# a cute paper: finding a most biased coin with fewest flips

I saw this paper on ArXiV a while back and figured it would be a fun read, and it was. Post-ISIT blogging may have to wait for another day or two.

Finding a most biased coin with fewest flips
Karthekeyan Chandrasekaran, Richard Karp
arXiv:1202.3639 [cs.DS]

The setup of the problem is that you have $n$ coins with biases $\{p_i : i \in [n]\}$. For some given $p \in [\epsilon,1-\epsilon]$ and $\epsilon \in (0,1/2)$, each coin is “heavy” ( $p_i = p + \epsilon$) with probability $\alpha$ and “light” ( $p_i = p - \epsilon$) with probability $1 - \alpha$. The goal is to use a sequential flipping strategy to find a heavy coin with probability at least $1 - \delta$.

Any such procedure has three components, really. First, you have to keep track of some statistics for each coin $i$. On the basis of that, you need a rule to pick which coin to flip. Finally, you need a stopping criterion.

The algorithm they propose is a simple likelihood-based scheme. If I have flipped a particular coin $i$ a bunch of times and gotten $h_i$ heads and $t_i$ tails, then the likelihood ratio is $L_i = \left( \frac{p+\epsilon}{p - \epsilon} \right)^{h_i} \left( \frac{ 1 - p - \epsilon }{ 1 -p + \epsilon} \right)^{t_i}$
So what the algorithm does is keep track of these likelihoods for the coins that it has flipped so far. But what coin to pick? It is greedy and chooses a coin $i$ which has the largest likelihood $L_i$ so far (breaking ties arbitrarily).

Note that up to now the prior probability $\alpha$ of a coin being heavy has not been used at all, nor has the failure probability $\delta$. These appear in the stopping criterion. The algorithm keeps flipping coins until there exists at least one $i$ for which $L_i \ge \frac{1 - \alpha}{\alpha} \cdot \frac{ 1 - \delta }{\delta}$
It then outputs the coin with the largest likelihood. It’s a pretty quick calculation to see that given $(h_i, t_i)$ heads and tails for a coin $i$, $\mathbb{P}(\mathrm{coin\ }i\mathrm{\ is\ heavy}) = \frac{\alpha L_i}{ \alpha L_i + (1 - \alpha) }$,
from which the threshold condition follows.

This is a simple-sounding procedure, but to analyze it they make a connection to something called a “multitoken Markov game” which models the corresponding mutli-armed bandit problem. What they show is that for the simpler case given by this problem, the corresponding algorithm is, in fact optimal in the sense that it makes the minimum expected number of flips: $\frac{16}{\epsilon^2} \left( \frac{1 - \alpha}{\alpha} + \log\left( \frac{(1 -\alpha)(1 - \delta)}{\alpha \delta} \right) \right)$

The interesting thing here is that the prior distribution on the heavy/lightness plays a pretty crucial role here in designing the algorithm. part of the explore-exploit tradeoff in bandit problems is the issue of hedging against uncertainty in the distribution of payoffs — if instead you have a good handle on what to expect in terms of how the payoffs of the arms should vary, you get a much more tractable problem.