Find the machine with the highest probability. Suppose I have multiple weighted coin machines. Each machine of a probability of giving a head. All the machines have a probability of 33% of turning out heads except for one. For example, I can have 3 machines that have probabilities of 33%, 33%, and 60% of giving head on a flip respectively. I'm trying to find a strategy that can find me the machine which has the highest probability of flipping heads while using as fewer flips as possible. I have a couple of ideas.

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*Using a predetermined number of coins on each machine: for example, 10 flips on each machine, choose the machine with the highest number of heads. The probability of a machine with a chance of 33% beating or drawing that of 60% is $1 - \sum_{x=0}^n P_{33\%}(X >= x) * P_{60\%}(X=x)= 1 - 0.152815186 = 84\%$ while using 30 coins.

*Randomly select a machine, and use coins until the probability of that machine being 60% is low enough. For example, on a 60% machine, the probability of me getting equal to or less than 3 heads is around 5.4%, this is enough for me to switch to another machine. I'm not quite sure what the expected number of coins I need to spend on a 33% machine is until I switch.

These are the only two strategies I can think of that might work well. Please suggest other strategies or point me to resources on similar problems. Thanks!
Follow Up: What if the high probability machine changes and we can only play on one machine at a time? I assume strategy 1 probably won't work well anymore since it is possible for the machine to change and mess up the statistic, but strategy 2 can adapt.
 A: This depends on what the other probability is. For example, if you have a machine with $0.34$ probability of heads, you will need to do a very large number of trials, regardless of the exact method you use.
Perhaps a useful way to quantify your problem is to work out the expected number of trials you need to be within a certain probability. I expect this will change based on the number of machines and the probability of the non 33% machine. For example, with your example using 3 machines with probability 33%, 33% and 60%, maybe you want to know how many trials are needed to be 95% sure. Your first strategy gives you a good way to find this, using an equal number of flips on each machine.
The reason I suggested needing to consider the expected number of trials, is that in your second strategy the outcomes of the flips and of the random choice of machine determine the number of trials you need.
For example, if you pick a machine, play it 5 times, and it returns heads each time, you can be pretty sure this is the one with the highest probability - the probability the 33% machines give 5 heads from 5 flips is $\frac{1}{3^5} \approx 0.4$%.
The first strategy you suggest is certainly the safe option - you can always work out a number of flips for which you can be sure with probability $p$ which machine has the best probability. On the other hand, a 'non symmetric' approach may end up being very efficient, or take significantly longer. I expect a good approach would be using the safe strategy and modifying your future turns based on the information you have so far. As I say, I suspect it also depends on the parameters involved.
A fun problem for sure - what motivated this question?
