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copied!<p>The probability calculation is more complicated than you think. The probability of winning is</p> <pre><code>P(win on 0th tick) + P(win on 1st tick) + ... + P(win on 10th tick) </code></pre> <p>Let's call the probabilities p(0) ... p(11). Then</p> <pre><code>P(win on 0th tick) = p(0) P(win on 1st tick) = (1-p(0)) * p(1) P(win on 2nd tick) = (1-p(0)) * (1-p(1)) * p(2) </code></pre> <p>etc. At each tick, the probability that you win on that tick is the probability that you didn't already win on any of the previous ticks, multiplied by the probability of winning <em>right now</em>.</p> <p>With the numbers you gave in your post, I think that your robot should win about 63.17% of the time (I am not sure why you are seeing about 30% - could this be a bug somewhere else in your program?)</p> <p>With the following numbers you should observe about a 90% success rate overall</p> <pre><code> 0 0 1 0.0068 2 0.0113 3 0.0188 4 0.0314 5 0.0524 6 0.0875 7 0.1459 8 0.2433 9 0.4059 10 0.6771 </code></pre> <hr> <p><strong>EDIT</strong></p> <p>How did I come up with these numbers? Trial and error. But we could invent a procedure that given any win probability, generates a suitable set of probabilities for each tick.</p> <p>Let's say that the total win probability is <code>Q</code>, so you want</p> <pre><code>P(Win on 0th tick) + ... + P(Win on 10th tick) = Q </code></pre> <p>Let's say we want there to be no chance of winning on the 1st tick, and a linearly increasing chance of winning on any tick after that. So the probabilities have to add up to Q, and the probability of the win coming at tick <code>i</code> is proportional to <code>i</code>. Therefore</p> <pre><code>P(Win on ith tick) = const * i </code></pre> <p>hence</p> <pre><code> c * 0 + c * 1 + c * 2 + ... + c * 10 = Q => 55 * c = Q => c = Q/55 </code></pre> <p>That gives us</p> <pre><code>P(Win on 0th tick) = 0 P(Win on 1st tick) = Q/55 P(Win on 2nd tick) = 2*Q/55 </code></pre> <p>etc. Now you use these to determine each of the <code>p(i)</code> using the formula at the top of the post. We have</p> <pre><code>p(0) = P(win on 0th tick) = 0 p(1) = P(win on 1st tick) / (1-p(0)) = Q/55 p(2) = P(win on 2nd tick) / (1-p(0)) / (1-p(1)) = 2*(Q/55) / (1-Q/55) </code></pre> <p>etc. Here's a Matlab routine that calculates the probabilities; it shouldn't be hard to translate it into C# or whatever you're using.</p> <pre><code>N = 10; Q = 0.9; p = zeros(N+1,1); for i = 1:N p(i+1) = i * Q/(0.5*N*(N+1)) / prod(1-p(1:i)); end </code></pre> <p>which gives this result</p> <pre><code> 0 0 1 0.0164 2 0.0333 3 0.0516 4 0.0726 5 0.0978 6 0.1301 7 0.1745 8 0.2416 9 0.3584 10 0.6207 </code></pre>

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