Hi. I read something on a post here, and it's been nagging me. He said that if you bet 12 random individual numbers for 37 spins, there is a possibility that they will NOT show up. (It is common for 12 numbers not to appear within 37 spins of the wheel)
HOWEVER, if you bet the 1st Dozen (or any dozen), it is almost impossible for this to lose 37 spins in a row. Even though BOTH bets are using 12 numbers!
FIRST QUESION: Is there any truth to this? If so, can someone explain how they would have different probabilities?
My guess would be that the 1st dozen is only up against 2 other competitors (and the other 2 dozens would hit more often to compensate for the 1st dozen that missed). Whereas the 12 straight up numbers are competing against 25 other numbers. And when those 12 S/U numbers don't hit, they are absorbed by the other 25 numbers.
If these probabilities ARE different, this leads me to my...
SECOND QUESION: If the 12 straight up numbers were placed on numbers #1 through #12 (same as the 1st dozen), Would these 12 numbers still have the same odds as the 12 random numbers from my First Question, or the same odds from the 1st Dozen?
If these 12 S/U numbers probabilities CHANGE, what causes it to change its probabilities from the first question asked? Because they are STILL competing against 25 other numbers! Hmmmmmm.
Hope I'm making sense. Thank you.