# The Bias of Odds & Evens Revisited

In an earlier *Digressions...* article I
wrote about the bias of Odds & Evens based on the
combinations of players' number selections. What I missed
was to examine the **permutations** of number
selections. Perhaps if I studied that, the game would work
out evenly?

Let's count.

1 | 2 | 3 | 4 | 5 | |

1 | E | O | E | O | E |

2 | O | E | O | E | O |

3 | E | O | E | O | E |

4 | O | E | O | E | O |

5 | E | O | E | O | E |

The *Odds* player wins 12 out of 25
permutations.

The *Evens* player wins 13 out of 25 permutations.

So Odds & Evens still looks biased for the
*Evens* player? Not quite.

Look at the above table. It assumes that both players
randomly put out a number. If that's true then yes, the
*Evens* player has an advantage, 52-48. **But
what if one player doesn't put out a number
randomly?** If the *Evens* player *always
plays an odd number* while the *Odds* player
puts out a number at random, the *Evens*
**advantage is 3 out of 5, or 60-40!** If the
*Odds* player *always plays an even number*
while the *Evens* player puts out a number at
random, the *Odds* advantage is also 60-40.

So a possible tactic is **if Evens, always
play an odd number; if Odds, always play an even
number**.

But what **if both players follow this
tactic**, do the advantages cancel out and it
reverts to a 50-50 advantage? NO! **The Odds
player wins 100% of the time!**

Recall, if an even number and an odd number are added, the
result is always odd. This is what happens if both the
*Evens* and *Odds* players follow the above
tactic.

The *strategy* of Odds & Evens should therefore
be to **always pick Odds and always play
even numbers.**

Maybe it's just better to flip a coin. [GRIN!]