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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)

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  • $\begingroup$ You you want to go 12 games while winning at least 10 (for your bonus question, 7 games while winning at least 5). Have you tried anything to solve this? Made any progress? $\endgroup$ – Morgan Rodgers Oct 17 '16 at 18:43
  • $\begingroup$ @MorganRodges my assumption is that it's 0.6^12*3 but I think I am adding the losses in a wrong way together. But you are interpreting it wrongly, you need to win 12 times and lose at most 2 times, so you play 12..14 games $\endgroup$ – sydd Oct 17 '16 at 18:45
  • $\begingroup$ Ahh I see, you have to actually win 12. $\endgroup$ – Morgan Rodgers Oct 17 '16 at 18:48
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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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You can either win 12 in a row, ($0.60^{12}$ probability), you can play 13 with one loss in the first 12 (probability $12\cdot0.60^{12}\cdot0.40$), or you can play 14 with two losses in the first 13 (probability ${13 \choose 2}\cdot0.60^{12}\cdot0.40^2$). Add these up and you have $$(1+12\cdot 0.40+78\cdot 0.40^{2})\cdot 0.60^{12}=0.03979$$ so about 4% chance.

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