Not true, I was using the math for one known fair dice, and one known biased dice.....and my math was correct in showing no more sevens are expected than normal distribution. EE's math trying to show that there would be more sevens than expectation is flat out wrong...period! Read Random Roller's post, it is spot on!
Absolutely wrong! You are either an idiot or a troll, or probably both! All you have to do to prove this is set one bias dice on a 4....leave it there! Roll another fair dice along side it.....you can roll that one fair dice till hell freezes over, and the odds of rolling a 7 combined with the bias 4 dice is always 1 in 6 since the 7 can only be rolled with a 3 on the fair dice, paired with the bias dice 4!
I am really enjoying this debate I have had several chuckles and smiles as I am reading the responses from every one I picture all of you guys hands and arms flailing reminds me of the old neighbor hood dagos disguising whose mothers sauce is the best....I do have a question who will win the car race tonight between GAS MONKEY and STREET OUTLAWS
rendom_roller Here is the BEEF ~ Unfortunately, random_ roller's Mathematics is Flawed for the reasons outlined below: 1) When dealing with that Pesky Extra 3, it can only combine with the #'s on the Second Dice, i.e. #'s 1 to 6, Yielding #'s (4), (5), (6), (7), (8), & (9), this appears to be agreed upon, by random_roller. 2) When dealing with Six Sided Dice, the following Dice Table Must be used to calculate things. I have reproduced it below: Total on dice for Pairs of dice, Probability 2 1+1,....1/36 = 3% 3 1+2,....2+1 2/36 = 6% 4 1+3,....2+2, 3+1 3/36 = 8% 5 1+4,....2+3, 3+2, 4+1 4/36 = 11% 6 1+5,....2+4, 3+3, 4+2, 5+1 5/36 = 14% 7 1+6,....2+5, 3+4, 4+3, 5+2, 6+1 6/36 = 17% 8 2+6,....3+5, 4+4, 5+3, 6+2 5/36 = 14% 9 3+6,....4+5, 5+4, 6+3 4/36 = 11% 10 4+6,..5+5, 6+4 3/36 = 8% 11 5+6,...6+5 2/36 = 6% 12 6+6,...1/36 = 3% 3) Now as only the #'s (4), (5), (6), (7), (8), & (9), can appear when combined with that Extra Pesky (3), a Dice table for that condition would eliminate the 2, 3, 10, 11, & 12 from the above table. 4) Add them up & you will see that this eliminates 9/36 of the chances, leaving 27/36. 5) Now those (6) #'s i.e. (4), (5), (6), (7), (8), & (9) will each appear at the same frequency on a full run of two Six Sided Dice! 6) As we now have 27, not the ORIGINAL 36 & an Equal distribution of the (6) #'s, we have a Frequency of 6/27 or .2222222 for each! 7) Frequency of (7's) was .1666666, for an INCREASE in (7's) of .2222222 - .1666666, or 0.0555556, or an increase of 5.555556% of (7's) Yes, the above is the TRUE Mathematics for this condition & is 100% CORRECT! eagleeye2
Why don I even bother? You can't count some of the rolls for some purposes and ignore other rolls for other purposes, which you are doing in your "meth" (typo intentional). The 2s (1), 3s (2), 10s (3), 11s (2), and 12s (1) in the 36 rolls unaffected by the "bias" remain part of the calculation. You don't get to eliminate those rolls from the calculation to suit your needs. They're only eliminated when the additional "pesky 3" on Die 2 ("Bias") is rolled (1x per 7 rolls in your example, or 6 rolls out of 42). Leave the bullshit math to the elected officials trying to fleece the taxpaying public. Below is the correct representation based upon 42 rolls, with the extra 3 per every 7 rolls (1,2,3,4,5,6,3) for the Die 2 ("Bias"), per your example. The orange color coded cells (rolls 1-36) reflect the expected distribution of 2 fair die rolled together. The green colored cells (rolls 37-42) are the results of the Die 1 (Fair) and Die 2 ("Bias"), when the "pesky 3" (your description) appears. The yellow colored cells have to do with the rolled number 7.
Several posters here that are very schooled in mathematics and probabilities, addressed this issue when EE first started proposing that all, and any, bias in the dice always produced more sevens than expectation. But EE just continues to deny that his analysis is flawed, always with an attack on the person offering the correct analysis...."Dumb as a Rock", etc. A typical response from someone that knows their position is wrong or weak, but can not accept that fact......cognitive dissonance.
Fact: There is no way a fair dice paired with a weighted dice will produce more sevens than normal expectation...none. With a weighted bias, the only way to have an increase in expected sevens is to have both dice weighted on opposite sides........1-6, 2-5, 4-3. I'll leave the five dice in the bowl selection for each shooter, as the "fly in the ointment" problem.
Onautopilot, You are an absolute IDIOT, any Weight Biased Die, when rolled with a FAIR die will produce more than the statistically indicated # of (7's), Period! random_roller's chart Verifies this Via the 5.5% Increase in 7's, as shown n the GREEN portion of the CHART in Post # 748. Only ones Blind as a BAT & Dumb as a ROCK (like Onautopilot & FredP) cannot see that the GREEN Portion off the table contains NO #'s Totaling (2), (3), (10), (11), or (12) & thus it's contribution to the #'s thrown VERIFIES an increase in the # of (7's) thrown! eagleeye2
Oh no! Eagleturd is wrong! Heaven forbid! What say you now Turdy? Wait I know: " RR you are blind as a bat and dumb as a rock". My math is 100% correct and cannot be denied! (even though It was proven wrong many times)".
This response could only come from a complete innumerate, or a Troll...or both! When faced with facts and mathematics that debunk EE's flawed math analysis of bias dice, he elects to respond with personal attacks, and rejection of a true analysis of the bias dice configurations he proposes. EE maintains that the Midnight Skullker, Random Roller, and several others schooled in math are wrong, and he is the only one that is a 100% right. The smart money is on the schooled mathematicians...at least mine is.
FredP, Your Stupidity Continues to Shine Through! randon_roller simply displayed the #'s that I outlined in My Post # 744 as they would appear fully laid out, in his Post # 748. randon_roller has NOT drawn ANY CONCLUSIONS from that display & such display is only different than the one I posted in Post # 744, by laying out the full set. It is obvious to all but those set in their ERROR filled ways that both Posts # 744 & # 748 say the same thing! Onautopilot & FredP are an absolute IDIOTS, any Weight Biased Die, when rolled with a FAIR die will produce more than the statistically indicated # of (7's), Period! random_roller's chart Verifies this Via the 5.5% Increase in 7's, as shown n the GREEN portion of the CHART in Post # 748. Only ones Blind as a BAT & Dumb as a ROCK (like Onautopilot & FredP) cannot see that the GREEN Portion off the table contains NO #'s Totaling (2), (3), (10), (11), or (12) & thus it's contribution to the #'s thrown VERIFIES an increase in the # of (7's) thrown! eagleeye2
Onautopilot, Don't you think that you ought to wait until random_roller draws a CONCLUSION as to the Increased # of (7's), before you go out on a limb like that??? eagleeye2
Random Roller: Your numbers match my post for a die bias where rolling the 3 has twice the probability of the other numbers. Not exactly that extra threes are thrown. The probability of rolling a 7 is the sum of the probabilities for each combination of numbers resulting in 7. Repeat this for all numbers as shown below. The three numbers at the end are in this order: biased prob----fair prob------difference P(sum =2 ) = (1/7)(1/6) = .0238 .0278 -.0040 P(sum = 3) = (1/7)(1/6) + (1/7)(1/6) = .0476 .0556 -.0080 P(sum = 4) = (1/7)(1/6) + (1/7)(1/6) + (2/7)(1/6) = .0952 .0833 +.0115 P(sum = 5) = (1/7)(1/6) + (1/7)(1/6) + (2/7)(1/6) + (1/7)(1/6) = .1190 .1111 +.0079 P(sum = 6) = (1/7)(1/6) + (1/7)(1/6) + (2/7)(1/6) + (1/7)(1/6) + (1/7)(1/6) = .1428 .1389 +.0039 P(sum =7) = (1/7)(1/6) + (1/7)(1/6) + (2/7)(1/6) + (1/7)(1/6) + (1/7)(1/6) + (1/7)(1/6) = .1666 .1666 0 P(sum =8) = (1/7)(1/6) + (2/7)(1/6) + (1/7)(1/6) + (1/7)(1/6) + (1/7)(1/6) =.1428 .1389 +.0039 P(sum =9) = (2/7)(1/6) + (1/7)(1/6) + (1/7)(1/6) + (1/7)(1/6) = .0952 .0833 +.0119 P(sum =10) = (1/7)(1/6) + (1/7)(1/6) + (1/7)(1/6) = .0714 .0833 -.0119 P(sum =11) = (1/7)(1/6) + (1/7)(1/6) = .0476 .0556 -.0080 P(sum =12) = (1/7)(1/6) = .0238 .0278 -.0040 You can see that the probability of rolling 2,3,10,11,12 is slightly reduced and the probability of rolling the 4,5,6,8, and 9 is slightly increased.The 7 is unaffected. In general, any combination of numbers that need a 3 on die one will have a slightly higher probability while the other combinations will have a lower probability.
