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motbob
United States12546 Posts
If you bet regularly on Dota2lounge, you are probably familiar with the fact that you need to take the length of a series into account. As the series gets longer, you should be willing to take odds that are more skewed towards the favorite. An example: today there was a match between Empire and Alliance. Empire has looked significantly better than Alliance recently, so the 60/40 odds look off at first glance... until you realize the match is a Bo1. OK, that's fine, Bo1 games are pretty high variance. The odds would probably be something like 65/35 if it were a Bo3.
But I worked out something interesting regarding the so-called "best of two" format where teams teams play two games (with the potential of a 1-1 tie). It turns out that, on Dota2lounge, you should treat Bo2 matches the same as a Bo5 for betting purposes. The favored team is about as likely to win you rares in a Bo2 as it would be to win a Bo5. Here's the math comparing a Bo3 and a Bo2:
Let's say Team A has a 60% chance of beating Team B in a given game.
In a Bo2, Team A has a .6*.6=.36 chance of going 2-0, while Team B has a .4*.4=.16 chance. The chance of a 1-1 is 0.6*0.4*2=0.48, but Dota2lounge returns your items in the case of a tie, so we need to focus only on the two cases where a team goes 2-0. Items are won/lost 52% of the time (.36+.16), and Team A goes 2-0 36% of the time, so we just take 36%/52%=0.69 to find out the proper odds for betting on Team A (69/31).
In a Bo3, Team A has a .36 chance of going 2-0. As mentioned above, the chance of a 1-1 is 0.48, and Team A has a 60% chance of winning the third game, so the chance of Team A winning in three games is .48*.6=0.288. Since either going 2-0 or winning the third game are the only two ways Team A can win the series, we add .36 and .288 to get the chance of Team A winning the series (0.65, so the proper odds for a Bo3 are 65/35).
69% for a Bo2 vs 65% for a Bo3. I won't do the math here, but the chance of Team A winning a Bo5 against Team B is also about 69%.
If you take issue with my math, I will be happy to defend it to the death in the thread below.
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Thoughts on fnatic vs NVMI?
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motbob
United States12546 Posts
On September 08 2014 10:05 Dubzex wrote:
Thoughts on fnatic vs NVMI?
Well, the odds are 67/33 right now. According to the OP, I should bet on Fnatic if I think they have about a 59% or greater chance of winning any given game.
I haven't watched Fnatic's games with the new roster, so I dunno. Looking at the two team's results since TI4, I think a bet on Fnatic might be justified but I am not personally going to bet on them having not seen their games.
EDIT: Actually, current key odds are 60/40 which I would judge to be worthy of a bet on Fnatic based on results alone.
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motbob
United States12546 Posts
You can do it if you want. It's up to you if you want to circulate this knowledge.
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Thanks! 
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Interesting, pretty cool. Did the math for some higher winrates to see if it stays true and apparently even 0.99% winrate has a worse chance of loosing at a best of two than a best of three. Which is somewhat counter-intuitive.
Might be worth noting that everything over 2/3 winrate means that the stronger team wins 2:0 is the most probable event.
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I can confirm this works, I posted a reddit thread about this a few months ago. It's worth pointing out that the potential losses and gains are both lower in a Bo2 bet because of the possibility of a 1-1 tie, so you should bet a bit more stuff (the more even the matchup is, the more you should bet) on Bo2s relative to Bo5s to get the same expected value.
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Shoot, I'm trying to think of how to solve this using statistics, and I feel really stupid right now, but I cannot remember.
My question is, what is the chance of the favored team winning a Bo7, if they have a 60% chance to win an individual game.
![[image loading]](http://i.imgur.com/juNf9o5.jpg)
Using the binomial distribution formula above, I would sub in x = 4 (as that is the victory condition), and take the sum of subbing in n=4,5,6,7, with the p value being 0.6 obviously. I'm getting wonky results though, and negative binomial distribution doesn't seem correct either (probably because it's used for something completely different).
Going through the distributions I've been taught: beta, chi-square, t, f, error, gamma, weibull, poisson, exponential (special case of gamma), or uniform... Definitely none of those are used to solve the problem. Any help would be much appreciated, it's bugging me too much.  (I'm not trying to solve this with the brute force method, as that is quite easy [but time consuming] with a little branched tree giving all possible events)
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1 - (.4^3 + .4^3 * .6 * 3 + .4^3 * .6^2 * 6)
Edit: 68.256%
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But what did you derive that formula from?
I'm looking for a general solution that I can apply for any situation, not how to solve this problem, that's not my intention.
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Interesting... The math makes sense to me, not that I know much, but I'm wondering: is it really worthwhile to bet on the favored team in a Bo2 series when the odds (before recalculating using motbob's technique) are 60% in their favor? Because while the new odds are quite favorable (69%), that's only looking at scenarios in which only one team wins any games. If the odds of the teams tying is 48%, then the odds of either the teams tying OR the less favored team winning should be 64%, right? So shouldn't this make Bo2s a low-risk environment in which to bet on the less-favored team?
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On September 10 2014 10:40 FiWiFaKi wrote:
But what did you derive that formula from?
I'm looking for a general solution that I can apply for any situation, not how to solve this problem, that's not my intention.
If you want a general solution just look up Binomial CDF, most statistical software and graphing calculators have the function. Here's the wikipedia link for the equation http://en.wikipedia.org/wiki/Binomial_distribution#Cumulative_distribution_function which was what you were doing. There is no 'one step' equation that replicates the Binomial CDF.
For the specific problem, take 100% - the probability of 0-3, - the probability of 1-3 (times the ways it can happen, 3 ways) - the probability of 2-3 (times the way it can happen)
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+ Show Spoiler +On September 10 2014 08:12 FiWiFaKi wrote:Shoot, I'm trying to think of how to solve this using statistics, and I feel really stupid right now, but I cannot remember. My question is, what is the chance of the favored team winning a Bo7, if they have a 60% chance to win an individual game. Using the binomial distribution formula above, I would sub in x = 4 (as that is the victory condition), and take the sum of subbing in n=4,5,6,7, with the p value being 0.6 obviously. I'm getting wonky results though, and negative binomial distribution doesn't seem correct either (probably because it's used for something completely different). Going through the distributions I've been taught: beta, chi-square, t, f, error, gamma, weibull, poisson, exponential (special case of gamma), or uniform... Definitely none of those are used to solve the problem. Any help would be much appreciated, it's bugging me too much. (I'm not trying to solve this with the brute force method, as that is quite easy [but time consuming] with a little branched tree giving all possible events)
Here's a few reasons why this solution doesn't work: First the n = 7 choose 4 solution includes several impossible scenarios (win-win-win-win-loss-loss-loss, win-win-loss-win-win-loss, etc.) for a Bo7. The n = 6 choose 4 and n = 5 choose 4 similarly include impossible results. Additionally, the probability calculation for an n = 7 choose 4 solution assumes impossible solutions for an actual Bo7. You can't have a loss-loss-loss-loss-loss-loss-loss Bo7, but the n = 7 choose 0 solution is assumed to be an alternate possibility for n = 7 and reduces the calculated probability for n = 7 choose 4. Also you can't sum the n = 7, n = 6, n = 5, n = 4 probabilities, because the n = 4 probability (win - win - win - win) is kind of redundant with a n = 5 possibility (win - win - win - win - loss) and so on.
I have no idea how to do this without brute-forcing though, sorry.
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Yes, I knew it wasn't right, but it seems like you explained well why that is the case, so thank you.
LSB, I know it needs cumulative binomial distribution, and I know how to solve problems with cumulative binomial distribution, but I simply don't know how to solve approach this one.
edit: Yay, I made solutions:
![[image loading]](http://i.imgur.com/G26RDxo.png)
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It's not 1-(Binomialcdf (7,4)). It is 1- (Binomialcdf (6,3) * 0.4)
The final game of the series is won be the team that is not favored to win, so you can just imagine it as the probability you will get 3 wins in a series of 6 * the probability of winning the final game.
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On September 11 2014 23:00 LSB wrote:
It's not 1-(Binomialcdf (7,4)). It is 1- (Binomialcdf (6,3) * 0.4)
The final game of the series is won be the team that is not favored to win, so you can just imagine it as the probability you will get 3 wins in a series of 6 * the probability of winning the final game.
a) If you're doing it from the perspective of the favored team (p>0.5) and trying to solve for the favored team: Binomialcdf(7,3) will give you the probability of the favored team winning 0,1,2, or 3, games, and that is exactly what we don't want, so all we have do to is simply:
P = 1 - binomialcdf(7,3), p=favored
b) If you're doing it from the perspective of the disadvantageous team (p<0.5), but still trying to solve for the favored team: Binomialcdf(7,3), which is the same thing as saying when the disadvantaged team wins 0,1,2, or 3 games, which is the same as the favored team winning, therefore solution is:
P= binomialcdf(7,3), p=disadvantaged
Intuitively this also makes sense, because naturally, we can create this identity, all it's saying is that the sum of probabilities is one:
binomialcdf(7,3)_a +binomialcdf(7,3)_b = 1
So if you have any of the two of the three above equations I sort of derived (but not really), you can get the last one. But in the end, method (a) and method (b) will both net you the same answers.
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I see what I made a mistake in, for some reason I was calculating best of 5.
For best of 7 it would be
1 - (.4^4 + .4^4 * .6 * 4 + .4^4 * .6^2 * 10 + .4^4 * .6^3 * 20)
But yes your solution is valid
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EPT
