login
Math/Statistics Help
-
No new posts
|
 2014-01-06 3:48 PM |
Sensei  Sin City |  Subject: Math/Statistics HelpI used to be decent of this but maybe too much chlorine to the brain has mad it foggy. I have 4 events. Each event has 2 outcomes. What are the total number outcomes by the four events? What I mean is. Event 1 can be A or B. Event 2 can be C or D, 3 is E or F, and 4 is G or H. So ONE possible outcome is A, C, E, G. Another is B, C, E, G.... I first thought it was 4 factorial = giving me 24 possible outcomes. That SEEMS high. Then I thought it was 2 possibilities x 2 possibilities x 2 x2.. So 16 possible outcomes. Any help? I guess by the time I wrote this and now wait for a responses, I can just list them all and see... |
| 2014-01-06 4:16 PM in reply to: Kido |
Elite  4564   Boise | Subject: RE: Math/Statistics Help |
| 2014-01-06 4:17 PM in reply to: JoshR |
Sensei Sin City | Subject: RE: Math/Statistics Help Originally posted by JoshR I believe it's 2^4. 2 out comes, 4 times. I agree. 4! seemed too high. |
| 2014-01-06 8:38 PM in reply to: Kido |
Expert 1456  Central New Jersey | Subject: RE: Math/Statistics Help My son the math whiz (only missed 2 questions on the SAT and finished college Calc 2 while still in HS) Says 16 is the correct answer It would only be 4 factorial if the first event had 4 outcomes, the 2nd had 3 etc. Lani |
| 2014-01-06 9:18 PM in reply to: 0 |
Pro 15655 | Subject: RE: Math/Statistics Help Lemme see here......i go with Kido and Josh, or the kid who only missed 2 on his SAT and has college Calc fresh in his head.. hmmmmmmm....... I say 16. Edited by Left Brain 2014-01-06 9:19 PM |
| 2014-01-06 10:35 PM in reply to: Left Brain |
Elite 4564 Boise | Subject: RE: Math/Statistics Help Originally posted by Left Brain Lemme see here......i go with Kido and Josh, or the kid who only missed 2 on his SAT and has college Calc fresh in his head.. hmmmmmmm....... I say 16. *cough* punch 2^4 into a calculator *cough* |
| 2014-01-06 11:34 PM in reply to: JoshR |
Pro 15655 | Subject: RE: Math/Statistics Help Originally posted by JoshR Originally posted by Left Brain *cough* punch 2^4 into a calculator *cough* Lemme see here......i go with Kido and Josh, or the kid who only missed 2 on his SAT and has college Calc fresh in his head.. hmmmmmmm....... I say 16. Cheater!! |
| 2014-01-07 2:46 AM in reply to: #4925486 |
Master 1970 Somewhere on the Tennessee River | Subject: RE: Math/Statistics Help 2 raised to the 4 th power is correct. |
| 2014-01-07 7:31 AM in reply to: Left Brain |
Elite 4564 Boise | Subject: RE: Math/Statistics Help Originally posted by Left Brain Originally posted by JoshR Originally posted by Left Brain *cough* punch 2^4 into a calculator *cough* Lemme see here......i go with Kido and Josh, or the kid who only missed 2 on his SAT and has college Calc fresh in his head.. hmmmmmmm....... I say 16. Cheater!! I'm a math dork so I knew the answer off the top of my head. |
| 2014-01-07 9:39 AM in reply to: MadMathemagician |
Pro 15655 | Subject: RE: Math/Statistics Help Originally posted by MadMathemagician 2 raised to the 4 th power is correct. Isn't that 16? |
| 2014-01-07 9:45 AM in reply to: Kido |
Champion 7558 Albuquerque, New Mexico | Subject: RE: Math/Statistics Help Each "event" is binomial (0 or 1, A or B, or whatever two outcomes you choose to assign) so multiply the quantity of possible outcomes for each event. Since the events all have the same number of possible outcomes, you can simplify 2*2*2*2 = 2^4 = 16. (The answer is the same whether you assign A-B, C-D, E-F, and G-H to the events or you assign A-B, A-B, A-B, and A-B to them.) |
| 2014-01-07 10:27 AM in reply to: McFuzz |
Pro 15655 | Subject: RE: Math/Statistics Help Josh? Paging Josh!!! |
| 2014-01-07 10:28 AM in reply to: Left Brain |
Elite 4564 Boise | Subject: RE: Math/Statistics Help Originally posted by Left Brain Josh? Paging Josh!!! What? Everything in this thread is correct. |
| 2014-01-07 10:31 AM in reply to: JoshR |
Pro 15655 | Subject: RE: Math/Statistics Help Originally posted by JoshR Originally posted by Left Brain What? Everything in this thread is correct. Josh? Paging Josh!!!
|
| 2014-01-07 10:56 AM in reply to: Left Brain |
Master 1970 Somewhere on the Tennessee River | Subject: RE: Math/Statistics Help Originally posted by Left Brain Originally posted by MadMathemagician 2 raised to the 4 th power is correct. Isn't that 16?
In my world it is....
|
| 2014-01-07 1:38 PM in reply to: MadMathemagician |
Extreme Veteran 3025 Maryland | Subject: RE: Math/Statistics Help it is also (2!)^4 BOOM MIND BLOWN |
| 2014-01-09 2:57 PM in reply to: Kido |
| Iron Donkey 38643 , Wisconsin | Subject: RE: Math/Statistics Help Originally posted by Kido I used to be decent of this but maybe too much chlorine to the brain has mad it foggy. I have 4 events. Each event has 2 outcomes. What are the total number outcomes by the four events? What I mean is. Event 1 can be A or B. Event 2 can be C or D, 3 is E or F, and 4 is G or H. So ONE possible outcome is A, C, E, G. Another is B, C, E, G.... I first thought it was 4 factorial = giving me 24 possible outcomes. That SEEMS high. Then I thought it was 2 possibilities x 2 possibilities x 2 x2.. So 16 possible outcomes. Any help? I guess by the time I wrote this and now wait for a responses, I can just list them all and see... Question: Are the events dependent or independent of each other or does order matter? Because event A C E G is not the same as C E G A when order is not important. If it requires that Event A or B is always picked first (by what I am assuming from your statement " Event 1 can be A or B. Event 2 can be C or D..."), then the total number of events is lower; however, if it doesn't matter if order is important, then the total number will be higher. List them out: So, yes, you are correct with 2 x 2 x 2 x 2. |
| 2014-01-09 4:06 PM in reply to: 1stTimeTri |
Champion 7558 Albuquerque, New Mexico | Subject: RE: Math/Statistics Help Originally posted by 1stTimeTri Originally posted by Kido I used to be decent of this but maybe too much chlorine to the brain has mad it foggy. I have 4 events. Each event has 2 outcomes. What are the total number outcomes by the four events? What I mean is. Event 1 can be A or B. Event 2 can be C or D, 3 is E or F, and 4 is G or H. So ONE possible outcome is A, C, E, G. Another is B, C, E, G.... I first thought it was 4 factorial = giving me 24 possible outcomes. That SEEMS high. Then I thought it was 2 possibilities x 2 possibilities x 2 x2.. So 16 possible outcomes. Any help? I guess by the time I wrote this and now wait for a responses, I can just list them all and see... Question: Are the events dependent or independent of each other or does order matter? Because event A C E G is not the same as C E G A when order is not important. If it requires that Event A or B is always picked first (by what I am assuming from your statement " Event 1 can be A or B. Event 2 can be C or D..."), then the total number of events is lower; however, if it doesn't matter if order is important, then the total number will be higher. List them out: So, yes, you are correct with 2 x 2 x 2 x 2. You didn't list the options if order were important!  |
| 2014-01-09 6:26 PM in reply to: McFuzz |
Master 1970 Somewhere on the Tennessee River | Subject: RE: Math/Statistics Help Originally posted by McFuzz Originally posted by 1stTimeTri You didn't list the options if order were important! Originally posted by Kido I used to be decent of this but maybe too much chlorine to the brain has mad it foggy. I have 4 events. Each event has 2 outcomes. What are the total number outcomes by the four events? What I mean is. Event 1 can be A or B. Event 2 can be C or D, 3 is E or F, and 4 is G or H. So ONE possible outcome is A, C, E, G. Another is B, C, E, G.... I first thought it was 4 factorial = giving me 24 possible outcomes. That SEEMS high. Then I thought it was 2 possibilities x 2 possibilities x 2 x2.. So 16 possible outcomes. Any help? I guess by the time I wrote this and now wait for a responses, I can just list them all and see... Question: Are the events dependent or independent of each other or does order matter? Because event A C E G is not the same as C E G A when order is not important. If it requires that Event A or B is always picked first (by what I am assuming from your statement " Event 1 can be A or B. Event 2 can be C or D..."), then the total number of events is lower; however, if it doesn't matter if order is important, then the total number will be higher. List them out: So, yes, you are correct with 2 x 2 x 2 x 2.
Order is not important in this case. |
| 2014-01-09 9:42 PM in reply to: McFuzz |
| Iron Donkey 38643 , Wisconsin | Subject: RE: Math/Statistics Help Originally posted by McFuzz . I was hoping you would. Originally posted by 1stTimeTri You didn't list the options if order were important! Originally posted by Kido I used to be decent of this but maybe too much chlorine to the brain has mad it foggy. I have 4 events. Each event has 2 outcomes. What are the total number outcomes by the four events? What I mean is. Event 1 can be A or B. Event 2 can be C or D, 3 is E or F, and 4 is G or H. So ONE possible outcome is A, C, E, G. Another is B, C, E, G.... I first thought it was 4 factorial = giving me 24 possible outcomes. That SEEMS high. Then I thought it was 2 possibilities x 2 possibilities x 2 x2.. So 16 possible outcomes. Any help? I guess by the time I wrote this and now wait for a responses, I can just list them all and see... Question: Are the events dependent or independent of each other or does order matter? Because event A C E G is not the same as C E G A when order is not important. If it requires that Event A or B is always picked first (by what I am assuming from your statement " Event 1 can be A or B. Event 2 can be C or D..."), then the total number of events is lower; however, if it doesn't matter if order is important, then the total number will be higher. List them out: So, yes, you are correct with 2 x 2 x 2 x 2. |
| 2014-01-10 10:44 AM in reply to: 1stTimeTri |
Sensei Sin City | Subject: RE: Math/Statistics Help I'll disclose!
A coworker had some notion that he could bet every possible outcome of the 4 games in a parlay and no matter the outcome, win money. That's because he could only think of 4 combinations. I told him that Vegas is not that dumb AND there were many more outcomes than he thought. He challenged me to come up with others. It took me 30 seconds to figure out 4 more than he had before I quit because I proved my point. I then wanted to know the mathematical way to figure it and narrowed it down to 4! or 2^4. I ruled out 4! right away leaving 2^4. I used BT to confirm. 16 possible outcomes. You can't bet all 16 because the winning ticket would not pay 16:1 to make up for the 15 outcomes that you lose.
|
| 2014-01-10 10:57 AM in reply to: Kido |
Regular 525 | Subject: RE: Math/Statistics Help Originally posted by Kido I'll disclose!
A coworker had some notion that he could bet every possible outcome of the 4 games in a parlay and no matter the outcome, win money. That's because he could only think of 4 combinations. I told him that Vegas is not that dumb AND there were many more outcomes than he thought. He challenged me to come up with others. It took me 30 seconds to figure out 4 more than he had before I quit because I proved my point. I then wanted to know the mathematical way to figure it and narrowed it down to 4! or 2^4. I ruled out 4! right away leaving 2^4. I used BT to confirm. 16 possible outcomes. You can't bet all 16 because the winning ticket would not pay 16:1 to make up for the 15 outcomes that you lose.
I'll take his action any time. |
| 2014-01-10 10:58 AM in reply to: Kido |
Regular 1023 Madrid | Subject: RE: Math/Statistics Help I used to bet $3 trifectas like this. I would look for 8 horse races and wheel 5 horses. I would go at it be eliminating the 3 horses I thought would not finish in the top 3. Over the course of 1 summer I hit 3 times.... |
| 2014-01-10 12:34 PM in reply to: Kido |
Champion 7558 Albuquerque, New Mexico | Subject: RE: Math/Statistics Help Originally posted by Kido I'll disclose!
A coworker had some notion that he could bet every possible outcome of the 4 games in a parlay and no matter the outcome, win money. That's because he could only think of 4 combinations. I told him that Vegas is not that dumb AND there were many more outcomes than he thought. He challenged me to come up with others. It took me 30 seconds to figure out 4 more than he had before I quit because I proved my point. I then wanted to know the mathematical way to figure it and narrowed it down to 4! or 2^4. I ruled out 4! right away leaving 2^4. I used BT to confirm. 16 possible outcomes. You can't bet all 16 because the winning ticket would not pay 16:1 to make up for the 15 outcomes that you lose.
Your friend is right...he would in fact win money betting on every outcome. What your friend fails to realize is that the expected payout will be lower than the cost to play because the house always takes a cut. If Denver has a 5% chance of winning and pays 15:1, the expected payout is 0.05*15=0.75. Your friend can repeat this calculation for each team and he's unlikely to find any team where the expected payout is greater than 1.0 (which is the cost to play). Now your friend might still collect more money than he pays, but only if the winning team pays out greater than 16:1, but he can't know this is going to happen when he buys in (if he knew this when he buys in, why would he purchase the non-winning tickets?) The question you originally posed doesn't really capture what your friend is proposing. The question you posed (4 events each with two possible outcomes) only describes the first round. You can look at it as there are only 8 outcomes (one of the 8 teams will win 3 times and the other 7 will lose once and it doesn't matter when they lose). Alternatively, you can look at it as (2^4)*(2^2)*(2^1)=2^7=128 outcomes if you care when a team loses (but does it really matter when a team loses?). |
| 2014-01-10 12:54 PM in reply to: McFuzz |
Sensei Sin City | Subject: RE: Math/Statistics Help Originally posted by McFuzz Originally posted by Kido Your friend is right...he would in fact win money betting on every outcome. What your friend fails to realize is that the expected payout will be lower than the cost to play because the house always takes a cut. If Denver has a 5% chance of winning and pays 15:1, the expected payout is 0.05*15=0.75. Your friend can repeat this calculation for each team and he's unlikely to find any team where the expected payout is greater than 1.0 (which is the cost to play). Now your friend might still collect more money than he pays, but only if the winning team pays out greater than 16:1, but he can't know this is going to happen when he buys in (if he knew this when he buys in, why would he purchase the non-winning tickets?) The question you originally posed doesn't really capture what your friend is proposing. The question you posed (4 events each with two possible outcomes) only describes the first round. You can look at it as there are only 8 outcomes (one of the 8 teams will win 3 times and the other 7 will lose once and it doesn't matter when they lose). Alternatively, you can look at it as (2^4)*(2^2)*(2^1)=2^7=128 outcomes if you care when a team loses (but does it really matter when a team loses?). I'll disclose!
A coworker had some notion that he could bet every possible outcome of the 4 games in a parlay and no matter the outcome, win money. That's because he could only think of 4 combinations. I told him that Vegas is not that dumb AND there were many more outcomes than he thought. He challenged me to come up with others. It took me 30 seconds to figure out 4 more than he had before I quit because I proved my point. I then wanted to know the mathematical way to figure it and narrowed it down to 4! or 2^4. I ruled out 4! right away leaving 2^4. I used BT to confirm. 16 possible outcomes. You can't bet all 16 because the winning ticket would not pay 16:1 to make up for the 15 outcomes that you lose.
I disagree. First of all, that's not how parlays actually work - Betting every outcome my HELP is changes to win, but it's far from a guarantee. In fact, betting the moneyline on every out come (say $100 a pop) costs you $1600 bucks. HOWEVER, if all 4 favorites win? You WIN $300. Net loss of $1300. However, if the four underdogs win, you get paid $12,700 and net $11,200 with the 15 losing tickets. Just playing around, you need at least TWO of the underdogs to win to get a winning parlay ticket that will cover the $1500 you lose on the other losing tickets. There is NO guarantee to win money. If I get time, maybe I will put all 16 possible outcomes into a parlay calculator assuming a $100 bet on each. The other wrinkle, maybe you can adjust you bet size based on the payout. For example, put $10 on the 4 teamer using the underdogs. That way, you invest less in the long shots so lose less when the favorites win... |
|
|
|
|
| |||
| |||
| Obviously I an NOT as smart as a 6th grader - MATH HELP Pages: 1 2 | ||
| |||
| Math Whizzes: Help!!!! Pages: 1 2 |
| ||||
|
| |||
|
| |||
|
|
View profile
Add to friends
Go to training log
Go to race log
Send a message
View album
CONNECT WITH FACEBOOK