In the online card game, Hearthstone, there is a play mode called Arena. Players who enter the arena build decks (commonly referred to as drafting) and play until they win 12 matches or lose 3 matches. Since the decks are built 'randomly,' and the matches are also random, we can assume that each player has a 0.5 chance of winning and a 0.5 chance of losing.

What would be the best way to calculate the probability of ending with exactly 0 wins, 1 win, 2 wins, ..., and 12 wins? Since the arena ends after any 3 loses there are branches of the binomial tree that would never occur.

ie Completing 3 matches have the following possible outcomes. WWW, WWL, WLW, WLL, LLL, LLW, LWL, LWW. Since LLL fulfills the 'lose 3 match requirement' LLLL and LLLW can not occur. Likewise during a 5th round, WLLLW, WLLLL, LWLLW, LWLLL, LLWLW, LLWLL, LLLWW, LLLWL, LLLLWL, LLLLL can not occur.

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    $\begingroup$ In reality its not 50/50. Building a deck and then using requires skill, which tilts the tables. $\endgroup$ – David says Reinstate Monica Aug 11 '14 at 19:11
  • $\begingroup$ one could also track wins over time and then calculate probabilities of # wins in the next arena based on that probability. for example, a player who averages 5 wins per arena has a 62.5% game win rate. of course, even that is skewed, because your probability of winning each game decreases as your record improves, because the game attempts to match you against players with the same record. $\endgroup$ – smcg Aug 11 '14 at 19:42
  • $\begingroup$ @Dgrin91 - That's true, but this is the realm of the Spherical Chicken. $\endgroup$ – Bobson Aug 12 '14 at 14:01
  • $\begingroup$ @Dgrin91 If we're assume all levels of players participate in the area (ranging the best of the best, 10 out of 10, to the worst, 1 out of 10), we would expect the average chance of winning chance to be 0.5 and as the number of games approaches infinity and the distribution to be pretty normal, right? $\endgroup$ – aso118 Aug 12 '14 at 17:03
  • $\begingroup$ @smcg I believe this takes into account for a player's skill level, right? One player might average 5 wins while another player might average 9 wins. If the player with 9 win achieves 11 wins one a single play through, that would be 'less' of an achievement then the player who averaged 5 wins achieving 11 wins. I believe, by calculating the probability of winning x rounds and losing y rounds, we will be able to predict a player's skill level without an incident number of trials. $\endgroup$ – aso118 Aug 12 '14 at 17:09

We look at the probability of $k$ wins, where $k\lt 12$.

This will happen if we play $k+3$ games, lose $2$ of the first $k+2$, and lose the last. The probability is $$\binom{k+2}{2}\left(\frac{1}{2}\right)^{k+2}\cdot\frac{1}{2}.$$

The case $k=12$ is a little different. One can find the probability by subtracting the sum of the above computed probabilities from $1$. Or else (better) we can divide into cases: (i) no losses: (ii) $1$ loss; (iii) $2$ losses. The first is very easy, and the analysis for the other two is as above.

The same idea will take care of situations where we win any game with probability $p$, and lose with probability $1-p$.

  • $\begingroup$ Would it be possible to show an example or two? Maybe when k = 7 and k = 12? $\endgroup$ – aso118 Aug 11 '14 at 18:04
  • $\begingroup$ When $k=7$ then from the formula we get $\binom{9}{2}\cdot\frac{1}{2^{10}}$, and $binom{9}{2}=36$. For $k=12$ there were three terms, the last one is $\binom{13}{2}\frac{1}{2^{13}}\cdot \frac{1}{2}$. You can try to find the other two, and if you leave a comment with them, I will confirm whether it is right. $\endgroup$ – André Nicolas Aug 11 '14 at 18:14
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    $\begingroup$ For $k=7$ we want to find the probability of $7$ wins. So we did not make it to the end, got $7$ wins and $2$ losses, in some order, and then sadly lost on the $10$th play of the game. The number of ways to choose **where** our $2$ losses were in the $9$ games is $\binom{9}{2}$. So the probability of $2$ losses in the first $9$ games is $\binom{9}{2}\frac{1}{2^9}$. Multiply by $\frac{1}{2}$ for the loss in the $10$th. $\endgroup$ – André Nicolas Aug 11 '14 at 18:18
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    $\begingroup$ Note that $\binom{9}{2}/2^{10}\approx 0.0351562$, in agreement with your calculation. This is not the binomial cdf at $2$, since the binomial cdf at $2$ is $\Pr(X=0)+\Pr(X=1)+\Pr(X=2)$, where $X$ has binomial distribution parameters $9,1/2$. Recall that cdf is cumulative distribution function. $\endgroup$ – André Nicolas Aug 12 '14 at 17:06
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    $\begingroup$ binomialpdf should give you a number that you then multiply by $1/2$ to get the right answer. $\endgroup$ – André Nicolas Aug 12 '14 at 17:28

Hint: Let's say you end up with exactly 5 wins. What does that mean? Well, you have won 5 games, and lost 3. Also, the 3rd loss came at the 8th game. So, we have to arrange 5W's and 2L's. In how many ways can it be done?

Similar analysis can be applied to n wins, where $0 \leq n \leq 12$


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