An elementary statistics problem from Hearthstone

In Hearthstone, a popular online card game, there is a minion called C’Thun which deals damage equal to its attack value, one damage at a time. This prompted a popular streamer named Trump (not the insufferable presidential candidate) to ask the following question once on his channel: when playing a C’Thun with 12 attack (hence, it will deal 12 damage one at a time), what is the probability that it will kill a 6 health minion when played, with no other minions on the board?

To those who don’t know how Hearthstone works, one can think of the question as follows: suppose you flip a coin 12 times in a row. Each time you flip the coin, you record down whether you got heads or tails. As long as you have fewer than 6 heads, the coin is fair and you have equal probability of getting heads or tails on your next flip. However, as soon as you hit 6 heads, then the coin becomes unfair and you can only get tails after. The question that Trump (real name Jeffrey Shih) asks is then equivalent to “what is the probability that you get 6 heads”.

To calculate this probability, we can ask the reverse question: what is the probability that we don’t get six heads? If we don’t get 6 heads, then the probability distribution is the same as if we always had a fair coin, and so the probability is the same. The probability of getting at most five heads when flipping a fair coin 12 times is given by

\displaystyle 2^{-12} \sum_{k=0}^5 \binom{12}{k}.

The well-known binomial theorem tells us that

\displaystyle 2^{-12} \sum_{k=0}^{12} \binom{12}{k} = 1.

This shows that the probability of getting at most 5 heads is equal to

\displaystyle \frac{1}{2} - 2^{-13} \binom{12}{6} = \frac{793}{2048}.

Therefore, the probability of getting 6 heads in the original game is equal to

\displaystyle 1 - \frac{793}{2048} = \frac{1255}{2048},

or roughly 61.28\%. Thus it is far more likely that C’Thun will kill a 6 health minion with only 12 damage than too many shots going to face.

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