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Topic closed. 12 replies. Last post 6 years ago by RL-RANDOMLOGIC.

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United States
Member #93947
July 10, 2010
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 Posted: September 1, 2010, 2:54 pm - IP Logged

For the believers in the Gambler's Fallacy mucking around in my Fooled By Randomness thread, this is going to be a real mind bender!

• I have four 6 sided FAIR dice, labeled A, B, C, D.

• They are numbered differently than normal dice, but each contain some combination of the set [0 1 2 3 4 5 6].

• If you pick any one of the 4, I can select one of the 3 remaining, and in a simple game of match, I can beat you,

on average, 2 out of 3 times.

• When you are convinced you are losing 2/3 of the time, you can then choose another of the 4 dice.

• Once you have chosen again, I will choose yet another from the 3 remaining, and...

• Proceed to beat you 2 out of 3 throws again!

• And so on...

Don't believe it?  Check here:

http://www.maa.org/mathland/mathtrek_04_15_02.html

Unfortunately for those who DO NOT TRUST Probability Theory, they will be forced to actually construct a set of

these dice, so they can do the simulation themselves, until they are convinced!

Zeta Reticuli Star System
United States
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January 17, 2006
10356 Posts
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 Posted: September 2, 2010, 7:37 pm - IP Logged

"• I have four 6 sided FAIR dice, labeled A, B, C, D.

• They are numbered differently than normal dice, but each contain some combination of the set [0 1 2 3 4 5 6]."

Who is going to get into a dice game with someone using a die with 0 on it? Or a blank die with no pips. Not me.

Those who run the lotteries love it when players look for consistency in something that's designed not to have any.

There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.

United States
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July 10, 2010
2180 Posts
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 Posted: September 3, 2010, 1:21 am - IP Logged

"• I have four 6 sided FAIR dice, labeled A, B, C, D.

• They are numbered differently than normal dice, but each contain some combination of the set [0 1 2 3 4 5 6]."

Who is going to get into a dice game with someone using a die with 0 on it? Or a blank die with no pips. Not me.

What did you think of the NON-Transitivity of the Probability Relationships among the 4 dice?

These dice are fun in a bar, playing for pennies.  Initially, most people think they're loaded.

When you let them keep picking a new one, they then suspect you have a trick way of throwing

them.  When you let one of their friends throw for you, and you still beat them, depending on

how much they had to drink, they get a very strange look on their face.  I wouldn't suggest doing

this with people who have a strong belief in the devil!

Zeta Reticuli Star System
United States
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January 17, 2006
10356 Posts
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 Posted: September 3, 2010, 6:48 pm - IP Logged

Yeah, good point, but I still would't get in such a game.

I forget exactly how it goes but there''s a bar room game of Solitaire for money, you pay \$52 for the deck and then try to win it back, and more. Guess who usually wins that one!

Those who run the lotteries love it when players look for consistency in something that's designed not to have any.

There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.

United States
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June 1, 2009
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 Posted: September 3, 2010, 8:55 pm - IP Logged

But what would happen when the chosen number containing the remaining ones were 2 out of 3 in the first and seconds throws?

Also, How would the bet be generated when it could've been based on an average probability theory not actual construct of a typical set?

Whether the 4 number selection was only 2/3 of the chosen 4 dice that remain, or whether the simulation was in fact a correct combination wouldn't matter because I'd rather have the ability to introduce the randomness delusion theory exceptions to the rule, and then  deem it debatable.

United States
Member #93947
July 10, 2010
2180 Posts
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 Posted: September 3, 2010, 9:53 pm - IP Logged

But what would happen when the chosen number containing the remaining ones were 2 out of 3 in the first and seconds throws?

Also, How would the bet be generated when it could've been based on an average probability theory not actual construct of a typical set?

Whether the 4 number selection was only 2/3 of the chosen 4 dice that remain, or whether the simulation was in fact a correct combination wouldn't matter because I'd rather have the ability to introduce the randomness delusion theory exceptions to the rule, and then  deem it debatable.

joker17,

http://www.maa.org/mathland/mathtrek_04_15_02.html

This isn't about a real dice game played for money.

It's about math and probability, which all gamblers should know cold!

--Jimmy4164

United States
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June 1, 2009
5345 Posts
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 Posted: September 4, 2010, 10:46 am - IP Logged

joker17,

http://www.maa.org/mathland/mathtrek_04_15_02.html

This isn't about a real dice game played for money.

It's about math and probability, which all gamblers should know cold!

--Jimmy4164

I understand that it's not about a real dice game, but my point was that the die which was the opposing picks retained prior to the elevation of the data dealing with variance, was the one that the probability component devised, and not the additional sets which can be determined by the factors preceding the nontransitive dice, but not to be confused with the magic square properties written about earlier when the proposal was made to triumph in 10 games or less.

In other words ,beating the second die, twice as much but not less than 12 times can be misconstrued to look as if the mistaken assumption was determined by the association with the same set of the 3 transitive dice paradox, which also constitutes a 4.7 degree of violation assesment when dealing with such paradoxes. Although, these equations may show the skewed elevations within the deviances soley based on the properties within the subset parameters of 5.0 or higher. But it is highly unlikely that they will.

Furthermore, the fourth die is 3 times as likely as the second and 1st die, to produce similar paradoxes when the receding rows of numbers prior to the original sets put forth in congruence, which BTW, was determined as invalid, therefore no amount of statistcal provision was calculated by the study to provide sufficient mean numbers and variants to come to the conclusion mentioned in the second paragraph above.

United States
Member #93947
July 10, 2010
2180 Posts
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 Posted: September 4, 2010, 12:49 pm - IP Logged

I understand that it's not about a real dice game, but my point was that the die which was the opposing picks retained prior to the elevation of the data dealing with variance, was the one that the probability component devised, and not the additional sets which can be determined by the factors preceding the nontransitive dice, but not to be confused with the magic square properties written about earlier when the proposal was made to triumph in 10 games or less.

In other words ,beating the second die, twice as much but not less than 12 times can be misconstrued to look as if the mistaken assumption was determined by the association with the same set of the 3 transitive dice paradox, which also constitutes a 4.7 degree of violation assesment when dealing with such paradoxes. Although, these equations may show the skewed elevations within the deviances soley based on the properties within the subset parameters of 5.0 or higher. But it is highly unlikely that they will.

Furthermore, the fourth die is 3 times as likely as the second and 1st die, to produce similar paradoxes when the receding rows of numbers prior to the original sets put forth in congruence, which BTW, was determined as invalid, therefore no amount of statistcal provision was calculated by the study to provide sufficient mean numbers and variants to come to the conclusion mentioned in the second paragraph above.

joker17,

Mr Peterson's description over at MathTrek makes a lot more sense to me than yours.  Of course, this could be a deficiency in my ability to comprehend [your] analysis. Are you going to be one of those members forced to craft a set of the dice so you can conduct the simulation yourself before you will be convinced?  On 2nd thought, based on what you wrote above, maybe I shouldn't jump to the conclusion that you are not convinced!

--Jimmy4164

United States
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June 1, 2009
5345 Posts
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 Posted: September 4, 2010, 3:23 pm - IP Logged

joker17,

Mr Peterson's description over at MathTrek makes a lot more sense to me than yours.  Of course, this could be a deficiency in my ability to comprehend [your] analysis. Are you going to be one of those members forced to craft a set of the dice so you can conduct the simulation yourself before you will be convinced?  On 2nd thought, based on what you wrote above, maybe I shouldn't jump to the conclusion that you are not convinced!

--Jimmy4164

I really can't explain it any clearer than the above explanations. But here are the equations I used to determine the problems facing the variance degradation and paradoxal curve coefficient flux parameters.

J(y,z)=\int_a^b$y''z' + xyz'' + z'''y^2$ dx Answer: {xz'' + (2y + 1)z''' = 0 {2y' + xy'' - 6y'y'' + (2y+1)y''' = 0..

(x-12) 2 + (y+9) 2 = 15 (y+9) 2 = 15 - (x-3) 2 y+9 = (15 - (x-12) 2 ) y =(15 - (x-12) 2 )-.. ( [3x - 3x^2 +1]^744 ) x ( [- 3x + 3x^2 +1]^745 ) ??

United States
Member #93947
July 10, 2010
2180 Posts
Offline
 Posted: September 7, 2010, 1:34 pm - IP Logged

For the believers in the Gambler's Fallacy mucking around in my Fooled By Randomness thread, this is going to be a real mind bender!

• I have four 6 sided FAIR dice, labeled A, B, C, D.

• They are numbered differently than normal dice, but each contain some combination of the set [0 1 2 3 4 5 6].

• If you pick any one of the 4, I can select one of the 3 remaining, and in a simple game of match, I can beat you,

on average, 2 out of 3 times.

• When you are convinced you are losing 2/3 of the time, you can then choose another of the 4 dice.

• Once you have chosen again, I will choose yet another from the 3 remaining, and...

• Proceed to beat you 2 out of 3 throws again!

• And so on...

Don't believe it?  Check here:

http://www.maa.org/mathland/mathtrek_04_15_02.html

Unfortunately for those who DO NOT TRUST Probability Theory, they will be forced to actually construct a set of

these dice, so they can do the simulation themselves, until they are convinced!

All!

Joker17 is having some fun here, but I really thought there would be sincere interest in this Paradoxical Phenomenon.

If you have aspirations of logically deducing a method to predict lottery draws, this is the kind of challenge you should be accepting in an effort to hone your skills!

--Jimmy4164

United States
Member #93947
July 10, 2010
2180 Posts
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 Posted: September 10, 2010, 10:10 am - IP Logged

Here's some more background for those interested...

http://mathforum.org/isaac/problems/prob1.html?

United States
Member #93947
July 10, 2010
2180 Posts
Offline
 Posted: September 10, 2010, 10:16 am - IP Logged

Here's some more background for those interested...

http://mathforum.org/isaac/problems/prob1.html?

Sorry for that "?"

http://mathforum.org/isaac/problems/prob1.html

United States
Member #59354
March 13, 2008
3986 Posts
Offline
 Posted: October 1, 2010, 2:11 am - IP Logged

For the believers in the Gambler's Fallacy mucking around in my Fooled By Randomness thread, this is going to be a real mind bender!

• I have four 6 sided FAIR dice, labeled A, B, C, D.

• They are numbered differently than normal dice, but each contain some combination of the set [0 1 2 3 4 5 6].

• If you pick any one of the 4, I can select one of the 3 remaining, and in a simple game of match, I can beat you,

on average, 2 out of 3 times.

• When you are convinced you are losing 2/3 of the time, you can then choose another of the 4 dice.

• Once you have chosen again, I will choose yet another from the 3 remaining, and...

• Proceed to beat you 2 out of 3 throws again!

• And so on...

Don't believe it?  Check here:

http://www.maa.org/mathland/mathtrek_04_15_02.html

Unfortunately for those who DO NOT TRUST Probability Theory, they will be forced to actually construct a set of

these dice, so they can do the simulation themselves, until they are convinced!

jimmy

no brainer, I let you select first and then I choose which I will use after you make your

selection.

RL

Working on my Ph.D.  "University of hard Knocks"

I will consider the opinion that my winnings are a product of chance if you are willing to consider

they are not.  Many great discoveries come while searching for something else

Trump / 2016 & 2020

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