Why You Should Always Call Even

by | November 1, 2016, 10:08am 11

Almost every game of ultimate starts with a flip to decide who pulls. Most players call even or odd seemingly on a whim, yet one of these choices is clearly better than the other. In this article I’ll show which choice is better, and why.

If the discs are perfectly fair – that is, they’re both 50% to be heads or tails – then even and odd are equally likely. But if the discs are skewed either way, they’re more likely to land even. I’m going to reveal the magic right away. This may not surprise you, but there is math behind it that I’m going to show you below. Ready? Okay. Here it the big disc flipping secret: Given that discs are not symmetrical on an up-down axis, they’re not equally likely to land heads or tails, so it’s better to call even.

Paul Illian wrote a great article about the flip way back in 2011. While he alluded to the math I’ll go over, he focused mostly on gathering empirical evidence about which way discs actually flip. I’m going to show that the correct choice is the same no matter what you believe about how likely a disc is to land either way.

The Math of the Flip

Okay, let’s dive into the math. I’ve noticed a strong correlation between ultimate prowess and advanced degrees in math and science, so for some of you this may be obvious. But if math is not your thing, don’t worry, it’s all fairly straightforward.

Let’s call the chance that the disc lands heads h and the chance it lands tails t. There are two ways to have the discs land even: two heads, or two tails. The chance of getting two heads is h*h which is the same thing as h2. The same is true for tails, and we add them together to get the chance of the discs being even:

Each disc has to land either heads or tails, which means in probabilistic terms that heads and tails have to sum to one. We can use this to say that t=1-h. Substituting into the above equation, we get:


Let’s graph it!


Notice that the graph bottoms out at .5. In other words, no matter what’s going on with the discs, by choosing even we guarantee ourselves at least a 50% chance of winning. And if the discs are skewed either way, we give ourselves a better chance by choosing even. In fact, the more skewed they are, the more likely even becomes. In game theory this is called a dominant strategy: even dominates odd because it has a significant upside and no downside.

Problems and Challenges

A key assumption that I didn’t state explicitly is that the chance of heads is the same for both discs. If you had two discs that skewed opposite directions – say, one that was 60% heads and one that was 40% heads – they’d actually be more likely to land odd.

I made that assumption because I believe the main reason for discs skewing one way or another is their unique shape, and all discs are basically the same shape (apart from surface imperfections). Ultimate players are notorious for rejecting any disc other than the Ultra-star model, even those that appear extremely similar.

Even if there are oddball discs out there, the above reasoning still holds as long as the discs are drawn randomly from the population. The only time a problem occurs is if you somehow know exactly one of the discs being used goes against the usual tendency. So, if you have a magical taco’d disc or a teammate with a bizarre flipping style, it might be better to call odd.


I’ve shown why you should always call even when you flip for the pull. There’s no downside to calling even and a very realistic upside. The edge you gain is likely to be small, but as ultimate becomes more and more competitive, every edge counts. Ultimate players may be odd, but more likely than not, the discs will land even.

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  • Kernal

    A fair flip, given any disc asymmetry, would be:
    1) Call up or down.
    2) Flip one disc.
    3) Flip a second disc.
    4) If discs come up different, use result of first disc. Otherwise, return to step (2)

    Regardless of likelihood of up vs. down, UD and DU will have equal probability, so the call will be fair.
    Another advantage: This flip can be performed with only one disc.

  • Andy Wong

    This can also be proven with fairly straightforward algebra and you don’t even need to use the fact that t = 1-h.

    Probability of evens is h^2 + t^2.
    Probability of odds is 2ht (because it could be HeadsTails or TailsHeads).
    Advantage to calling evens over odds: h^2 + t^2 – 2ht.
    So is this positive, zero, or negative?
    Ans: Always non-negative because it’s (h-t)^2 and anything squared will be non-negative.

    QED or something

  • Tommy Newell

    If any engineers have matlab, here is a script that simulates 100,000 flips based off of probabilities h and t for heads and tails.

    %% ODD OR EVEN

    h = 0.5; %Probability of head
    t = 0.5; %Probability of tails
    %% Simulation
    evenwins = 0; oddswins = 0;
    for flip = 1:100000
    D1 = rand(1); D2 = rand(1);
    headcount = 0; tailcount = 0;
    if D1 > t
    headcount = headcount + 1;
    tailcount = tailcount + 1;

    if D2 > t
    headcount = headcount + 1;
    tailcount = tailcount + 1;

    if headcount > 1
    evenwins = evenwins + 1;
    elseif tailcount > 1
    evenwins = evenwins + 1;
    oddswins = oddswins + 1;

    • Tommy Newell

      Feel free to yell at me for using a loop. I’m feeling lazy.

      • John

        Why would you simulate 100k flips when you could just determine the probabilities analytically?

  • Dave Klink

    WHY WOULD YOU PUBLISH THIS ARTICLE? Masters players like me need to hold on to every incremental advantage we can, and you are just giving away this information to the masses for free?

  • Morrison Luke Smith

    Is this a troll? Your graph is just showing that unfair coins tend to be more likely to lead to even results(.6*.6 is >.5*.5, likewise, (1-.3)^2 > 0.5^2). Your explanation of statistics for independent observations of a fair coin is incorrect: sample space: since your fair flip outcomes assuming independence are HH, HT, TH, TT, and P(H) = P(T) = 0.5. P(HH) = 0.5*0.5. P(HT) = 0.5*0.5. P(HT) = 0.5*0.5. P(TT) = 0.5*0.5. And total probability is 4*.25 = 1. P(HH) + P(TT) / P(HH or HT or TH or TT) = .5/1 = .5. For andy wong, its h^2 + t^t – 2ht; since h=t, it is 2h^2 – 2h^2 = 0. No statistical difference assuming flips are independent of each other. The headness of a coin is irrelevant — HT has as much meaning in the sample space as HH. Your math holds if you use real numbers for ht or hh; h*t = h*h.

    • Nate Solon

      It’s slightly tongue-in-cheek because of course it doesn’t really matter much what you call on the flip, but no, I wouldn’t call it a troll.

      “But it’s a very poor explanation in that it suggests that given a fair coin even is more likely.” In fact I state the opposite. For example, check out the very first sentence of the second paragraph: “If the discs are perfectly fair – that is, they’re both 50% to be heads or tails – then even and odd are equally likely.” This idea is reiterated several times throughout the article.

      “Your explanation of statistics for independent observations of a fair coin is incorrect.” What explanation are you referring to?

      • Morrison Luke Smith

        fair point. i got lost in the weeds.

  • Assume: discs are not fair.
    Thusly: discs landing even is more likely.
    Thusly: calling even results in more flip wins.
    Assume: flip wins result in advantage in winning the actual game of Ultimate.
    Thusly: the team calling the disc flip has an advantage in winning the actual game of Ultimate.

    Question: how do we determine which team is given the advantage in calling the disc flip?

    • Andy Wong

      Flip for it.