0
$\begingroup$

If someone could please point me in the right direction on these. I get lost on how to think about them.

In a game there are four holes with values 0, 1, 2, and 4. You are given 6 balls to shoot into the holes.

What is the sample outcome: This is $4^6$. Any ball can go into any hole.

Probability that at least one of the holes has three or more balls? First off, what is the complement here. I know it might not be used, but I would like to know. I thought it might be none of the holes have three or more balls. So all the holes have less than 3 balls, so at most 2. This is where I get lost on how think about where to put the balls into the holes.

Prob at least one of 4 holes has no ball? Here I said prob of 1 hole having no balls is $1/6$. Prob of 2 holes is $1/36$. Prob of 3 holes having no balls is $1/6^3$. Lastly prob of 4 holes having no balls is 0. So I add up these correct?

If score 20, win another game. What is Prob of a new game? Here I said, 6 balls into 4, get 24. 5 balls in 4 get 20, so have one ball left can go into any 3 holes, so have 3 ways. Lastly 4 balls in 4, need 2 balls in 2. So there are 5 chances of a new game. So $5/6^4$ I said, which seems very small. Order does not matter in this case correct?

$\endgroup$
2
$\begingroup$

There are $4^6$ equally probable microstates in the sample space.   Each distinct ball can have $4$ distinct states.

Probability that at least one of the holes has three or more balls?

The complement of this event is that all holes contain less than three balls. Which will be:

  • (Case $a$) 3 holes each containing 2 balls, and 1 hole empty, or
  • (Case $b$) two holes with two balls and two holes with one ball.

$$1-\dfrac{{4\choose 3}{6\choose 2,2,2}+{4\choose 2}{6\choose 2,2,1,1}}{4^6}$$


Prob at least one of 4 holes has no ball?

Use the principle of inclusion and exclusion, to count the ways at least one hole can be empty.

Count the ways to select one hole, and put the balls in the other three, minus the ways to select 2 holes and put the balls into the other two, plus the ways to select 3 holes and put the balls into the remaining hole.

$$\frac{{4\choose 1}3^6-{4\choose 2}2^6+{4\choose 3}1^6}{4^6}$$

We use the Principle of Inclusion and Exclusion to avoid over counting intersections.

The set of arrangements where hole 1 is empty intersects with set of arrangements where hole 2 is empty. The intersection is the set of arrangements where holes 1 and 2 are empty. Et cetera.

So let $E_x$ represent the arrangements where hole $x$ is empty, $E_{x,y}$ those where hole $x$ and $y$ are both empty and likewise $E_{x,y,z}$.

$$\begin{align} |\bigcup_x E_x| & = \sum_x |E_x| - \sum_{x<y} |E_{x,y}| + \sum_{x<y<z} |E_{x,y,z}| & \text{P.I.E.} \\ & = 4 |E_\ast| - {4\choose 2}|E_{\ast,\ast\ast}| + {4\choose 3} |E_{\ast,\ast\ast,\ast\ast\ast}| & \text{each set of the same arity has the same cardinality} \\ & = 4\cdot 3^6 - 6\cdot 2^6 + 4\cdot 1^6 \end{align}$$


If score 20, win another game. What is Prob of a new game?

Yes. Considering the cases needed to score 20 or more gives:

  • Six balls score $4$.
  • Five balls score $4$, one balls scores $0$, $1$, or $2$.
  • Four balls score $4$, two balls each score $2$.

Can you complete?


Note Do not use stars and bars, or counting visible outcomes, as we must assume each ball is distinct for the purpose of evaluating probabilities.

Although the balls appear to be indistinguishable they are in fact distinct entities; they have hidden-distinctness. Several arrangements of such entities can result in the same outcome. To measure the probabilities of an event it is not sufficient to merely count the number of distinct outcomes, we must actually measure the probabilities of each distinct outcome by considering the arrangements of hidden-distinct entities that contribute to them.

As demonstrated below:

If I take 2 marked balls and toss them into 3 distinct holes, the $3^2$ equally possible outcomes are:

$$\begin{array}{|c|c|c|}\hline AB & & \\\hline & AB & \\\hline & & AB \\\hline A & B & \\\hline B & A & \\\hline A & & B \\\hline B & & A \\\hline & A & B \\\hline & A & B \\\hline \end{array}$$

All these outcomes are equally likely if each ball has an independent $1/3$ probability of landing in any given hole.

So for example, the probability that the balls land in two different boxes is: $2/3$.

Now, should a coloured sheet of plastic be placed between you and the hole so that the marks become indistinct, then the visible outcomes become: $$\begin{array}{|c|c|c|}\hline XX & & \\\hline & XX & \\\hline & & XX \\\hline X & X & \\\hline X & & X \\\hline & X & X \\\hline \end{array}$$

However, the probabilities that the balls land in two different boxes does not magically become $1/2$. The individual balls still have the same probabilities of landing where they may even if you can no longer tell them apart.

$\endgroup$
7
  • $\begingroup$ could you please provide an explanation for why it is not 6^4. Is it because each time assuming that have 6 balls to shot in each hole when that is not always the case. But each ball can be shot into any four hole. Not any four hole have 6 total balls at a time? $\endgroup$ – Jack Armstrong Mar 3 '15 at 22:05
  • $\begingroup$ @Jack It's $4^6$ because each ball can be in only one hole at a time. There's $4$ ways one ball can land, $4^2$ ways two balls can land, et cetera. Also the balls have hidden distinctiveness. If we were tossing two identical coins then we'd still calculate probabilities of those events by treating the coins as distinct, even though we can't tell them apart. $\endgroup$ – Graham Kemp Mar 3 '15 at 22:05
  • $\begingroup$ In reading this and understanding it I will ask questions along the way. In case b, why do you have 4 choose 2? I get 4 choose 3 in case a. Why not 4 choose 2,2 because of how we pick holes. I know we end up with the same number, but. $\endgroup$ – Jack Armstrong Mar 3 '15 at 22:27
  • $\begingroup$ that and I do not understand why you use inclusion-exclusion principle because it is at least one of 4 has no ball. Which you normally add to find the other aspects. I see how you need the combination part. I have not done the arithmetic to add it up. $\endgroup$ – Jack Armstrong Mar 3 '15 at 22:42
  • $\begingroup$ @Jack You use PIE to avoid over counting events in intersections. The set of ways to select hole 1 and to distribute the balls into the other holes intersects with the set of ways to select hole 2 and to distribute balls into the other holes. This intersection is the set of ways to select holes 1 and 2, and to distribute the balls to the other holes. $\endgroup$ – Graham Kemp Mar 3 '15 at 23:02
0
$\begingroup$
  1. This is the classic stars and bars problem, so there are ${6+4-1\choose6}=210$ arrangements.
  2. The complement here is that all of the holes have 2 or less balls.
  3. Reuse the stars and bars but eliminate one of the holes (the one that has no balls) and multiply this by the 4 ways there are of choosing that hole.
  4. Your numerator looks right, your denominator is $210$. So $\frac{1}{42}$.
$\endgroup$
1
  • $\begingroup$ No. "Stars and bars" counts distinguishable outcomes, but does not give weight to their occurrence. The distinguishable outcomes are not equally likely. $\endgroup$ – Graham Kemp Mar 3 '15 at 3:31
-2
$\begingroup$
  1. The following configurations and respective permutations are possible:

    • (6,0,0,0): can occur $\frac{4!}{3!} = 4$ ways.
    • (5,1,0,0): can occur $\frac{4!}{2!} = 12$ ways.
    • (4,2,0,0): can occur $\frac{4!}{2!} = 12$ ways.
    • (4,1,1,0): can occur $\frac{4!}{2!} = 12$ ways.
    • (3,3,0,0): can occur $\frac{4!}{2!2!} = 6$ ways.
    • (3,2,1,0): can occur $4! = 24$ ways.
    • (3,1,1,1): can occur $\frac{4!}{3!} = 4$ ways.
    • (2,2,2,0): can occur $\frac{4!}{3!} = 4$ ways.
    • (2,2,1,1): can occur $\frac{4!}{2!2!} = 6$ ways.

There are 84 possible arrangements.

  1. The complement is "none of the holes has 3 or more balls" which occurs with the (2,2,2,0) and (2,2,1,1) configurations. So, the probability of the complement is 10/84. As a result the probability that at least one of the holes has three or more balls is 74/84.

  2. Probability that at least one of 4 holes has no ball is given by the complement of the event "all four holes have balls". This event is represented by the (3,1,1,1) and (2,2,1,1) configurations and its probability is 10/84. As a result the probability of your event of interest is 74/84.

  3. The 5 arrangements that produce a score of 20 or above seems to be correct. The probability of a new game is 5/84.

$\endgroup$
2
  • $\begingroup$ This is misleading. You can not just count distinct outcomes, you must properly weigh their probabilities. There are $4$ ways pattern $(5,1,0,0)$ can be permutated, but there are $6$ ways to select a ball to be the lone ball. There are $24$ equally probable ways exactly five balls can land in a hole. $\endgroup$ – Graham Kemp Mar 3 '15 at 3:26
  • $\begingroup$ @GrahamKemp I found out the mistake in my approach. I realized that indeed I forgot to weigh the probabilities of each configuration. I ran a MATLAB script and the results were exactly as predicted by your answer. Thanks for pointing it out. $\endgroup$ – tiagotvv Mar 3 '15 at 11:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.