# Probability win/lose balls

1. Aug 31, 2013

1. The problem statement, all variables and given/known data
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?
2. Relevant equations

3. 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 cant seem to find it.

2. Aug 31, 2013

### vela

Staff Emeritus
You need to incorporate the probability of drawing the L ball.

3. Aug 31, 2013

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. Aug 31, 2013

### vela

Staff Emeritus
I'm saying your answer to (b) is wrong because you didn't incorporate the probability of selecting an L.

5. Aug 31, 2013

Youre talking about who selecting the L?Me or the other guy?

6. Sep 1, 2013

### Ray Vickson

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: Sep 1, 2013
7. Sep 3, 2013

well i did, i think

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. Sep 3, 2013

### Ray Vickson

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 any one 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!

9. Sep 4, 2013

### vela

Staff Emeritus
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.

10. Sep 4, 2013

ah no problem, lol should have made it clearer i guess.

11. Sep 4, 2013

### verty

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. Sep 4, 2013

### Ray Vickson

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
$$p_n ={n-1 \choose 2} \frac{1}{1140} = \frac{(n-1)(n-2)}{2} \frac{1}{1140}, \; n = 3, 4, \ldots, 20.$$

13. Sep 5, 2013

### verty

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.