and so on...

Stick =55.28%^8 into any spreadsheet, and you'll find that the chance of hitting blanks on your next 8 hands is 0.87%, about one in 115.

Here are some values (that'll probably format miserably):

streak 1 2 3 4 5 6 7 8 9 10 11 12
chance 55.28% 30.56% 16.89% 9.34% 5.16% 2.85% 1.58% 0.87% 0.48% 0.27% 0.15% 0.08%
1 in … 1.8 3.3 6 11 19 35 63 115 207 375 679 1228

BUT, those are the probabilities of hitting blanks on your first "n" hands or your next "n" hands.   If you're wondering about, say, losing 12 straight hands some time during a 1000-hand play, then the probabilities would be much higher because you've got 988 possible starting points for your bad streak.

--Dunbar


---In vpFREE@yahoogroups.com, wrote :

Good afternoon folks.


Say we play 1000 hands, what is the theoretical loss %?   9/6 DDB. I'm thinking if average is ,say, 55% loss then  550 of 1000 would be losers.  Thats assuming I'm thinking right.


What is unusual number of losses in row, same game?  Would 8, 10 or 12 losing games in row be unusual? 


Hoping someone has the info at their finger tips?

 


Cheers.....Jeep

__._,_.___
Posted by: "APEppInk@yahoo.com" <APEppInk@yahoo.com>
Reply via web post • Reply to sender • Reply to group • Start a New Topic • Messages in this topic (7)
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.

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Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

 

The 1000 as opposed to the 'n' aspect of it.

Thx.

Sent from my LG K10, an AT&T 4G LTE smartphone

------ Original message------
Date: Mon, Feb 13, 2017 16:44
To: h_dunbar@hotmail.com [vpFREE];
Cc:
Subject:Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

"BUT, those are the probabilities of hitting blanks on your first "n" hands or your next "n" hands.   If you're wondering about, say, losing 12 straight hands some time during a 1000-hand play, then the probabilities would be much higher because you've got 988 possible starting points for your bad streak.

--Dunbar"

Can I possibly get you to lay that out mathematically with possibly a little accompanying explanation? I just can't get a handle on it.

Thx.

------ Original message------
From: h_dunbar@hotmail.com [vpFREE]
Date: Sun, Feb 5, 2017 05:27
Cc:
Subject:[vpFREE] Re: what's % of losing hands. DDB 9/6

 

"What is unusual number of losses in row, same game [9/6DDB]?  Would 8, 10 or 12 losing games in row be unusual?"


The chance of getting "nothing" in the 9/6 DDB pay table is 55.28%.   

The chance of getting nothing twice in a row is 55.28% x 55.28% = 30.56%
The chance of getting nothing 3 times in a row is 55.28% x 55.28% x 55.28% = 16.89%
and so on...

Stick =55.28%^8 into any spreadsheet, and you'll find that the chance of hitting blanks on your next 8 hands is 0.87%, about one in 115.

Here are some values (that'll probably format miserably):

streak 1 2 3 4 5 6 7 8 9 10 11 12
chance 55.28% 30.56% 16.89% 9.34% 5.16% 2.85% 1.58% 0.87% 0.48% 0.27% 0.15% 0.08%
1 in … 1.8 3.3 6 11 19 35 63 115 207 375 679 1228

BUT, those are the probabilities of hitting blanks on your first "n" hands or your next "n" hands.   If you're wondering about, say, losing 12 straight hands some time during a 1000-hand play, then the probabilities would be much higher because you've got 988 possible starting points for your bad streak.

--Dunbar


---In vpFREE@yahoogroups.com, wrote :

Good afternoon folks.


Say we play 1000 hands, what is the theoretical loss %?   9/6 DDB. I'm thinking if average is ,say, 55% loss then  550 of 1000 would be losers.  Thats assuming I'm thinking right.


What is unusual number of losses in row, same game?  Would 8, 10 or 12 losing games in row be unusual? 


Hoping someone has the info at their finger tips?

 


Cheers.....Jeep

__._,_.___
Posted by: "APEppInk@yahoo.com" <APEppInk@yahoo.com>
Reply via web post • Reply to sender • Reply to group • Start a New Topic • Messages in this topic (6)
Have you tried the highest rated email app?
With 4.5 stars in iTunes, the Yahoo Mail app is the highest rated email app on the market. What are you waiting for? Now you can access all your inboxes (Gmail, Outlook, AOL and more) in one place. Never delete an email again with 1000GB of free cloud storage.

.

__,_._,___

Re: [vpFREE] Re: what's % of losing hands. DDB 9/6

 

"BUT, those are the probabilities of hitting blanks on your first "n" hands or your next "n" hands.   If you're wondering about, say, losing 12 straight hands some time during a 1000-hand play, then the probabilities would be much higher because you've got 988 possible starting points for your bad streak.

--Dunbar"

Can I possibly get you to lay that out mathematically with possibly a little accompanying explanation? I just can't get a handle on it.

Thx.

------ Original message------
From: h_dunbar@hotmail.com [vpFREE]
Date: Sun, Feb 5, 2017 05:27
Cc:
Subject:[vpFREE] Re: what's % of losing hands. DDB 9/6

 

"What is unusual number of losses in row, same game [9/6DDB]?  Would 8, 10 or 12 losing games in row be unusual?"


The chance of getting "nothing" in the 9/6 DDB pay table is 55.28%.   

The chance of getting nothing twice in a row is 55.28% x 55.28% = 30.56%
The chance of getting nothing 3 times in a row is 55.28% x 55.28% x 55.28% = 16.89%
and so on...

Stick =55.28%^8 into any spreadsheet, and you'll find that the chance of hitting blanks on your next 8 hands is 0.87%, about one in 115.

Here are some values (that'll probably format miserably):

streak 1 2 3 4 5 6 7 8 9 10 11 12
chance 55.28% 30.56% 16.89% 9.34% 5.16% 2.85% 1.58% 0.87% 0.48% 0.27% 0.15% 0.08%
1 in … 1.8 3.3 6 11 19 35 63 115 207 375 679 1228

BUT, those are the probabilities of hitting blanks on your first "n" hands or your next "n" hands.   If you're wondering about, say, losing 12 straight hands some time during a 1000-hand play, then the probabilities would be much higher because you've got 988 possible starting points for your bad streak.

--Dunbar


---In vpFREE@yahoogroups.com, wrote :

Good afternoon folks.


Say we play 1000 hands, what is the theoretical loss %?   9/6 DDB. I'm thinking if average is ,say, 55% loss then  550 of 1000 would be losers.  Thats assuming I'm thinking right.


What is unusual number of losses in row, same game?  Would 8, 10 or 12 losing games in row be unusual? 


Hoping someone has the info at their finger tips?

 


Cheers.....Jeep

__._,_.___
Posted by: "APEppInk@yahoo.com" <apeppink@yahoo.com>
Reply via web post • Reply to sender • Reply to group • Start a New Topic • Messages in this topic (5)
Have you tried the highest rated email app?
With 4.5 stars in iTunes, the Yahoo Mail app is the highest rated email app on the market. What are you waiting for? Now you can access all your inboxes (Gmail, Outlook, AOL and more) in one place. Never delete an email again with 1000GB of free cloud storage.

.

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[vpFREE] Re: what's % of losing hands. DDB 9/6

 

Hi Dunbar


Looking at post, can i deduct;.... On average every 115 hands will have one losing streak of 8 losses in row, in long run.   So in 1150 hands we would see approximately ten  8 game losing streaks in 1150 hands. I am guessing might have to play 11,500 hands to get a close actual average?

Anxious to hear if I figured the info right.  Math is not my best suit.

Cheers.....Jeep

__._,_.___
Posted by: whitejeeps@yahoo.com
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[vpFREE] Re: A Probability Question

 

Out of curiosity, can you add in cash equivalent promotions (making the game "playable") ... say 2.5%, in the case of 7/5 bp?  (equivalent to asking the p that play return > 97.5% after 1 mil hands))



---In vpFREE@yahoogroups.com, <mpeck1@...> wrote :

I wrote a little program to do this type of computation some time ago. If I entered all the numbers correctly the probability of being even or better after 1 million hands of 7/5 BP are about 3.3 x 10^-5, or in other words the odds against are a little worse than 30,000:1. That's with perfect play.

Mike

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Posted by: harry.porter@verizon.net
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