von duck, What you mean is that you will not be MAKING UP any more of your B.S. tonight! What a HOOT you are... eagleeye2

Vig on win only. Had my first issue already. Set the 4 as the point so I took full odds and left the buy up. It hit 1/3 both times. Who'd of thunk that could happen?

If one die always shows 6 than there will be one 6/1 every six rolls. Period. No EE2 can not be right, he has never been right, he uses biased dice to hide the fact he can’t shoot. This has been known 10-15 years now.

I'm not hoping anybody believes anything. I'm looking for the correct answer to the problem, it's that simple. You're looking at the problem with a preconceived notion of what the answer should be, and so far it's proving to be wrong. The other problem is, that you do not have the gumption, to admit that your theory, may be flawed. This is science Man, the pursuit of the truth, it's for not those that " can't handle the truth". There is no mathematics, that you are capable of understanding. Hell man, you can't even understand why I doubled the sample size.

von duck, You sure as HELL do not understand Biased Dice, or the Mathematics involved therein, DUH. All @ Your Peril, assuming you Play Casino CRAPS! eagleeye2

To speed up the process I changed the bias to 3 extra ones every 36 rolls and reduced the other numbers by equal amounts. Would you reduce the other numbers equally, take it all off the 6, or something different?

In my example, I biased the 6 and made up all of the difference on the one. I did this this for simplicity, and it just seems that a 6 biased die, would almost have to produce less ones.

I don't think so...but is there any "Monty Hall paradox" type issue that is being overlooked or anything overlooked as Night Attack noted?

Eagleeye2 has long enough been touting his asinine theory about how “any Weight Unbalance whatsoever, favors the Casino, in that MORE than the Statistically Correct # of SEVENS will automatically occur, whenever there is a Weight UNBALANCE in the DICE!”. Here’s proof that that’s a crock. You can analyze the effects of biased dice using the following example steps and calculations. 1) Assign a bias value to the numbers on one or both dice, e.g. die 1 bias allows the probability of the number 3 to roll two times more than the probability of the other numbers; die 2 is unbiased and rolls all numbers with equal probably. 2) A probability distribution must add up to 1 (this means one of the numbers is sure to roll). Therefore the distributions for the dice will be: Die 1 Die 2 # Probability # Probability 1 p 1 1/6 2 p 2 1/6 3 2p 3 1/6 4 p 4 1/6 5 p 5 1/6 6 p 6 1/6 3) The die 1 sum is set to 1 to find the value of p. 7p=1 p= 1/7 4) The distribution for die 1 is now: # Probability 1 1/7 2 1/7 3 2/7 4 1/7 5 1/7 6 1/7 5) Now consider all the possible ways to roll a 7with two dice: 1-6, 2-5, 3-4, 4-3, 5-2, and 6-1. The probability of each of these events occurring is found by multiplying the probability of each part of the event. For instance, to find the probability of rolling a 1- 6, you must multiply the probability of rolling a 1 by the probability of rolling a 6. The probability of rolling a 7 is the sum of the probabilities for each combination of numbers resulting in 7. 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) = .0238 + .0238 + .0476 + .0238 + .0238 + .0238 P = .166 The unbiased dice value for P is 1/6 = .166 So here’s a case where a bias has no effect on the probability of rolling a 7, which is a direct contradiction to eagleeye2’s “theory”. In fact, if you do a more general calculation you’ll find that any bias of one die together with a fair die results in the same probability of rolling a 7 which is 1/6. (I’ll be glad to show this more general calculation if needed.) When considering a bias on both dice it gets more complicated. You can have biases which increase the probability of rolling a 7, e.g. if die 1 has a bias for rolling more 5s and die 2 has a bias for rolling more 2s. And you can have biases which actually reduce the number of 7s, e.g. if both dice produce a 4 three times more likely than the other numbers. There’re infinite ways to assign biases and infinite results in which more or less 7s are rolled. Note also that dice having opposite sides adding up to 7 doesn’t even enter the picture. To be somewhat crude and insulting (like eagleeye2), this proves he’s full of shit, and since he is a dunce that never got past 6th grade he won’t understand a bit of this and will reference his trash posts in rebuttal. BTW: eagleeye2 was banned on the Wizard of Vegas site for trolling and multiple ids. Surprise?

OK boys, it's time to tackle the 5/9 combinations, and remember the bias is as defined in posts #s 45 & 50 of this thread, and the numbers are for 72 come-out rolls. With true dice, there are 16 combinations of 5/9 with 8 of each. We would win 6.4 hands and lose 9.6 hands for a net loss of 3.2 units. Our biased dice would produce 9, nines, and 7, fives, same total (16) but again weighted, to the high number. 7, fives would result in 7X7/19 = 49/19 = 2.579 winners, and 4.421 losers for a net loss of 1.842 units on the point of five. 9, nines would result in 9 X 9/21 = 81/21 = 3.857 winners and 5.143 losers, for a net loss of1.286 units, on point of nine. 1.842 + 1.286 = 3.128 units lost on the 5/9 combinations, which once again is actually a little bit better for the player, than straight dice. Again, I am very surprised with the answer, and I'll bet you are too. And keep in mind, this is a very biased die. Next up, we'll tackle the 5/6 combinations, and then tally everything, to see just HOW wrong EE2 has been all these years. I too, had some wrong theories, about the results. Hopefully I'll get to it this evening.

Exactly. To benefit the house on the pass side you would have to pick up two dice with opposite bias. The casinos present 5 dice to choose from. If three were biased to the 1, and 2 were biased to the 6 there's a 60% chance you will pick two dice that the combined bias will result in fewer sevens. If people want to believe in bias dice and choose not to play because of that it could be a good thing. That person will lose less money. If you're living in a car and blaming biased dice instead of your poor life choices, there probably isn't much hope for you.

Harley... You are trolling just like your buddy James. You offer no facts to discredit the math that the posters have presented. That's because the math can not be discredited. Honestly I was surprised and amazed by Von Duck's post and the math he provided. Clearly someone that has done his homework and understands MATH. Yea yea I know... Play craps @my own PERIL !. There no need to reply. I did it for you Harley. G

Sorry little buddy, I never have drank the biased dice koolaid and I'm doing just fine. I understand normal variance and the bankroll requirements to play craps with an advantage. Maybe you should take up new endeavor. My wife can teach you how to play advantage bingo if you wish.

Here's a good article about the effect of more or less 7s on the game. http://www.rgtonline.com/article/wh...475?CategoryName=Gaming Tips&SubCategoryName=

First 720 rolls. I decided to post results every 720 rolls so we can see what variance does even in a rigged game. First picture is verses normal dice, second is compared to what random results would be with the skew, third picture is the bank, and last picture is how I configured.

The fundamental basis of the Monty Hall problem is that it involves multiple phases between which the contestant can change his/her decision based on additional information.

For as many times as I've read the Monty Hall problem....I still can't figure out why. I know the math makes it so, still tough to accept.

What in the mortal hell is a "Monty Hall paradox issue"? And by the way his name is JAMES, not Monty.