# A Challenging Question

• Jul 31st 2009, 12:13 AM
gtaplayr
A Challenging Question
An urn contains 17 blue balls, and 3 red balls. You and an opponent take turns drawing a ball from the urn (and keeping it). The winner of the game is the one who draws the third red ball. Find the probability of winning if you draw (a) first, (b) second.
• Aug 1st 2009, 10:55 PM
matheagle
Let's take the case of picking first.

The probability of winning on YOUR second pick meant that you picked red, your opponent picked red
and you picked red again.

$\displaystyle P({\rm win \; on\; your\; second\; pick})=P(RRR)= \biggl({3\over 20}\biggr) \biggl({2\over 19}\biggr) \biggl({1\over 18}\biggr)$

The probability of winning on YOUR third pick meant that on the first four selections there were two red and two blues
AND then you picked the third red on the fifth overall selection.

$\displaystyle P({\rm win\; on\; your\; third\; pick})={4\choose 2} P(RRBBR)$

$\displaystyle ={4\choose 2} \biggl({3\over 20}\biggr) \biggl({2\over 19}\biggr) \biggl({17\over 18}\biggr)\biggl({16\over 17}\biggr)\biggl({1\over 16}\biggr)$

$\displaystyle P({\rm win\; on\; your\; fourth\; pick})={6\choose 2} P(RRBBBBR)$

$\displaystyle ={6\choose 2} \biggl({3\over 20}\biggr) \biggl({2\over 19}\biggr) \biggl({17\over 18}\biggr) \biggl({16\over 17}\biggr) \biggl({15\over 16}\biggr) \biggl({14\over 15}\biggr) \biggl({1\over 14}\biggr)$

and so on.....

These fractions reduce, but I wanted to show the pattern.

They reduce to

$\displaystyle {2n\choose 2} \biggl({3\over 20}\biggr) \biggl({2\over 19}\biggr) \biggl({1\over 18}\biggr)$

where n=1,2,..9, since your last pick is the 19th and we need that to be red and have conditions placed on the first 18 balls. So my answer seems to be

$\displaystyle \sum_{n=1}^9{2n\choose 2} \biggl({3\over 20}\biggr) \biggl({2\over 19}\biggr) \biggl({1\over 18}\biggr) = \biggl({3\over 20}\biggr) \biggl({2\over 19}\biggr) \biggl({1\over 18}\biggr)\sum_{n=1}^9{2n\choose 2}$
• Aug 2nd 2009, 07:09 PM
gtaplayr
Thanks that helps a lot.

Now for the second part:

If the probability of winning from drawing first is P(A) and the probability of winning from drawing second is P(B).

Then it must be that P(B) = 1 - P(A) ??
• Aug 2nd 2009, 08:15 PM
matheagle
I think that's correct, but it's always a good idea to see if those two sum to one.
• Aug 3rd 2009, 03:33 PM
gtaplayr
Worked through them both and they do add to 1.

But why is it that P(A) is less then P(B), when you draw first you can win on the third pick, which is earlier that when you draw second, so shouldnt P(A) be greater than P(B)?
• Aug 3rd 2009, 03:55 PM
matheagle
I would think P(A)>P(B), but I didn't add them up.
• Aug 3rd 2009, 10:49 PM
gtaplayr
I think now it is because to win on your second pick when drawing first it has to be RRR, but to win on your second pick when drawing second it can go RRBR or BRRR or RBRR.

So the smaller prob comes from there being more combinations available to the person drawing second.

Is this right?
• Aug 3rd 2009, 11:04 PM
matheagle
That does make some sense, BUT I would need to see the numbers.
I didn't calculate the probabilities because I feel that you need to work through these problems.
Students don't learn by having the entire problem solved for them.
• Aug 4th 2009, 02:05 AM
Flyingdutchman
also if you draw second there are less balls left, which would increase the chance of picking the last red one
• Aug 13th 2009, 12:28 AM
Lomofun
I was wrong.
• Aug 15th 2009, 06:03 PM
gtaplayr
To win you must pick out the last red ball, so the combinations are made up of 1 less than the amount of balls taken out, because the last must be red