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POChoose between multiple options with defined probability
 

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    <p>I have a scenario where I need to show a different page to a user for the same url based on a probability distribution,</p> <p>so for e.g. for 3 pages the distribution might be</p> <pre><code>page 1 - 30% of all users page 2 - 50% of all users page 3 - 20% of all users </code></pre> <p><strong>When deciding what page to load for a given user, what technique can I use to ensure that the overall distribution matches the above?</strong></p> <p>I am thinking I need a way to choose an object at "random" from a set X { x1, x2....xn } except that instead of all objects being equally likely the probability of an object being selected is defined beforehand. </p> <hr> <p>Thanks for the input everyone, after doing some prototyping, this is what I ended up using</p> <pre><code>private static int RandomIndexWithPercentage(Random random, int[] percentages) { if (random == null) { throw new ArgumentNullException("random"); } if (percentages == null || percentages.Length == 0) { throw new ArgumentException("percentages cannot be null or empty", "percentages"); } if(percentages.Sum() != 100) { throw new ArgumentException("percentages should sum upto 100"); } if (percentages.Any(n =&gt; n &lt; 0)) { throw new ArgumentException("percentages should be non-negative"); } var randomNumber = random.Next(100); var sum = 0; for (int i = 0; i &lt; percentages.Length; ++i) { sum += percentages[i]; if (sum &gt; randomNumber) { return i; } } //This should not be reached, because randomNumber &lt; 100 and sum will hit 100 eventually throw new Exception("Unexpected"); } </code></pre>
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    1. COHow strict is your use of the term "ensure" here? Obviously if we just pick numbers at random you're not ensuring that 30% of your users will land on Page1 - you're merely stacking the odds in favor of that outcome. Is the complexity here that you have to guarantee that for any set of N users, the page distribution will absolutely be within a very small margin of error of your target percentages?
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      1. USAndrew Anderson
    2. COThat's a good point Andrew, this test runs for a couple of days so I don't need a tight margin for any set of N users. Is there an algorithm that will provide minimum variation from this distribution for all values of N?
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    3. COYou could adapt the algorithm to bias itself depending on which users are on which pages, but it would be a lot more overhead for very little payoff. The best way to do that would be not use random numbers at all - if 3 users are viewing page one, 4 are viewing page two, and 2 are viewing page three, place the next user on page two (obviously). If another user comes, place him in page two as well, because it's the largest pool - so a single additional user there would impact your overall distribution the least.
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