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i tried several searches on google to find this; alas no luck.

say player 1 wins 6 in a row on a best out of 13 game. then player 2 wins 6 in a row. everything else being equal, what are the chances of player 2 winning the 7th game? i've heard he has better chances, but have yet to see the math.

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Like James22 said, if the games are independent of the previous games, it's a coin toss. However, since statistically speaking it is highly unlikely to have 6 wins followed by 6 losses, we can all but rule out the chance of independence (granted, with a small sample size, the p-value of such a hypothesis test would be somewhat higher than what is typically considered strong enough to rule out a null hypothesis, but let's go with it anyway). The examination shows that it is likely that something occurred between the 6th and 7th rounds (in a physical game, probably an injury; in a mental game, perhaps exhaustion, or player 2 found a new strategy) that would suggest that player 2 has the better chance of winning.

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Actually, player two does have a higher chance of winning the 7th game.See the spoiler to find out why.

Actually, the chances of the 2nd player winning the 7th game is 66%. The key is the fact that it is the 7th game being asked for, not the 13th game. Then we see the following sequence

????????????? - these represent the outcomes of the games, x for player one, and o for player two, we know that games occur in the order xxxxxxoooooo (6 x and 6 o), which gives the four possibilities of

x-xxxxxxoooooo

o-xxxxxxoooooo

xxxxxxoooooo-x

xxxxxxoooooo-o

which give equal chances of winning the 7th game. However, the first option is invalid, because it makes player one win the game automatically at game 7, and no further games would be played, therefore we get the three possibilities (this time written with the 7th game hyphenated)

oxxxxx-x-oooooo

xxxxxx-o-oooooo

xxxxxx-o-ooooox

Therefore, player 2 has a 2/3 chance of winning the 7th game. The previous replies make the mistake of assuming the final (and hence 13th) is the one being debated, when it specifically asked for the 7th game. Sorry if this explanation was too long though!

Edited by Palustrius
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Actually, player two does have a higher chance of winning the 7th game.See the spoiler to find out why.

Actually, the chances of the 2nd player winning the 7th game is 66%. The key is the fact that it is the 7th game being asked for, not the 13th game. Then we see the following sequence

????????????? - these represent the outcomes of the games, x for player one, and o for player two, we know that games occur in the order xxxxxxoooooo (6 x and 6 o), which gives the four possibilities of

x-xxxxxxoooooo

o-xxxxxxoooooo

xxxxxxoooooo-x

xxxxxxoooooo-o

which give equal chances of winning the 7th game. However, the first option is invalid, because it makes player one win the game automatically at game 7, and no further games would be played, therefore we get the three possibilities (this time written with the 7th game hyphenated)

oxxxxx-x-oooooo

xxxxxx-o-oooooo

xxxxxx-o-ooooox

Therefore, player 2 has a 2/3 chance of winning the 7th game. The previous replies make the mistake of assuming the final (and hence 13th) is the one being debated, when it specifically asked for the 7th game. Sorry if this explanation was too long though!

we're given that it is a series of 13 and that both players have won six in a row and that player 1 won his six first. Regardless of how the OP is worded, I'm sure phil is asking for the odds of the 13th game (the 7th win). I also agree that it depends on the style of game being played and whether there is any influence from previously won games (and whether that data is quantifiable).

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Molly Mae, while I am nervous to argue with a moderator (me being a new member and all), I must state that you are killing the puzzle. To put it simply, it is not fair to assume additional factors for a given puzzle. If I could do that in every puzzle I came across, I could solve any of them. For example, In the logic puzzles between the logic master and his three apprentices (I'm sure everyone has seen these on the logic puzzle page), I am NOT allowed to say there is a mirror in the room that only one of the apprentices can see. It just takes the fun out of it all.</p><p><br></p><p>Essentially, what you just stated was, "regardless of how the puzzle was told, it isn't true, and I can assume anything about the puzzle I so wish". DON'T DO THAT PLEEEAAASE!!!

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Molly Mae, while I am nervous to argue with a moderator (me being a new member and all), I must state that you are killing the puzzle. To put it simply, it is not fair to assume additional factors for a given puzzle. If I could do that in every puzzle I came across, I could solve any of them. For example, In the logic puzzles between the logic master and his three apprentices (I'm sure everyone has seen these on the logic puzzle page), I am NOT allowed to say there is a mirror in the room that only one of the apprentices can see. It just takes the fun out of it all.</p><p><br></p><p>Essentially, what you just stated was, "regardless of how the puzzle was told, it isn't true, and I can assume anything about the puzzle I so wish". DON'T DO THAT PLEEEAAASE!!!

I love heated points in my direction.

1. I'm not really a moderator--My custom title is "Moderator" not the actual group I'm in.

2. I'm not assuming constraints on the puzzle, I'm assuming that phil is asking for the odds of the 7th win, not the 7th game (which has already been played). I'm not assuming anything beyond the constraints of the riddle, I just interpret it in a different way (especially the word "then").

Wording that might be better assuming that your interpretation is correct: "say player 1 wins 6 in a row on a best out of 13 game. At some point after that, player 2 wins 6 in a row. everything else being equal, what are the chances that player 2 won the 7th game?"

Wording that might be better assuming that my interpretation is correct: "say player 1 wins 6 in a row on a best out of 13 game. Immediately after, player 2 wins 6 in a row. everything else being equal, what are the chances of player 2 winning the 13th game?"

I don't believe that I'm assuming any more or less than you are.

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Thank you for taking it so well Molly. In which case, we must either assume that there is something about this puzzle we don't know, or that the chances of winning are 50/50. Is there nothing else to be added phi?

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Thank you for taking it so well Molly. In which case, we must either assume that there is something about this puzzle we don't know, or that the chances of winning are 50/50. Is there nothing else to be added phi?

I take criticism very well and just figured there was an interpretation difference. =P

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Wait... phi, is that exactly as you heard it? What did you actually hear about this. From your last statement the wording could be extremely CRUCIAL! (jumping up and down in excitement)

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I'm pretty sure that you just want to know the chances of winning a "next game". But the way you phrased it allows me to edge in my previous formula again. In this case, we consider the chances that, in the whole 13 game series, player two wins the last game. In this case the three possibilities are again

xxxxxxoooooo-o

xxxxxxoooooo-x

oxxxxxxooooo-o

with the exclusion of xxxxxxxooooo-o, because it gives one the victory first. Once again, the chance of victory for player two is 66% in the final match.

And finally, how this actually solves the initial question. Your question stems from the fact that someone (I don't know who) noticed a correlation between winning streaks and victories in the fashion you mentioned. However, from the previous logic, you must realize that the winning streak is not the CAUSE of the final victory. It just so happens that when two winning streaks occur one after the other, the later streak is more likely to encompass the final game. That is probably the source of the apparent correlation you are speaking of. Of course, this can still be debated endlessly based on circumstantial evidence. But I believe this is the most logically sound answer you are going to get. The odds of winning will still be 50/50 (assuming a fair game) when considering the following game. I hope this is somewhat helpful to you.

Edited by Palustrius
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unfortunately i disagree with your logic.

in the case of OXXXXXXOOOOOO-O player 2 would have to win the 13th and final game in order to have 6 in a row.

we're given that player 1 won 6 in a row, directly followed by player 2 winning 6 in a row. and ask what is the probability of player 2 winning the final game.

also, there would be no difference in the streak. by your argument, you should have a 66% chance of winning in any come from behind victory.

no, i think it's more like the following.

probability of 7 game winning streak in 13 games. 1-(1-1/2^7)^7 = .05 or 5%

probability of 6 game winning streak. 1-(1-1/2^6)^8 = .11

probability of 7 game winning streak given a 6 game winning streak? .11/2/.05 = 1.1 ??

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yeah, that's how i heard it.

here's an interesting paper i found on a similar concept.

http://www.retroshee...ningStreaks.pdf

From that link...the first sentence is: "I'm interested in whether the "spirit" of a team has any effect on its play..."

To me, this is a no-brainer in that of course the spirit of a team (aka their momentum) will impact their performance. Think of any time you have played any type of competitive sport (or even watched it on TV). If your team is on a "hot-streak", your spirit is up and you play better and better (unless you get cocky and give up trying because you think you're that good). On the same hand, the other team's morale is lowered, and they are more likely to realize they can't win/have tried their best and lost X-games in a row.

So, with regards to this specific question, if you are looking MATHEMATICALLY at the answer to your question, there is a 50% chance for each team to win the final game GIVEN that they each play equally as well as they have in the past, and you don't take morale/spirit/whatever into account. HOWEVER, if you do take that into account, I agree the team that just won 7 games in a row is more likely to win the last game...however, I don't think there is any way to actually quantify the percentage of this advantage.

There are also many other reasons the second team could have a better chance. For example, team 1 could have injured their best player in game 7, which is what caused team 2 to have the competitive advantage. I'm just presenting this to emphasize that if you look at REAL WORLD data, then yes, team 2 will show a more likely chance of winning the last game, but if everything is equal and you take out all unseen factors, then it is completely even.

Hope this made sense and I didn't ramble on too long.

Edited by Pickett
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