Probability: An urn contains 20 balls

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bry32321

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An urn contains 16 balls marked LOSE and 4 balls marked WIN.You and your friend take turns randomly picking a ball from the urn. The person who picks the fourth WIN ball wins the game. It does not matter who picked the first three WIN balls.


If your opponent draws first, what is the probability that you win? HINT: You could win on your second, third, fourth,..., or tenth draw but not on your first.


I have pretty much no idea how to go about this problem.
All I know is that it should be something along the lines of
the sum from k=2 to 10 for (insert some generalized formula)


Could someone explain this problem to me in terms of how to solve it and why you would solve it in such way?

Please and thank you!
 
bry32321 said:
An urn contains 16 balls marked LOSE and 4 balls marked WIN.You and your friend take turns randomly picking a ball from the urn. The person who picks the fourth WIN ball wins the game. It does not matter who picked the first three WIN balls.
If your opponent draws first, what is the probability that you win? HINT: You could win on your second, third, fourth,..., or tenth draw but not on your first.
Click to expand...

You left out a key bit of information!

After each draw is the ball discarded or returned to the urn?
That determines the nature of the probability function.
 
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Sorry about that.
It is without replacement, which is why when the fourth WIN ball is pulled, a winner is determined.
 
Hello, bry32321!

I made an exhaustive List.

An urn contains 16 balls marked LOSE and 4 balls marked WIN.
You and your friend take turns randomly picking a ball from the urn.
The person who picks the fourth WIN ball wins the game.
It does not matter who picked the first three WIN balls.

If your friend draws first, what is the probability that you win?
Click to expand...

Draw the balls one at a time and place them in a row.
There are: \(\displaystyle {20\choose4,16} = 4845\) possible outcomes.

You can win on your second draw
if the first four draws are \(\displaystyle WWWW.\)
There is 1 way this can happen.

You can win on your third draw
if the first 5 draws are \(\displaystyle \{W,W,W,L,L\}\) in some order
and you get \(\displaystyle W\) on your third draw.
There are: \(\displaystyle {5\choose3,2}\cdot1 = \color{blue}{10}\) ways.

You can win on your fourth draw
if the first 7 draws are \(\displaystyle \{W,W,W,L,L,L,L\}\) in some order
and you get \(\displaystyle W\) on your fourth draw.
There are: \(\displaystyle {7\choose3,2}\cdot1 = \color{blue}{35}\) ways.

You can win on your fifth draw
if the first 9 draws are \(\displaystyle \{W,W,W,L,L,L,L,L,L\}\) in some order
and you get \(\displaystyle W\) on your fifth draw.
There are: \(\displaystyle {9\choose3,2}\cdot1 = \color{blue}{84}\) ways.

. . . and so on.


The number of ways in which you win is:

\(\displaystyle 1 + {5\choose3} + {7\choose3} + {9\choose3} + {11\choose3} + {13\choose3} + {15\choose3} + {17\choose3} + {19\choose3}\)

. . \(\displaystyle =\;1 + 10 + 35 + 84 + 165 + 286 + 455 + 680 + 969 \)

. . \(\displaystyle =\;2685\)


Therefore: .\(\displaystyle P(\text{You win}) \;=\;\dfrac{2685}{4845} \;=\;\dfrac{179}{323} \;=\;0.554179567
\)

You will win about 5 times in 9 games.
 
Thank you so very much! I know that exhaustive list was probably also pretty exhausting to make so thank you for your time and your help!


But I have one more question, why is it always (2k-1 Choose 3) as in how did you get the 3?
 
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Thank you! I can see that the second picker has a better chance at winning than the person who picks first.

I'm having problems understanding how to deduce the generalized formula for the answer. From the help of soroban, I was able to see that the formula to derive the answer from is the sum from k=2 to 10 for (((2k-1) choose 3)/ (20 choose 4)). However, I don't know how you would figure out that the (2k-1) choose 3 should have the 3 in there.
I can see that logically having that 3 there is correct, but it doesn't seem obvious to me and I'm trying to understand it in a way that would allow me to see that obviously it should be (2k-1) choose 3.
 
Let me try to explain a little differently (but pretty much the same). Maybe it will help.

There are only 4 W's. To win on your turn (the 2nth draw), the first three W's should be drawn already before this turn, so 3W's must be present in draws 1,...,2n-1.

\(\displaystyle P(\text{win on 2}n^{th}\text{ draw}) = P(\text{[first } 2n-1\text{ draws contain 3 Ws] } \cap [2n^{th} \text{ draw is a W.}])\)

\(\displaystyle P(\text{first } 2n-1\text{ draws contain 3 Ws })\cdot P( 2n^{th} \text{ draw is a (the last) W.})\)

\(\displaystyle \left(\dfrac{{4\choose 3}{16\choose 2n-4}}{20\choose 2n-1}\right)\cdot \dfrac{1}{20-(2n-1)}\)

Then the solution is

\(\displaystyle \sum_{n=2}^{10} \left(\dfrac{{4\choose 3}{16\choose 2n-4}}{20\choose 2n-1}\right)\cdot \dfrac{1}{21-2n}\)
 
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