Link for the description: http://us.battle.net/hearthstone/en/blog/20324471/introducing-heroic-tavern-brawl-10-17-2016

Lest assume that you build a very good deck with 60% winrate.

The problem worded more mathematically:
You have $60\%$ chance to win a game. If you lose $ 3 $ times you are out. What are the chances of winning $12$ games while having having $2$ or less losses between the wins? (Order if losses does not matter). So you will play 12 or 13 (if you lose once) or 14 (if you lose twice) games.

Bonus question: lets assume that the $10$ dollars entry fee is worth it (you can get more rewards than just buying them from the shop) if you win $7$ or more games while losing $2$ or less times. What are the chances that your $10$ dollars was worth it? (Again assuming 60% win chance)


1 Answer 1


Let us make the following simplification: We continue to play games after reaching the twelve-win maximum or the three loss maximum.

Recognize then that to have reached twelve wins, then in the first fourteen games you play you will have lost at most two times.

$\binom{14}{0}0.6^{14}0.4^0 + \binom{14}{1}0.6^{13}0.4^1 + \binom{14}{2}0.6^{12}0.4^2 \approx 0.03979$

Similarly, to have reached at least 7 wins, within the first nine games you will have lost at most two times.

$\binom{9}{0}0.6^{9}0.4^0+\binom{9}{1}0.6^80.4^1+\binom{9}{2}0.6^70.4^2\approx 0.231787$


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