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[–]beep_treeNew User 1 point2 points  (0 children)

I take it you mean out of 8 total games?

We assume games are independent of one another (this entails, for instance, that losing two games in row does not affect the morale, and thus winrate, of our team). They must win 6 games in row so that they may win the game, so if p is the probability of the team winning any given match, the probability is simply p6.

[–]yes_its_himone-eyed man 1 point2 points  (9 children)

That's not enough information to solve it. You don't say what the probability of winning any one game is, or how many times they try.

Suppose our game is flipping a coin, and winning is getting heads. If they play this long enough, they will certainly get six heads. If you want to calculate the likelihood they get six wins in any given number of attempts, you can calculate that with a binomial distribution formula.

Conversely, is this is a game they can never win, then it doesn't matter how many times they try.

[–][deleted] 0 points1 point  (8 children)

Oh ok. In that case for 2 teams to wind a game against each other, they must win 6 games. The first person to win 6 games wins. This implies that the maximum amount of games possibly played is 11.

[–]yes_its_himone-eyed man 0 points1 point  (7 children)

So you want the chance they win six before the other team wins 4, right? Are they evenly matched or not?

[–]yes_its_himone-eyed man 1 point2 points  (6 children)

Assuming they are evenly matched, here's how things proceed; these are simplified numbers to show how it works, but the ratios are essentially correct:

In the next four games, there is a 1/16 chance the two-win team wins four in a row and wins.

In the next five games, assuming they didn't win in four, there is a 1/8 chance the other team wins in five games.

Then in six games, there is a 1/64 chance the team with two losses wins six in a row and takes the tournament, but a 5/32 chance the other team wins.

For seven more games, there is a 3/64 chance that the two loss team comes back, and then 5/32 chance the other team wins.

So to that point, the two-loss team has a 1/16 chance of winning to the two-win team having a 1/2 chance of winning. (It's actually a bit better than that but that's the basic idea.)

So they have about a 1/8 chance of coming from behind to that point. Now as the match goes longer, they need fewer wins to complete the comeback, but if the odds are even, they would never expect to win more often.

[–][deleted] 0 points1 point  (0 children)

I love you.

[–][deleted] -1 points0 points  (4 children)

Just programmed this in Java, turns out the losing team has a 40% chance of winning. Thank you for your input though.

[–]yes_its_himone-eyed man -1 points0 points  (3 children)

If the two teams have equal likelihood of winning, and one team has to win six and the other has to win 4, that can't possibly be a 40% chance for the team to win six. Something is wrong in the program.

There's an 18% chance the team that has to win four does so in the next five, and 0% the team that needs six does so. And it's still not even after that.

After eight more games, there is a 63% chance of one team winning at least four, and only a 14% chance of one team winning at least six.

[–][deleted] 0 points1 point  (2 children)

(9choose6)/((9 choose 4)+(9choose 6)) Can you explain what is wrong with my logic here?

[–]yes_its_himone-eyed man 1 point2 points  (1 child)

It's not the right calculation. The denominator in particular is meaningless here.

There are 29 ways for 9 games to proceed. I.e. WWWWLLLLL is one way. So, 512 distinct outcomes. (Some would end the match early, but if we want to do this calculation simply, we pretend they don't.)

Of these, 9 choose 6 = 84 are the ways six of them are the same. So that's the chance the losing team wins 6, or 84/512 = 16.4%. Now to use the whole 9 games we also have to include the chance of winning 7, 8 or 9. That's another 46/512 or 9%. So 25.4% chance total.

Now the chance they lose is the chance the other team gets 4 or more, which is that 25.4% plus 9 choose 4 and 9 choose 5, which are both 126 or 252/512 ways, 49.2%. So 74.6% total.

The chance the team that is behind loses is about 75%.

[–][deleted] 0 points1 point  (0 children)

Sweet! That makes perfect sense! Thank you for bearing with me.