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NFT: I'm not understanding a basic function of math right now.

Mike in Long Beach : 4/26/2015 3:26 pm
Help me out here. First off, let me say, I know what the right answers are. At least I think I do. What I understand is how it logically works.

Let's use sports as the medium:

1) Team A is trailing 3 games to 2 in a best of 7 series. Team A needs to win both games 6 and 7 in order to advance to the next round.

Team A has a 1/2 chance of winning game 6 and also has a 1/2 chance of winning game 7. 1/2 x 1/2 = 1/4, or 25% chance of advancing.

2 Team B is trailing 3 games to 2 in a best of 7 series. Team B needs to win both games 6 and 7 in order to advance to the next round.

Team B has a 2/5 chance of winning game 6 and a 3/5 chance of winning game 7. 2/5 x 3/5 = 6/25, or 24%

So the net is, Team A has a 25% chance of advancing when team B only has a 24% chance of advancing. I don't understand this. (1/2 + 1/2) is = to (2/5 + 3/5). To me that means that both teams should have an equal chance to advance... but obviously they don't.

Someone help me out here. Feel free to call me idiot as well, as long as you help.

Meant to say I *don't understand the logic  
Mike in Long Beach : 4/26/2015 3:27 pm : link
.
Since both teams  
OBJXIII : 4/26/2015 3:32 pm : link
need to win both games, and Team B only has a 2/5 chance at winning the first game, while Team A has a 1/2 chance, Team B has tougher slightly tougher odds.
2ndscenario  
Les in TO : 4/26/2015 3:34 pm : link
There is less than 50 percent chance team B will advance so harder to get to a game 7 and therefore slightly lower overall chance of winning series
RE: Since both teams  
Mike in Long Beach : 4/26/2015 3:35 pm : link
In comment 12250046 OBJXIII said:
Quote:
need to win both games, and Team B only has a 2/5 chance at winning the first game, while Team A has a 1/2 chance, Team B has tougher slightly tougher odds.


That doesn't compute to me. In the 2nd game then, Team A has a 50% chance of winning where Team B has a 60% chance of winning then. So that should equally make up for the discrepancy for each teams' game 6s.
Conditional probability.  
BrettNYG10 : 4/26/2015 3:35 pm : link
Game 7 only happens if they win game 6. You'd sdd if they're ere mutually exclusive.
They were.*  
BrettNYG10 : 4/26/2015 3:36 pm : link
.
Add not sdd.  
BrettNYG10 : 4/26/2015 3:36 pm : link
This Blues game is distracting.
RE: 2ndscenario  
Mike in Long Beach : 4/26/2015 3:36 pm : link
In comment 12250048 Les in TO said:
Quote:
There is less than 50 percent chance team B will advance so harder to get to a game 7 and therefore slightly lower overall chance of winning series


That still doesn't work to me. Flip flop it then. Give Team B the 60% chance in game 6 and the 40% chance in game 7. Their odds will still equate to 24%, but now they'd have the more likely opportunity to advance when compared to Team A.
(Idiot.)  
BrettNYG10 : 4/26/2015 3:36 pm : link
.
what bret said,  
OBJXIII : 4/26/2015 3:39 pm : link
If you flip a coin 100 times, Team B makes it to the second game 40% of the time and Team A makes it 50%. If you are determining the total odds of each team winning BOTH GAMES, team B has it tougher and less of a chance because they HAVE to win the first game.
RE: Conditional probability.  
Mike in Long Beach : 4/26/2015 3:40 pm : link
In comment 12250053 BrettNYG10 said:
Quote:
Game 7 only happens if they win game 6. You'd sdd if they're ere mutually exclusive.


That sorta makes sense to me, but I still struggle with it.

I used sports as the vehicle here, but theoretically, you could run the same scenario where the 7th game didn't depend on the 6th outcome. Like if you were running a simultaneous simulation and needed two things to occur to be the same... like a slot machine.
RE: RE: 2ndscenario  
BrettNYG10 : 4/26/2015 3:40 pm : link
In comment 12250058 Mike in Long Beach said:
Quote:
In comment 12250048 Les in TO said:


Quote:


There is less than 50 percent chance team B will advance so harder to get to a game 7 and therefore slightly lower overall chance of winning series



That still doesn't work to me. Flip flop it then. Give Team B the 60% chance in game 6 and the 40% chance in game 7. Their odds will still equate to 24%, but now they'd have the more likely opportunity to advance when compared to Team A.


Right, but it's why adding the probabilities together is irrelevant.
More importantly,  
Tim in Capital City : 4/26/2015 3:40 pm : link
that's not the way probabilities work. If a team has a 1/10 shot of winning the first game and a 9/10 shot of winning the second, they only have a 9% chance of advancing. It doesn't matter that 1/10 + 9/10 = 1/2 + 1/2.
RE: RE: Conditional probability.  
BrettNYG10 : 4/26/2015 3:41 pm : link
In comment 12250064 Mike in Long Beach said:
Quote:
In comment 12250053 BrettNYG10 said:


Quote:


Game 7 only happens if they win game 6. You'd sdd if they're ere mutually exclusive.



That sorta makes sense to me, but I still struggle with it.

I used sports as the vehicle here, but theoretically, you could run the same scenario where the 7th game didn't depend on the 6th outcome. Like if you were running a simultaneous simulation and needed two things to occur to be the same... like a slot machine.


If they were mutually exclusive, you'd add them, so the math is different.
Mike it has been explained correctly.  
section125 : 4/26/2015 3:42 pm : link
Team B only has a 40% chance in winning game 6 while Team A has 50% chance to win.

Means Team A has a 10% greater chance to win game 6. Cannot get to game 7 unless they win game 6.
RE: Mike it has been explained correctly.  
Mike in Long Beach : 4/26/2015 3:43 pm : link
In comment 12250071 section125 said:
Quote:
Team B only has a 40% chance in winning game 6 while Team A has 50% chance to win.

Means Team A has a 10% greater chance to win game 6. Cannot get to game 7 unless they win game 6.


Section, then flip flop it.

Give team B the 60% chance in game 1. I shouldn't have assigned specific values to each game. That's my bad. What I meant was Team B has a 60% chance to win one of the games and a 40% chance to win the other.
RE: RE: Mike it has been explained correctly.  
Jim in Fairfax : 4/26/2015 3:57 pm : link
In comment 12250074 Mike in Long Beach said:
Quote:

Section, then flip flop it.

Give team B the 60% chance in game 1. I shouldn't have assigned specific values to each game. That's my bad. What I meant was Team B has a 60% chance to win one of the games and a 40% chance to win the other.

Lower odds in either game negatively impacts the overall odds.

Extreme example:

Team needs to win game 6 & 7 to advance. They have a 100% chance of winning game 6 and a 0% chance of winning game 7.

By the way you calculate the odds, their chances are 50% of advancing. But the reality is their chances are 0%.
I don't think that's accurate...  
Mike in Long Beach : 4/26/2015 4:01 pm : link
Quote:
Extreme example:

Team needs to win game 6 & 7 to advance. They have a 100% chance of winning game 6 and a 0% chance of winning game 7.

By the way you calculate the odds, their chances are 50% of advancing. But the reality is their chances are 0%.


That would be 0/100 x 100/100

100 x 0 = 0, so the math would correctly calculate a 0% chance.
But has  
Tim in Capital City : 4/26/2015 4:06 pm : link
he demonstrated the fallacy in your 0/100 + 100/100 = 1/2 + 1/2 thought process?
RE: I don't think that's accurate...  
Jim in Fairfax : 4/26/2015 4:15 pm : link
In comment 12250106 Mike in Long Beach said:
Quote:

That would be 0/100 x 100/100

100 x 0 = 0, so the math would correctly calculate a 0% chance.

Exactly. But In your example above, you were using addition rather than multiplication.
To understand why...  
Dan in the Springs : 4/26/2015 4:20 pm : link
think about how probabilities are calculated. The probability of an event (A) happening is equal to (A) divided by the total outcomes in the set.

To calculate multiple probabilities we multiply the denominators to find the total outcomes in the new set. Sometimes a decision tree helps you see it best.

Let's start with team A.

Game 1 outcome Game 2 outcome Combined Outcome
P(win)=1/2 P(win)=1/2 P(win,win)=1/4
Win Loss WL
Win WW
Loss Loss LL
Win LW

Total outcomes = 4 (2 sets of 2 outcomes) and total events where A wins = 1.

Team B
Game 1 outcome Game 2 outcome Combined Outcome
P(win)=2/5 P(win)=3/5 P(win,win)=1/4
1. Win Loss WL
2. Loss WL
3. Win WW
4. Win WW
5. Win WW
6. Win Loss WL
7. Loss WL
8. Win WW
9. Win WW
10. Win WW
11. Loss Win LW
12. Win LW
13. Win LW
14. Loss LL
15. Loss LL
16. Loss Win LW
17. Win LW
18. Win LW
19. Loss LL
20. Loss LL
21. Loss Win LW
22. Win LW
23. Win LW
24. Loss LL
25. Loss LL

There are 25 possible outcomes (five possible outcomes in the first game, with two of them representing wins and three representing losses, and five for each of those five possible outcomes in the second game, with three of those representing wins and 2 representing losses). Of those 25 outcomes only 6 are WW, so you have P(WW)=6/25, or 24%.
Dan.  
Mike in Long Beach : 4/26/2015 4:55 pm : link
Thanks for laying out like that. Makes more sense now, and Tim:

Quote:
In comment 12250111 Tim in Capital City said:
[quote] he demonstrated the fallacy in your 0/100 + 100/100 = 1/2 + 1/2 thought process?


Very true.
This is an example of  
danshave : 4/26/2015 5:17 pm : link
the difference between the arithmetic mean and geometric mean.

The arithmetic means of [1/2,1/2] and [3/5,2/5] are the same, but their geometric means are not.
http://en.wikipedia.org/wiki/Geometric_mean - ( New Window )
Reading  
Big Al : 4/26/2015 5:57 pm : link
the opening statement, it seemed so basic I could not understand why you were confused. Then I saw a long number of posts which were not incorrect but did not seem to be relevant to your confusion until danshave finally got to it.
You dont mulitpe the chances of team B wining  
Kevin999 : 4/26/2015 6:31 pm : link
You mulitply the changes of them NOT winning, them subtract from 100%
Say they have a 50% chance of winning each game.
They have a 50% of not winning times a 50% of not winning game 7. So a 25% of losing both. There a 75% of winning either game.
To the OP...  
whozzat : 4/27/2015 6:38 am : link
reconsider your original scenario this way... two contestants are about to run an obstacle course.

Contestant A will make two passes... both times, he will have rocks flung at him by bystanders... giving him a 50% chance of surviving each pass, or 25% of survivng both.

Contestant B will make two passes... one time he will have nerf balls shot at him (95% chance of survival), the other time it will be live ammo (5% chance of survival) - so a 4.75% chance of surviving both.

The fact that he gets essentially a free pass against the nerf balls isn't going to help him against the bullets.
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