Check out the First Part of this to get to speed.

If we have four coins (and flip them all at once) what are all the ways they can be flipped?

TTHH | ||||

THTH | ||||

THHH | HTTH | TTTH | ||

HTHH | THHT | TTHT | ||

HHTH | HTHT | THTT | ||

HHHH | HHHT | HHTT | HTTT | TTTT |

Ok so that is 16 different outcomes and you can see that six of the sixteen outcomes is half heads and half heads.

TTHH | ||||

THTH | ||||

THHH | HTTH | TTTH | ||

HTHH | THHT | TTHT | ||

HHTH | HTHT | THTT | ||

HHHH | HHHT | HHTT | HTTT | TTTT |

1 | 4 | 6 | 4 | 1 |

Well the odds on any one of these out comes is 1/16 =(1/2)(1/2)(1/2)(1/2) so if there are six ways to get to tails then the odds of flipping two tails is 6/16 or 37.5%.

TTHH | ||||

THTH | ||||

THHH | HTTH | TTTH | ||

HTHH | THHT | TTHT | ||

HHTH | HTHT | THTT | ||

HHHH | HHHT | HHTT | HTTT | TTTT |

1 | 4 | 6 | 4 | 1 |

1/16 = 6.25% | 4/16 = 25% | 6/16 = 37.5% | 4/16 = 25% | 1/16 = 6.25% |

Well what if someone hands you four loaded coins, which come up heads 60% of the time. Let’s look at the last table again this time with the odds of each result coming up.

TTHH | ||||

THTH | ||||

THHH | HTTH | TTTH | ||

HTHH | THHT | TTHT | ||

HHTH | HTHT | THTT | ||

HHHH | HHHT | HHTT | HTTT | TTTT |

1 | 4 | 6 | 4 | 1 |

12.96% = (.6)(.6)(.6)(.6) | 8.64% = (.6)(.6)(.6)(.4) | 5.76% = (.6)(.6)(.4)(.4) | 3.84% = (.6)(.4)(.4)(.4) | 2.56% = (.4)(.4)(.4)(.4) |

1 x 12.96% = 12.96% | 4 x 8.64% = 34.56% | 6 x 5.76% = 34.56% | 4 x 3.84% = 15.36% | 1 x 2.56% = 2.56% |

As you can see the odds of the coins coming up head is shifted to the right. But in WH40K things are not this easy, I often have to roll 20 to 40 dice in one go, and it would take me a large dedicated effort to make a table of all the outcomes to come up with these odds, but have no fear, there a formula (Binomial Theory) for this and that what we will talk about next.

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