So, I'm trying to figure out how to get an equal probability of any one of six outcomes just by flipping a single coin. I found some good information here and here, but I'm not that math-literate and some of it went over my head.
What I did gather, though, is that there's a simple method of turning a coin into a functional six-sided die. If you flip three times in a row, there are eight possible outcomes (THH, HTH, etc.). So assign the values 1 to 6 to six of the outcomes; assign "null" to two of them. When you get a "null" outcome, ignore it and flip 3X again. This lets you simulate rolling a dice pretty well, but there's a 2-in-8 (or 0.25) chance you'll have to flip the coin six times or more, and the process could in theory go on for a long time.
Before learning this, I came up with an idiosyncratic method that I think also works:
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Flip one time, assigning tails to odd numbers in the interval 1 through 6, and heads to even numbers in said interval (or vice versa). This yields either the set {1, 3, 5} or the set {2, 4, 6}.
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Flip three times, and make the three flips correspond to the smallest, middle and largest number of the set selected in step 1, in that order. Heads is "keep"; tails is "discard."
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If only one number from the set was "kept", that number is your answer from the "die." If no number from the set was "kept," repeat step 2. If two or three numbers from the set were "kept," flip again for each number, starting with the lowest, and repeat until only one number is "kept."
I'm guessing that my method takes more flips to select a number, on average, than the simpler method. That would make it a worse method. However, I don't know this for a fact, and I'd like to find out. How do I find the average number of flips, roughly, for each method?
Submitted January 27, 2018 at 12:24AM by browsing_moose http://ift.tt/2DPKFhl