In yesterday's Iowa Caucus, Hillary Clinton beat Bernie Sanders in six out of six tied counties by a coin-toss*. I believe we would have heard the uproar about it by now if this was somehow rigged in her favor, but I wanted to calculate the odds of this happening, assuming she really was that lucky, and assuming she rigged various numbers of the tosses.

* As many people have pointed out already, this turned out to be a selective data set - Sander's won just about as many coin tosses as Mrs. Clinton did. Read on if you still care about the problem.

At first I calculated the odds using simple rules for the probabilities of independent events: $$ P(6\text{H})=6*P(\text{H})=\left( \frac{1}{2} \right)^{6}= \frac{1}{64} \approx 1.56\% $$ i.e. naively, there was a 1.56% chance it was fair.

But I vaguely remembered from reading about Bayesian inference that we can make a more educated statement about whether or not this was fair using Bayes' Theorem, and assuming various numbers of the coin tosses were rigged.

I tried it out myself, and here's what I came up with, but I'm fairly positive I worked this out incorrectly, so here's hoping you wonderful people can help. Here's my shot at it:

Example assuming it was fair (0% chance it was rigged): $$ P(6\text{H}) = \underbrace{P(6\text{H}|\text{fair})}_{1/64}\underbrace{P(\text{fair})}_{1} + \underbrace{P(6\text{H}|\text{not fair})}_{1}\underbrace{P(\text{not fair})}_{0} = \frac{1}{64} $$ and by Bayes Theorem: $$ P(\text{fair}|6\text{H}) = \frac{P(6\text{H}|\text{fair})P(\text{fair})}{P(6\text{H})} = \frac{(1/64)(1)}{(1/64)}=1 $$ (obviously).

Assuming $n$ of the tosses were rigged: $$ P(6\text{H}) = \underbrace{P(6\text{H}|\text{fair})}_{\left(\frac{1}{2}\right)^{6}}\underbrace{P(\text{fair})}_{1-\frac{n}{6}} + \underbrace{P(6\text{H}|\text{not fair})}_{1}\underbrace{P(\text{not fair})}_{\frac{n}{6}} = \frac{6-n}{384} + \frac{n}{6} = \frac{63n+6}{384} $$ and by Bayes' Theorem: $$ P(\text{fair}|6\text{H}) = \frac{P(6\text{H}|\text{fair})P(\text{fair})}{P(6\text{H})} = \frac{\left(\frac{1}{64}\right)\left(\frac{6-n}{6}\right)}{\left(\frac{63n+6}{384}\right)}=\frac{6-n}{63n+6} $$

Here's a plot of the probabilities that the coin tosses were fair given an assumption of $n$ unfair coins:



  1. I'm pretty sure some of my assumptions for probabilities were off in various parts of this - if so, where did I go wrong?
  2. On the off chance I carried this out correctly, what can be made of these results? For example, is it most probable that there were 0, 1, or 2 coin tosses that were unfair, as making the assumption that there were $n<3$ unfair coins gives a probability $P(\text{fair}|6\text{H})$ greater than the $1/64$ chance it was fair?


@Eric Wofsey Informed me that I was calculating the wrong probability. What I really wanted to calculate was $P(0|6H)$, the probability of 0 coins being rigged, considering an outcome of 6 heads. What I learned (I'm new to Bayesian inference) is that it all depends upon your prior guess as to the probability that n of the coins were rigged. As he pointed out: $$ P(0|6H) = \frac{P(0)}{\sum_{i=0}^{6}2^iP(i)} $$ where $$ P(n) = {6 \choose n}p^n(1-p)^{6-n} $$ and $p$ is the prior probability that each coin toss was rigged. Here's what $P(0|6H)$ looks like fully expanded (assuming the prior $p$ is the same for each $P(n)$):

Expanded P(0|6H)

As I learned, the prior probability is arbitrarily chosen, and represents your belief/guess as to the likelihood that the coins were rigged.

I was interested in looking at what the distribution of $P(0|6H)$ looked like for values of $p$ from 0 to 1 (0 meaning you believe there's no possibility the coins were rigged, 1 meaning you're certain the coins were rigged). Here's the plot:

P(0|6H) as fn of p

I may be going way off the reservation here, but if this graph represents values of $P(0|6H)$ for prior probabilities of having rigged coins, wouldn't the integral of this from $p=0$ to $1$ represent the total probability of 0 rigged coins, considering an outcome of 6 heads, with each prior $p$ weighted equally? Whether or not I'm abusing the maths, the integral evaluates to: $$ \int_{0}^{1} P(0|6H)(p) \ dp = 0.0822\dots $$

Note: I'm thinking in retrospect that the prior $p$ should probably be different for every $P(n)$ and assuming they're the same for each $P(n)$ is likely problematic, but I thought I'd share my process anyway.


On further thought, it seems like what I really want to compute is the integral: $$\int{\int{\int{\int{\int{\int{\int P(0|6H) \ d p_0 \ d p_1 \ d p_2 \ d p_3 \ d p_4 \ d p_5 \ d p_6}}}}}}$$ where $$ P(0|6H) = \dfrac{(1-p_0)^6}{(1-p_0)^6 + 12p_1(1-p_1)^5 + 60p_2^2(1-p_2)^4 + 160p_3^3(1-p_3)^3 + 240p_4^4(1-p_4)^2 + 192p_5^5(1-p_5) + 64p_6^6} $$ and $$ p_0 + p_1 + p_2 + p_3 + p_4 + p_5 + p_6 = 1 $$ and $p_n$ is the prior probability that $n$ coins are rigged.

I have absolutely no idea how one would go about even thinking about evaluating this integral - it seems as though there are a range of values for the integral anyway, depending on the choices of $p_n$. It seems it is definitely possible given specific choices for $p_n$ and maybe even a distribution for the $p_n$s, dependent on n, such that the distribution is still normalized, like a weighted decaying distribution or something (gets less likely as n increases that that number of coins was rigged).

Happy Tuesday

  • 2
    $\begingroup$ Not that it matters for the math but "The initial 6-for-6 report, from the Des Moines Register missed a few Sanders coin-toss wins. (There were a lot of coin tosses!) The ratio of Clinton to Sanders wins was closer to 50-50, which is what we'd expect." (washingtonpost.com/news/the-fix/wp/2016/02/02/…) $\endgroup$
    – Ant
    Commented Feb 2, 2016 at 23:29
  • 1
    $\begingroup$ "Naively" is the right word there. Tossing 6 fair coins, we get 6 heads with a probability of 1.56% ... events less likely that that happen to you every day. $\endgroup$
    – GEdgar
    Commented Feb 2, 2016 at 23:29
  • 2
    $\begingroup$ While the news coverage did report Clinton partisans winning six of six coin tosses, according to NPR's evening news coverage today (All Things Considered), there were other coin tosses, some of which were won by Sanders partisans. So your data makes for an interesting problem to analyze, but it turns out to be more of a selective data set than a problem of unfair coin tosses. $\endgroup$
    – hardmath
    Commented Feb 2, 2016 at 23:37
  • $\begingroup$ @Ant didn't catch that, that's good to hear. I guess my first lesson is to get news from multiple sources. At this point though, I've invested so much time in the problem I'm FAR more interested in the maths than the politics :) $\endgroup$
    – D. W.
    Commented Feb 2, 2016 at 23:37
  • $\begingroup$ Where are you getting that $P(fair)=1-n/6$? $\endgroup$ Commented Feb 3, 2016 at 0:10

1 Answer 1


Your computation doesn't make any sense. Assuming that $n$ of the tosses were rigged doesn't mean that you're assigning a prior probability of $n/6$ to "not fair". If you're saying the only possibilities are "fair" and "not fair" and "not fair" means a 100% of Clinton winning all $6$ tosses (which is what your computation of "$P(6H)$" implies), then that means either all the coins are rigged or none of them are, with $P(\text{not fair})$ being the prior probability that they are all rigged.

The computation that you seem to be trying to do when you're computing "$P(6H)$" is not $P(6H)$ but $P(6H|n\text{ rigged coins})$, i.e. the probability of getting $6$ heads assuming exactly $n$ of the coins were rigged. This is very easy to compute: it's just the probability of the $6-n$ non-rigged coins coming up heads, which is $1/2^{6-n}$. Note that if you allow this possibility, you are no longer saying the only options are "fair" and "not fair"; rather the options are "$n$ rigged coins" for each $n$ between $0$ and $6$ (with $n=0$ being "fair" and $n=6$ being what you called "not fair"). You then get that $$P(6H)=P(6H|0)P(0)+P(6H|1)P(1)+\dots+P(6H|6)P(6)=\frac{P(0)}{2^6}+\frac{P(1)}{2^5}+\dots+ P(6),$$ where I am abbreviating the event "$n$ rigged coins" as simply "$n$". You then get that $$P(0|6H)=\frac{P(6H|0)P(0)}{P(6H)}=\frac{P(0)}{P(0)+2P(1)+\dots+2^6P(6)}$$ is the probability that all the coins were fair, given that they all came up heads. Note here that $P(0),P(1),\dots,P(6)$ are prior probabilities: the probability (before you knew the outcome of the coin tosses) with which you believed that $n$ of the coin tosses were rigged. You don't get a value for $P(0|6H)$ until you plug in values for these priors. If, for instance, you believe that each coin toss independently had a prior probability of $p$ of being rigged, then $P(n)=\binom{6}{n}p^n(1-p)^{6-n}$. However, this is probably not a reasonable assumption (you wouldn't expect the riggedness of each toss to be independent--if one of them is being rigged, then that makes it more likely there is a conspiracy which means more of them will be rigged).

  • $\begingroup$ Thank you so much - I definitely mixed up what probability I was looking for. Just to clarify, in the last step, $P(6H|0)=\frac{1}{2^6}=1/64$ correct (the probability of getting 6H assuming 0 were rigged)? $\endgroup$
    – D. W.
    Commented Feb 3, 2016 at 0:42
  • $\begingroup$ Yes, that's correct. $\endgroup$ Commented Feb 3, 2016 at 0:42
  • $\begingroup$ Just a few more questions: is our choice of $p$ in $P(n)$ really arbitrary? Is there any way to use the results of $P(0|6H)$ to determine the likelihood that the coin tosses were, in any/some way, rigged? $\endgroup$
    – D. W.
    Commented Feb 3, 2016 at 0:47
  • $\begingroup$ The values of $P(n)$ must be values you could compute before any of the coins were tossed. They represent your level of belief that exactly $n$ of the coins were going to be rigged, before they were tossed. So you cannot assign their values based on anything you learned in this computation. $\endgroup$ Commented Feb 3, 2016 at 0:50
  • $\begingroup$ So, the results of $P(0|6H)$ depend inherently on our assumptions about how many coins were rigged - i.e. we're really just left to make arbitrary guesses as to what we think the probabilities of having $n$ rigged coins are? Is there any better probabilistic technique for making an educated statement/judgement as to the likelihood that any number of the coins were rigged, without knowing how many actually were rigged ahead of time? $\endgroup$
    – D. W.
    Commented Feb 3, 2016 at 0:54

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