rmani said:Also Phoen$x I don't know if she was hot because the 1st rounds are just a telephone interview......I must admit though she sounded pretty cute.
ok...so what's the probability that she <B>IS</B> hot???
Can anyone answer this probability question?
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ok...so what's the probability that she <B>IS</B> hot???
sob.....it's nice to know that you're right for once in your life.....
for those that didn't get the right answer, don't sweat it.....
i didn't get it by chance...it took hard work for 4 years studying as an undergrad and grad in electrical engineering by going through tons of calculus, statistics and EE hardcore math.
u can't imagine how many brain cells have been fried at those times....
as for the casino, the chance of winning at each game is probably already less than 50% due to the statistics of card games....yes, some may argue that you either win or lose a card game, but the chances of you obtaining winning cards is a totally different calculation.....52 possibilities in various combinations and permutations.
i'm outta here before i lose more brain cells.....
for those that didn't get the right answer, don't sweat it.....
i didn't get it by chance...it took hard work for 4 years studying as an undergrad and grad in electrical engineering by going through tons of calculus, statistics and EE hardcore math.
u can't imagine how many brain cells have been fried at those times....
as for the casino, the chance of winning at each game is probably already less than 50% due to the statistics of card games....yes, some may argue that you either win or lose a card game, but the chances of you obtaining winning cards is a totally different calculation.....52 possibilities in various combinations and permutations.
i'm outta here before i lose more brain cells.....
The rationale in this thread may appear to logical @ firstsight but when in fact it's illogical. Let's take a practical example for illustration:
A preganacy tester is 95% (or any number for that matter) accurate, what is the probability of actual pregnacy if a woman takes 2/4/6 pregancy tests.
Well, judging by the rationale introduced earlier, the probability actually decreases as more tests are administered. Therefore showing a negative correlation between the likelihood of pregnacy and the number of tests conducted When in fact the probability of a positive pregnacy should be heightened with the addtional tests. I think I have the answer to this one, just need to dig up the actual fomulas/logics.
A preganacy tester is 95% (or any number for that matter) accurate, what is the probability of actual pregnacy if a woman takes 2/4/6 pregancy tests.
Well, judging by the rationale introduced earlier, the probability actually decreases as more tests are administered. Therefore showing a negative correlation between the likelihood of pregnacy and the number of tests conducted When in fact the probability of a positive pregnacy should be heightened with the addtional tests. I think I have the answer to this one, just need to dig up the actual fomulas/logics.
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why don't we reason through this one as well.....
you've changed factors already....a 95% "accuracy" means 19/20 times it will be correct and thus
let's make some numbers again... 1-19 = positive; 20 = negative
(positive is pregnant)
for 1 pregnancy test, chances are 19/20
for 2 tests, you need to give more specifics
1 out of 2 tests positve = pregnant???? OR
2 out of 2 tests positive = pregnant???
Let's calculate both just for fun
test 1 test 2
1 1
2 2
3 3
. .
. .
. .
20 20
There are a total of 400 possible combinations
if only 1 test passed is needed, then chances are 399/400 = 99.75%
if both tests need to pass, then chances are 361/400 = 90.25%
just use excel and build yourself a 2-D 20X20 matrix and X out all of the possibilities...build a 3-D 20X20 matrix for 3 tests, etc....
just for fun....
you've changed factors already....a 95% "accuracy" means 19/20 times it will be correct and thus
let's make some numbers again... 1-19 = positive; 20 = negative
(positive is pregnant)
for 1 pregnancy test, chances are 19/20
for 2 tests, you need to give more specifics
1 out of 2 tests positve = pregnant???? OR
2 out of 2 tests positive = pregnant???
Let's calculate both just for fun
test 1 test 2
1 1
2 2
3 3
. .
. .
. .
20 20
There are a total of 400 possible combinations
if only 1 test passed is needed, then chances are 399/400 = 99.75%
if both tests need to pass, then chances are 361/400 = 90.25%
just use excel and build yourself a 2-D 20X20 matrix and X out all of the possibilities...build a 3-D 20X20 matrix for 3 tests, etc....
just for fun....
fangtl said:
This is simple.
The chance for error is 0.05 (5%) if only one test is conducted.
The chance for two consecutive test errors is (5%)^2=0.0025 = 0.25% so the outcomes of two consecutive positive tests have a 99.75% chance of being correct, which is higher than the 95% accuracy of just doing one test.
The probability for accuracy increases exponentially as additional tests are conducted.
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hee....u guys don't think i get paid to come to work and surf prime all day, do u?
fangtl said:
That's not how statistic is derived. A 95% accuracy really means that if you repeat the experienment many times, then 95% of the time, you get the desired result. Just like flipping a coin. If you flip a fair coin once and get a tail, then the probability of getting a tail is 100%, you can repeat the experienment 5 times, and you may get a string of 4 tails and 1 head, still have a very high probability of getting a tail. But if you repeat the experienment many times, let's say 100 times, than the probability of getting a tail will likely be only 50%. This is what we mean when saying there is a 50% chance of getting a tail in a coin flip, not that each flip will produce a tail 50% of the time. Statistics only works when the number of experienment is large, in other words the sample population is large.
Like I said, this problem is a simple binomial distribution problem. You have an experienment--playing a basketball game, with two outcomes--win or lose (tie is not given as a outcome for the problem, so we ignore it). The two outcomes are mutually exclusive and collectively exhaustive(means their probability adds up to 1). Thus this is a binary experienment. We repeat this experienment many times and it would become binomial. If you want to formula for calculating binomial probabilty, here it is:
P(x)=[n!/x!*(n-x)!]*[Pi^(x)]*[(1-Pi)^(n-x)]
where:
x=number of successes (the desired outcome)
n=number of experienments
Exactly, maomao. In the scenarios discussed here, the outcome of one test does not affect the outcome of the next one. So the likelyhood that you'll beat MJ doesn't go up as you play more games. You still have 50% chance of winning each game regardless of how many games you play, but the likelihood that you'll win a game goes up with more games played.
No idea why I am jumping in on this one, but I will.
In the way the way the question was worded, the only safe answer is to play one game. Why?
"You can play a 1 on 1 game against Jordan and have a 50% chance of winning 1 game."
It does not say you have a 50% chance of winning every game. It just says you have a 50% chance of winning one game.
If you choose a one game series, you have a 50% chance of winning that game.
If you choose a series of 3/5/7/etc., there are no set rules for the outcomes of any game after 1. It is possible that your chances of winning 1 game is 50%, while your chances of winning additional games in 10%. True, your chance of winning additional games could be 90%, but without knowing, the safe bet, is 1 game.
And perhaps here is where knowing your opponent is Michael Jordan is useful.
In the way the way the question was worded, the only safe answer is to play one game. Why?
"You can play a 1 on 1 game against Jordan and have a 50% chance of winning 1 game."
It does not say you have a 50% chance of winning every game. It just says you have a 50% chance of winning one game.
If you choose a one game series, you have a 50% chance of winning that game.
If you choose a series of 3/5/7/etc., there are no set rules for the outcomes of any game after 1. It is possible that your chances of winning 1 game is 50%, while your chances of winning additional games in 10%. True, your chance of winning additional games could be 90%, but without knowing, the safe bet, is 1 game.
And perhaps here is where knowing your opponent is Michael Jordan is useful.
i dont think im totally understanding the initial story problem.
it would seem if each game is a 50/50 chance that your best likliehood of winning 'One' game would be to repeat the games multiple times(five games being ideal). Repitition increases the likliehood of the 50% statistic averaging out. But i dont think she really meant 'One' game. Even if you only had a 20% chance of winning one game, youd want to play one hundred times and youd likely win 20times. If you only played twice youd almost certainly not win one game if you only had a 20% chance. Was the goal to win at least one game, or to win the most games possible and beat Jordan in general?
it would seem if each game is a 50/50 chance that your best likliehood of winning 'One' game would be to repeat the games multiple times(five games being ideal). Repitition increases the likliehood of the 50% statistic averaging out. But i dont think she really meant 'One' game. Even if you only had a 20% chance of winning one game, youd want to play one hundred times and youd likely win 20times. If you only played twice youd almost certainly not win one game if you only had a 20% chance. Was the goal to win at least one game, or to win the most games possible and beat Jordan in general?
still going? you guys are more efficient than the energizer bunny!
huckster said:
The goal was to win the series, ie win 1 out of 1 game, 2 out of 3 games, or 3 out of 5 games. If our interpretation of the question is correct, and the 50% odds of winning applies to every game played, then it doesn't matter how long of a series you pick, your odds of winning are still 50% overall.
Does anyone else think we've gone waaaaaay too far in analyzing this problem already?
<B>cxr344</B>: we're way beyond analyzing this...
I dont understand how the probability of you winning could decrease with more games. If you restate the question as you have a 50% chance of losing you'd get the same answer.... that just don't make no sense at all.
Also, playing Jordon we'd be lucky if we had a 5% chance of winning. They should have reworded the question as "racing a boxster against your NSX".... would have made more sense.
I dont understand how the probability of you winning could decrease with more games. If you restate the question as you have a 50% chance of losing you'd get the same answer.... that just don't make no sense at all.
Also, playing Jordon we'd be lucky if we had a 5% chance of winning. They should have reworded the question as "racing a boxster against your NSX".... would have made more sense.
UPDATE
Hey guys just wanted to let you know I recieved an email from that trading firm...looks like whatever the answer was to this probabilty question I answered it ok...because I made the 2nd round of interviews! Just thought I'd share, wish me luck!
BTW if anyone knows any applied probabilty questions feel free to let me know so I can practice more of them. Thanks!
Hey guys just wanted to let you know I recieved an email from that trading firm...looks like whatever the answer was to this probabilty question I answered it ok...because I made the 2nd round of interviews! Just thought I'd share, wish me luck!
BTW if anyone knows any applied probabilty questions feel free to let me know so I can practice more of them. Thanks!
Oh no, you started it again!
Actually, congrats! Have a look here:
http://www.wilmott.com/310/messageview.cfm?catid=16&threadid=5297
Actually, congrats! Have a look here:
http://www.wilmott.com/310/messageview.cfm?catid=16&threadid=5297
If the chances of you winning each game are 50 percent, then the chances of either of you winning a majority of any odd number of games is also 50 percent, assuming you haven't started playing yet, and assuming that a win in one game doesn't affect the chances of winning any game after that.
However, that statement is only true for 50 percent. If the chances of you winning each game are less than 50 percent, then the chances of you winning a series of games goes down the longer the series is.
Let's say your chances of winning each game are 20 percent, and work through the math.
Obviously, the chances of you winning 1 game out of 1 is 20 percent.
Let's say you play best 2 out of 3. The chances of you winning the first two games are 4 percent (.2 * .2), and the series is over at that point. The chances of you losing the first game and winning the last two games are 3.2 percent (.8 * .2 * .2). The chances of you winning the first game, losing the second game, and winning the third game are also 3.2 percent (.2 * .8 * .2). Thus, the chances of you winning a best 2 out of 3 series are 10.4 percent.
Now let's work through the math using the formula that George posted (and thanks, George):
First, I would like to insert an additional set of parentheses to make one thing clearer, because the formula is likely to be interpreted the wrong way as originally written:
P(x)=[n!/(x!*(n-x)!)]*[Pi^(x)]*[(1-Pi)^(n-x)
This formula is correct but with two conditions: (1) it assumes that all "n" games (experiments) are played, and (2) it calculates the chances that you win exactly x times, not at least x times.
(If you're wondering where this formula comes from, it takes into account all the different combinations that the wins and losses can occur - for example, winning exactly two out of three games can happen in three different combinations: losing the first game, losing the second game, or losing the third game.)
Thus, the chances of winning exactly two games out of 3, according to this formula, as long as all 3 games are played, are:
P(2)=[3!/(2!*(3-2)!)]*[.2**2]*[(1-.2)**(3-2)]
=[3]*[.04]*[.8]=.096 or 9.6 percent
And the chances of winning exactly three games out of 3, according to this formula, as long as all 3 games are played, are
P(3)=[3!/(3!*(3-3)!)]*[.2**3]*[.8**0]
=[1]*[.008]*[1]=.008 or 0.8 percent
so that, again, the chances of you winning a best 2 out of 3 series are 10.4 percent.
If you play a best of five series, your chances of winning are P(3)+P(4)+P(5), calculated as follows:
P(3)=[5!/(3!*(5-3)!)]*[.2**3]*[(1-.2)**(5-3)]
=[10]*[.008]*[.8**2]=.0512 or 5.12 percent.
P(4)=[5!/(4!*(5-4)!)]*[.2**(4)]*[(.8)**(5-4)]
=[5]*[.0016]*[.8]=.00064 or .064 percent.
P(5)=[5!/(5!*(5-5)!)]*[.2**(5)]*[(.8)**(5-5)]
=[1]*.000128*[1]=.000128 or .0128 percent.
Thus the chances of winning a best of five series are 5.1968 percent.
As you can see, if your chances of winning a single game are less than 50 percent, then your chances of winning a series of games go down, the longer the series is going to be.
However, that statement is only true for 50 percent. If the chances of you winning each game are less than 50 percent, then the chances of you winning a series of games goes down the longer the series is.
Let's say your chances of winning each game are 20 percent, and work through the math.
Obviously, the chances of you winning 1 game out of 1 is 20 percent.
Let's say you play best 2 out of 3. The chances of you winning the first two games are 4 percent (.2 * .2), and the series is over at that point. The chances of you losing the first game and winning the last two games are 3.2 percent (.8 * .2 * .2). The chances of you winning the first game, losing the second game, and winning the third game are also 3.2 percent (.2 * .8 * .2). Thus, the chances of you winning a best 2 out of 3 series are 10.4 percent.
Now let's work through the math using the formula that George posted (and thanks, George):
maomaonsx said:
First, I would like to insert an additional set of parentheses to make one thing clearer, because the formula is likely to be interpreted the wrong way as originally written:
P(x)=[n!/(x!*(n-x)!)]*[Pi^(x)]*[(1-Pi)^(n-x)
This formula is correct but with two conditions: (1) it assumes that all "n" games (experiments) are played, and (2) it calculates the chances that you win exactly x times, not at least x times.
(If you're wondering where this formula comes from, it takes into account all the different combinations that the wins and losses can occur - for example, winning exactly two out of three games can happen in three different combinations: losing the first game, losing the second game, or losing the third game.)
Thus, the chances of winning exactly two games out of 3, according to this formula, as long as all 3 games are played, are:
P(2)=[3!/(2!*(3-2)!)]*[.2**2]*[(1-.2)**(3-2)]
=[3]*[.04]*[.8]=.096 or 9.6 percent
And the chances of winning exactly three games out of 3, according to this formula, as long as all 3 games are played, are
P(3)=[3!/(3!*(3-3)!)]*[.2**3]*[.8**0]
=[1]*[.008]*[1]=.008 or 0.8 percent
so that, again, the chances of you winning a best 2 out of 3 series are 10.4 percent.
If you play a best of five series, your chances of winning are P(3)+P(4)+P(5), calculated as follows:
P(3)=[5!/(3!*(5-3)!)]*[.2**3]*[(1-.2)**(5-3)]
=[10]*[.008]*[.8**2]=.0512 or 5.12 percent.
P(4)=[5!/(4!*(5-4)!)]*[.2**(4)]*[(.8)**(5-4)]
=[5]*[.0016]*[.8]=.00064 or .064 percent.
P(5)=[5!/(5!*(5-5)!)]*[.2**(5)]*[(.8)**(5-5)]
=[1]*.000128*[1]=.000128 or .0128 percent.
Thus the chances of winning a best of five series are 5.1968 percent.
As you can see, if your chances of winning a single game are less than 50 percent, then your chances of winning a series of games go down, the longer the series is going to be.
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As I understood the question the goal was to win 1 game and the odds of winning was 50% (odds are normally stated for one play/game). I would pick 5 games, 5 chances to win one game looks like the best bett to me. Thinking about commodity trading there are really only 2 out comes, Make money (win) and loose money on each trade and that that stays the same no mater how many trades you make. So the more trades you make the more likelly you will win 1.
Has any one heard the old saying "If you don't at frist succeed try try again"
Has any one heard the old saying "If you don't at frist succeed try try again"
Briank said:
Nope. The question was, the odds of winning a majority of the games. If you're only looking at the chances of winning at least one game, then it's obvious that the more games, the better the chances. But that wasn't the question.
Briank said:
I don't think my objection in investment trading would be to have at least one trade that made money, regardless of how many trades you lost on, or how much money you lost.
But again, that wasn't the question.
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