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We have a die. Success event (S) is 1-4, failure event (F) is 5 and 6 (in other words chances for success are ~0.67 or 2/3 for one roll). We roll the die, say, a billion or n times.

What is the probability of the success events number being higher than the failure events number?

An approximate result, or a clue about the (possible?) tendency for S to go higher/lower as n increases, will suffice!

Extra question if you feel like answering: What is the probability of the number of S being higher than the probability of the number of S in an "alternate universe" (let's call it Sx), where chances of success and failure for the die are equal? (Same n)

Mystic_Quest
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1 Answers1

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The intuitive thing is that if you have one event that is more likely (your $S$) it will overwhelm the other (your $F$) in the long run. In your case, because the probability difference is huge, it will be very soon. For $n$ rolls you should expect $\frac {2n}3$ successes and $\frac n3$ failures. If we use the normal approximation, the standard deviation of the number of successes is $\sqrt {n\frac 13 \cdot \frac 23}=\frac 13\sqrt{2n}$. At the $3\sigma$ level we want $\frac {2n}3-\sqrt {2n} \gt \frac {n}3+\sqrt {2n}$ which requires $\frac n3 \gt 2\sqrt {2n}$ or $n \gt 72$ so if you roll $72$ times (in this approximation) you have $99.7\%$ chance to have more successes than failures. We have ignored the chance of a tie, but for $72$ rolls that is not so likely.

Ross Millikan
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  • Wow, didn't expect the probability it to escalate that fast. Thanks, was not even sure it will be overwhelmed, only by intuition - not familiar with these approximation formulas (or probability theory at all), will look them up for some extra insight. Thanks. – Mystic_Quest Feb 27 '18 at 04:33
  • Look at Wikipedia on the normal approximation to the binomial distribution. Your underlying rolls are a binomial distribution. For large numbers of rolls the number of successes follows (approximately) the normal distribution. – Ross Millikan Feb 27 '18 at 04:37