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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!]