Probability win/lose balls

In summary: These can be answered by looking at specific cases and computing the probability of the events. You don't need a general formula.My n is different, it is the number of balls remaining once the last WIN ball is chosen. So it relates to a win on draw 20-n. A win on draw 3 means n here is 17, therefore 1/1140.One comment, when I said a formula is not required to answer the 3 given questions, I meant a general formula is not required. These can be answered by looking at specific cases and computing the probability of the events. You don't need a general formula.Yes, you are right, I made a mistake when I computed probabilities
  • #1
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Homework Statement


17 balls named LOSE
3 balls named WIN
no replacements
person who selects 3rd win ball wins the game, regardless of who selected the other 2 WIN balls

a)If i draw 1st, find the probability that I win the game on my second draw
b) If I draw 1st, find the probability that the opponent wins the game on his second draw
c)If I draw 1st, what is the probability that I win? HINT:you can win on 2nd,3rd,4th draw etc, but not on the 1st draw.
d) Would you prefer to draw 1st or 2nd?

Homework Equations





The Attempt at a Solution


a)WWW
(3/20)(2/19)(1/18)=1/1140

b)WWLW
LWWW
WLWW

3*(1/1140)

c) and d) I'm not sure... the question seems to be hinting about using 1-(complement)..but I can't seem to find it.
 
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  • #2
thoradicus said:

Homework Statement



17 balls named LOSE
3 balls named WIN
no replacements
person who selects 3rd win ball wins the game, regardless of who selected the other 2 WIN balls

a) If i draw 1st, find the probability that I win the game on my second draw
b) If I draw 1st, find the probability that the opponent wins the game on his second draw
c) If I draw 1st, what is the probability that I win? HINT:you can win on 2nd,3rd,4th draw etc, but not on the 1st draw.
d) Would you prefer to draw 1st or 2nd?


Homework Equations





The Attempt at a Solution



a)WWW
(3/20)(2/19)(1/18)=1/1140

b)WWLW
LWWW
WLWW

3*(1/1140)
You need to incorporate the probability of drawing the L ball.


c) and d) I'm not sure... the question seems to be hinting about using 1-(complement)..but I can't seem to find it.
 
  • #3
So I need to calculate the total probability of me losing when I draw first?
Like,
WWLW
LWWW
WLWW
WLLLLLWW
up until the 20th draw
etc?

something seems to be wrong lol.

What abt my other 2 answers? Are they right?
 
  • #4
I'm saying your answer to (b) is wrong because you didn't incorporate the probability of selecting an L.
 
  • #5
Youre talking about who selecting the L?Me or the other guy?
 
  • #6
thoradicus said:
Youre talking about who selecting the L?Me or the other guy?

Your value for P(LWWW or WLWW or WWLW) is incorrect: it forgets to include the fact that one 'L' is also drawn.

Anyway, if you know the hypergeometric distribution, it is not hard to find the probability that somebody wins on draw n: the first n-1 draws come from 17Ls and 3Ws and must result in exactly n-3 Ls and 2 Ws (which will then be followed by the final W). Getting the probability for the first n-1 is just an example of the hypergeometric distribution; see, eg., http://en.wikipedia.org/wiki/Hypergeometric_distribution .

So, just compute these for all relevant n and sum them up.
 
Last edited:
  • #7
well i did, i think :confused:

for example, (3/20)(2/19)(17/18)(1/17) WWLW, last win ball drawn by opponent, probability is still 1/1140?

alright, ill try the above method.
 
  • #8
thoradicus said:
well i did, i think :confused:

for example, (3/20)(2/19)(17/18)(1/17) WWLW, last win ball drawn by opponent, probability is still 1/1140?

alright, ill try the above method.

OK: P{2nd wins on draw 4} = 3/1140 = 1/380, with that factor of 3 arising, of course, from the fact that the L can be in anyone of three places, and all three places give the same probability of 1/1140. The '3' in your expression is the number of W balls---so is a 'different' 3 from the one that counts the places for L!
 
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  • #9
thoradicus said:
well i did, i think :confused:

for example, (3/20)(2/19)(17/18)(1/17) WWLW, last win ball drawn by opponent, probability is still 1/1140?
Yes, you were right. Sorry. It's just looked like you took the answer to part (a) and multiplied it by 3 to answer (b), which turns out to yield the right answer but for the wrong reason.
 
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  • #10
ah no problem, lol should have made it clearer i guess.
 
  • #11
Another way to get a formula is to think of the balls as placed in a line. The tail of the list is a WIN ball followed by n LOSE balls. The probability of this is 3 * 17!/(17 - n)! * (20-(n+1))!/20! = 1/2280 * (19-n)!/(17-n)! = (18 - n)(19 - n)/2280.

However, a formula is not required to answer part c) (nor the others).
 
  • #12
verty said:
Another way to get a formula is to think of the balls as placed in a line. The tail of the list is a WIN ball followed by n LOSE balls. The probability of this is 3 * 17!/(17 - n)! * (20-(n+1))!/20! = 1/2280 * (19-n)!/(17-n)! = (18 - n)(19 - n)/2280.

However, a formula is not required to answer part c) (nor the others).

This formula does not work: it gives 0 when n = 18 or 19.

A correct formula for the probability that somebody wins on draw n is
[tex] p_n ={n-1 \choose 2} \frac{1}{1140} = \frac{(n-1)(n-2)}{2} \frac{1}{1140}, \; n = 3, 4, \ldots, 20.[/tex]
 
  • #13
Ray Vickson said:
This formula does not work: it gives 0 when n = 18 or 19.

A correct formula for the probability that somebody wins on draw n is
[tex] p_n ={n-1 \choose 2} \frac{1}{1140} = \frac{(n-1)(n-2)}{2} \frac{1}{1140}, \; n = 3, 4, \ldots, 20.[/tex]

My n is different, it is the number of balls remaining once the last WIN ball is chosen. So it relates to a win on draw 20-n. A win on draw 3 means n here is 17, therefore 1/1140.

One comment, when I said a formula is not required to answer the 3 given questions, I meant a general formula is not required.
 

1. What is the definition of "Probability win/lose balls"?

Probability win/lose balls is a mathematical concept used to determine the likelihood of a certain event occurring by counting the number of possible outcomes and dividing it by the total number of outcomes.

2. How are probability win/lose balls calculated?

To calculate probability win/lose balls, you need to determine the total number of possible outcomes and the number of desired outcomes. The probability is then calculated by dividing the number of desired outcomes by the total number of possible outcomes.

3. What is the difference between theoretical and experimental probability?

Theoretical probability is the probability of an event occurring based on mathematical calculations, while experimental probability is based on actual observations and data collected from experiments.

4. How does the number of win/lose balls affect the probability?

The number of win/lose balls has a direct impact on the probability. The more win/lose balls there are, the lower the probability of winning and vice versa. This is because the more possible outcomes there are, the less likely a specific outcome will occur.

5. How can probability win/lose balls be applied in real-life situations?

Probability win/lose balls can be applied in various real-life situations, such as in gambling, insurance, and weather forecasting. It can also be used to make informed decisions in business, sports, and other areas where predicting outcomes is important.

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