Di test, valid or not ?

Look at each individual dice. All number from 1 to 6 have equal chance. Then we should take a baby step to test to see if someone can influence the dice. If we set the dices 66,55,11,22. If someone can maintain straight axis and hit the back wall and landed without showing a 4 and 3 then will you call him that he can influence the dices ? Mathematically 3/4 will show up at 25% by 4 faces / 16faces = .25 = 25%
How about :
5% 3 and or 4 show up ?
10% 3 and or 4 show up ?
15% 3 and or 4 show up ?
20% 3 and or 4 show up ?

say at 100 rolls ? 1000 rolls ? 10,000 rolls ?
Will this test help to prove if one can Di ?

Quote: ideas2product

Look at each individual dice. All number from 1 to 6 have equal chance. Then we should take a baby step to test to see if someone can influence the dice. If we set the dices 66,55,11,22. If someone can maintain straight axis and hit the back wall and landed without showing a 4 and 3 then will you call him that he can influence the dices ? Mathematically 3/4 will show up at 25% by 4 faces / 16faces = .25 = 25%
How about :
5% 3 and or 4 show up ?
10% 3 and or 4 show up ?
15% 3 and or 4 show up ?
20% 3 and or 4 show up ?

say at 100 rolls ? 1000 rolls ? 10,000 rolls ?
Will this test help to prove if one can Di ?

Your math is off -- the chances of a 3 or 4 appearing at least once in your scenario under the assumption of uniform faces is 5/9 (55.56%).

But if someone can actually maintain a straight axis after hitting the wall and coming to rest, that's definitely influence. Even doing that one time in ten rolls on average will give you a big edge. I've never seen a video of someone able to do this. If the dice go off-axis, bounce around, and then go *back* to the axis, that doesn't count; that's going to happen anyway.

Math quote (Your math is off -- the chances of a 3 or 4 appearing at least once in your scenario under the assumption of uniform faces is 5/9 55.56%)

Maybe you are misunderstand what i am trying to say. If you set 1 dice at 6512 then you only have 3 and 4 on either left or right. Just roll 1 dice only, assume one can throw a straight axis ,hit the back wall and the dice show only 6,5,2 or 1. Would i be wrong if 3 have 1/6 chance and 4 have 1/6. 3 and 4 together have 1/3 chance to show up mathematically. Now combine by 2 dices 3 and 4 have total 4 faces. 6521 have 8 faces. Total both dices have 12 faces. So 4 faces of 3/4 divide by 12 faces = 33% or 1/3 to be exactly.

I try to find a test ( not perfect) where we can prove or disprove Di

So test for yourself see if you can throw with setting 66551122 to see how many dice show face # 3 ( not 1+2) or face #4 ( 2+2 or 1+3)show up.
If you can throw without show up at less than 33% then you are have the ability to influence the dice. Of course, the lower % the better you are.

Math quote (Your math is off -- the chances of a 3 or 4 appearing at least once in your scenario under the assumption of uniform faces is 5/9 55.56%)

Maybe you are misunderstand what i am trying to say. If you set 1 dice at 6512 then you only have 3 and 4 on either left or right. Just roll 1 dice only, assume one can throw a straight axis ,hit the back wall and the dice show only 6,5,2 or 1. Would i be wrong if 3 have 1/6 chance and 4 have 1/6. 3 and 4 together have 1/3 chance to show up mathematically. Now combine by 2 dices 3 and 4 have total 4 faces. 6521 have 8 faces. Total both dices have 12 faces. So 4 faces of 3/4 divide by 12 faces = 33% or 1/3 to be exactly.

I try to find a test ( not perfect) where we can prove or disprove Di

So test for yourself see if you can throw with setting 66551122 to see how many dice show face # 3 ( not 1+2) or face #4 ( 2+2 or 1+3)show up.
If you can throw without show up face #3 or face #4 at less than 33% then you are have the ability to influence the dice. Of course, the lower % the better you are.

Quote: ideas2product

Maybe you are misunderstand what i am trying to say. If you set 1 dice at 6512 then you only have 3 and 4 on either left or right. Just roll 1 dice only, assume one can throw a straight axis ,hit the back wall and the dice show only 6,5,2 or 1. Would i be wrong if 3 have 1/6 chance and 4 have 1/6. 3 and 4 together have 1/3 chance to show up mathematically. Now combine by 2 dices 3 and 4 have total 4 faces. 6521 have 8 faces. Total both dices have 12 faces. So 4 faces of 3/4 divide by 12 faces = 33% or 1/3 to be exactly.

No, I understand exactly what you're trying to say and your math is off. Write out each combination of two dice and then circle those with either a 3 or 4 in them. What fraction are circled? It's far more than 1/3.

Quote: MathExtremist

No, I understand exactly what you're trying to say and your math is off. Write out each combination of two dice and then circle those with either a 3 or 4 in them. What fraction are circled? It's far more than 1/3.

OH M.E. Now you've done it. You've got me screaming "At least one of the dice is a two." It took me weeks of therapy to get that out of my head.