Odds and Evens

indbnm exWhat should you call when you flip two discs to start the game? Evens.

And I can prove it*. (Of course, if you really hate maths, you’ll just have to take my word for it. But it’s not complicated.)

Let’s call the two possible landing positions for a disc heads and tails.

First, let’s consider the case where the disc is entirely ‘fair’ – so that it’s exactly 50/50 which way it lands. Then the four results – HH, HT, TH, TT – are all equally likely, and there’s two odd and two even. So it makes no difference whether you call odds or evens.

Now there’s only one other possibility – that the disc is not fair; that either heads or tails happens more often, on average. That’s equivalent to saying that the probabilities h and t are not equal; or, more usefully:

h-t ≠ 0   [h and t are not the same, so one minus the other won’t equal zero]

Now we can play with the maths a bit:

(h-t)² > 0   [The square of any non-zero number – positive or negative – is larger than zero]

(h-t)(h-t) > 0   [Same thing written out in full]

hh – ht – th + tt > 0   [Just multiplying out the brackets]

hh + tt > ht + th   [Adding ht + th to both sides, which won’t change which side is bigger]

And look what’s happened – the two ‘evens’ possibilities added together are more probable than the two ‘odds’ possibilities. You’ve got to call evens.

Remember we didn’t specify whether heads or tails was the more likely** – just that they weren’t equal – but it doesn’t matter. If the discs have a bias either way, you need to call evens. And even if they don’t, if it’s straight 50/50, calling evens is just as good as calling odds. Evens is never*** the wrong call to make.

*To be fair, this proof is due to Colin Johnstone, an astrophysicist who discovered it while trying to prove the opposite. My wishy-washy intuitive grasp of calling evens didn’t persuade him, but in trying to point out my error he accidentally proved my point. Doubtless he’s still miffed.
** Paul Illian’s spectacularly in-depth article on this topic suggests that which way a disc is biased can depend on the surface it lands on. But it doesn’t matter, when flipping two discs, as long as both discs are biased the same way – evens is always best. Paul also has a nice graph (Figure 8) showing visually that odds is never better than evens. In fact, Paul’s article is just flat-out comprehensive, but I really like the concise 5-line proof given here so I posted it anyway.
*** If the discs are in fact biased in opposite directions (one is more likely tails, one more likely heads) then a similar argument would show that you should call odds. But no matter how scuffed the discs, I don’t think there are many situations where you could reasonably expect two discs with the same shape to behave in opposite ways. Perhaps if you’re throwing on two different surfaces, Paul’s results would suggest that might be the case – but the chances are very high that you’ll be on the same field. Another remote possibility I can imagine is where one flipper is facing upwind, and the other downwind, so that the discs are flipping in opposite directions relative to the breeze. It’s possible that the wind getting under the hollow side of the disc could affect the likelihood of heads/tails differently for the two discs – that would take a lot of experimenting to find out! Instead, just make sure you’re facing the same way and always call evens… 😉
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17 Responses to Odds and Evens

  1. Anonymous says:

    Are we going to have to involve a flip to decide who calls the flip now?


    • Actually, probably not. (And yes, I know this was a joke, but I’m replying anyway…) It’s never that far from fair – Paul shows in his article that even if the discs are biassed 55% to 45% one way or the other, the advantage for evens is only 50.5% to 49.5%. You need a hugely biased disc – which we would have noticed by now – to move the probabilities very far from 50%. But nevertheless, evens is guaranteed to be the slightly better call.


  2. mrkopo says:

    Ok, we call Evens
    Then what? We choose O or D or sides with wind?


    • Answering that would be another whole article…


    • GBrell says:

      Assuming weather is not a factor, you should always call O.

      D seem to be better because of the possibility of the double-break, but assuming the game is hard-capped, whichever side gets more break always wins.

      The only time the choice matters is if the two sides get the same number of breaks. There are three situations where this can happen (O gets more breaks in the first half, D gets more in the first half, the two sides have the same number in the first half). The only one that the D wins is where they get more breaks in the first half (e.g., D up 1 break is 7-8 at half, O gets 1 break after to make it 8-8, D wins 14-15). So in two of the three scenarios O wins.
      *This is a simplification of the math, but I’ve run some experiments on it and the D win % given equal breaks never gets much over ~40%.

      Choosing wind is a separate question and entails an evaluation of team skills vs. weather conditions.


  3. Anonymous says:

    Your assuming both discs are weighted with the same bias.


    • They don’t require the exact same amount of bias, but simply to be biased in the same direction. The maths is very slightly more complicated, but the result is the same. See the note at the end for when they have opposite bias.


    • OK, so now I’m back home from work, I can write it out. Disc one has probabilities of h and t, and disc two has probabilities of h‘ and t‘. So we replace line 3 of the original proof with:

      (h-t) (h‘-t‘) > 0

      Under the assumption that they’re both biased the same way, either a) both of (h-t) and (h’-t’) are positive, so the product is larger than zero; or b) they’re both negative, so the product is again larger than zero. It’s only zero if one or both of the discs is unbiased (so evens is still not wrong), and it would only be less than zero if the discs had opposite bias (note that this is the basis of the proof mentioned elsewhere – oppositely-biased discs means you should call odds).

      Continuing to multiply out the brackets as in the original proof, we get:

      hh‘ – ht‘ – th‘ + tt‘ > 0; and then
      hh‘ + tt‘ > ht‘ + th

      The only assumption required to get here is that both discs are biased in the same direction. I think the non-symmetric shape of the disc, if it does cause any bias, is likely to be a bigger effect than any tiny bias from scuffing or slight warping or whatever, so this assumption will generally hold.


  4. I am all for learning and sharing. This kind of knowledge is not to be shared. This takes away our advantage over those who will choose odds or evens randomly. This type of knowledge is only to be discussed in… well… the gatherings we’re not supposed to talk about.


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  6. Wayne says:

    Not to rubbish the maths here, but I once went a whole tournament getting the other captains to call the flip. Seven times they called “evens” and seven times the discs were odd. Even winning the actual tournament was a secondary pleasure after that.


  7. John says:

    The bias of the dis is negligible compared to the spin you give it. With a little practice you can make it land one way or the other.

    This takes the chance out of it and the flip becomes more like paper rock scissors – what will your opponent throw?


  8. Dani says:

    Until now, I was convinced by your argument of having the slight bias being evens, until I saw this video: https://www.youtube.com/watch?v=AYnJv68T3MM

    According to this, there is an intrinsic bias towards a single flip ending up either the same or different(1) from how you release it. When two people flip, they will have the same bias, and so probably the best choice is to look at how the discs were released by either flipper and if call odds or evens according to whether both were released with the same side up or not.

    Lots of fun experiments to be done here, to figure out which of the two effects is bigger 🙂

    (1) For coins there is a 51%/49% bias towards the same release, but disc flips probably don’t quite lie in the same realm as coin flips, and it would need someone to actually work it out for discs to know which it is in this case.


    • Nice.

      I would anticipate the bias due the the disc shape as being larger than .51 – a coin is much more symmetric than a disc, where all the weight is on one side, and I’d expect the effect of that to be larger than the effect of initial orientation. Paul Illian’s article above suggests a larger effect due to surface (which has to be down to the way the asymmetric disc hits different surfaces) – though for sure you’re right that we’d need a bigger data set to be sure!


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