[Math] vase with blue and red balls

conditional-expectationexpectationprobability

At first I hope this is not a duplicate post. I tried to find it but I have not found it.

I hope that someone could help me with understanding the exercise.

This question is about a vase with r red balls en b blue balls.

I have the following two questions:

In each of the questions we have different vase with r red balls and b blue balls. So everything you will do at each question does not matter for the other question.

You will pick randomly balls from the vase, till you pick one blue ball. Proof that the total amount of balls that is picked is equal to
$$ \frac{b+r+1}{b+1} $$
What is expected amount of remaining balls if you keep removing balls from the vase till all the balls in the vase has the same color?

I have trouble of understanding how I can do this question. I know that the +1 in the formula is because you have b-1 balls, so you know that 1 blue ball is picked.

My thoughts about question 2 is as follows:

Fist I calculated the probability that the only remaining balls are red.

I defined X for the remaining red balls after the last blue has been picked.

Then I numbered every red ball 1 < j < b, to find the probability that ball j has been picked before all the blue balls.
We need to calculate the value of P. This means that I have to calculate what the probability is that every ball has been picked except for the blue ball j has been picked. So what is the chance that blue b j is picked at last?

This is $$ \frac{1}{1+r} $$

Now I can calculate the expected value of only remaining red balls left, which is: $$ EX[x] = \frac{b}{r+1} $$

Now I only have to calculate the expected amount of the only remaining blue balls and add this to the expected value of only remaining red balls and divide that by 2. Like this:

$$ EX[X] = \frac{\frac{b}{r+1}+ \frac{r}{b+1}}{2} $$

I am not sure if any of this is correct. Both of the questions are without replacement.

if anything is unclear, please ask.

Best Answer

Question 1: It is easiest to prove this by induction. If $r = 0$ (there are no red balls) and $b \geq 1$, then clearly the first ball will be blue. That is, the number of balls $N$ that we need to draw in order to draw a blue ball has expected value $E(N) = 1$. Note that we can write this as

$$ E(N) = \frac{b+1}{b+1} = 1 $$

in accordance with the desired formula. Now, consider any other value of $r$, and $b \geq 1$ still. Suppose that we already have shown that for $r-1$ red balls and $b$ blue balls, the expected number of balls until we draw a blue ball is $\frac{b+r}{b+1}$, in accordance with the desired formula. We now draw a ball from an urn with $r$ red balls and $b$ blue balls. With probability $\frac{b}{b+r}$, the first ball is blue, and $N = 1$. With probability $\frac{r}{b+r}$, the first ball is red, and we now have an urn with $r-1$ red balls and $b$ blue balls. By the premise, we already know what the expected number of additional balls that need to be drawn in order to produce a blue ball; it is $\frac{b+r}{b+1}$. Therefore, for the urn with $r$ red balls and $b$ blue balls, we have

$$ \begin{align} E(N) & = \frac{b}{b+r} \cdot 1 + \frac{r}{b+r} \cdot \left(1+\frac{b+r}{b+1}\right) \\ & = \frac{b}{b+r} + \frac{r}{b+r} + \frac{r}{b+1} \\ & = 1 + \frac{r}{b+1} = \frac{b+r+1}{b+1} \end{align} $$

and we are done.

Question 2: By symmetry, the expected number of remaining balls left when only balls of one color (either red or blue) remain is equal to the number of balls of the same color we draw at the beginning. (To see this, draw all the balls out from the urn, and lay them out on a table, in sequence. The sequence from left to right is exactly as probable as the sequence from right to left.)

With probability $\frac{r}{b+r}$, the first ball is red. We are then left with $r-1$ red balls and $b$ blue balls, and we want to know how many balls we draw before (not until) we draw the first blue ball. This is one less than the result from Question 1. To this we have to add the first red ball. So when the first ball is red, our answer is

$$ N_{red} = \frac{b+r}{b+1}-1+1 = \frac{b+r}{b+1} $$

With probability $\frac{b}{b+r}$, the first ball is blue, and by symmetry, the number of consecutive blue balls at the start is

$$ N_{blue} = \frac{b+r}{r+1} $$

Thus, the expected number of consecutive balls drawn of the same color (which is also the answer to the original Question 2) is

$$ E(N) = \frac{r}{b+r} \cdot N_{red} + \frac{b}{b+r} \cdot N_{blue} = \frac{r}{b+1} + \frac{b}{r+1} $$

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[Math] Probability question involving simulations of picking balls from a bag

All ${100 \choose 30}$ patterns are equally likely. There is probably a better approach to your first question, but this produces a nice answer:

Suppose that $R(r,b)$ is the expected number of runs given that the first ball is red and $B(r,b)$ is the expected number of runs given that the first ball is blue, in each case when there are $r$ red balls and $b$ blue balls. Then $R(r,0)=1$ and $B(0,b)=1$ and by symmetry $R(r,b)=B(b,r)$. More generally we have the recurrence $$R(r,b)=\frac{r-1}{r+b-1} R(r-1,b) + \frac{b}{r+b-1} (1+R(b,r-1))$$ which has the solution $R(r,b)= \dfrac{2rb+r-1}{r+b-1}$ with $R(1,0)=1$. So the expected number of runs overall when there are $r$ red balls and $b$ blue balls is $$\frac{r}{r+b}R(r,b)+\frac{b}{r+b}R(b,r)=1+\frac{2rb}{r+b}.$$

In your question you had $r=70$ and $b=30$, giving an expected number of runs of $43$.

It is slightly unfortunate that $43$ is odd, as it complicates the distribution of runs (e.g. there might be $21$ or $22$ red runs with probabilities $0.3$ and $0.7$).

Added: In general, if there $n$ balls of a particular colour and they are distributed between $k$ positive runs, then a simple stars and bars argument says that the probability that a given run is of length $M$ is $$\Pr(M=m|n,k)= \dfrac{\displaystyle {n-m-1 \choose k-2}}{\displaystyle {n-1 \choose k-1}}$$ the each of the $k$ runs has a identical (but not independent) distribution. So for example if you have $70$ red balls in $22$ runs or $30$ blue balls in $21$ runs, the probability distribution for the length $M$ in a red run or of a blue run would be about

m   red 70,22   blue 30,21
1   0.3043478   0.6896552
2   0.2148338   0.2216749
3   0.1507043   0.0656814
4   0.1050363   0.0176835
5   0.0727174   0.0042440
6   0.0499932   0.0008842
7   0.0341224   0.0001538
8   0.0231152   0.0000210
9   0.0155364   0.0000020
10  0.0103576   0.0000001
11  0.0068466   not possible
12  0.0044857   
13  0.0029118   
14  0.0018718   
15  0.0011912   
16  0.0007500   
17  0.0004670   
18  0.0002874   
19  0.0001747   
20  0.0001048   
21  (smaller)

But we can go further than that, and see the distribution of runs unconstrained by how many runs there are. As an example consider two red balls and one blue ball: the patterns $R_2B_1$, $R_1B_1R_1$, and $B_1R_2$ are equally likely and so with repeated experiments over time you expect to see as many runs of $R_2$ as $R_1$ in this particular case. More generally, if there are $r$ balls and $b$ blue balls, the expected proportion of red runs which would be of length $m$ in that sense, is

$$ \dfrac{b}{r} \dfrac{ {r \choose m} }{ {r+b-1 \choose m}}$$

where reversing $r$ and $b$ would give the expected proportion of blue runs of length $m$ in that sense. With $r=70$ and $b=30$ this would give about

m   red runs    blue runs
1   0.3030303   0.7070707
2   0.2133581   0.2092352
3   0.1495706   0.0603978
4   0.1043878   0.0169869
5   0.0725221   0.0046490
6   0.0501482   0.0012364
7   0.0345106   0.0003191
8   0.0236323   0.0000798
9   0.0161011   0.0000193
10  0.0109130   0.0000045
11  0.0073571   0.0000010
12  0.0049326   0.0000002
13  0.0032884   (smaller)
14  0.0021795   
15  0.0014359   
16  0.0009402   
17  0.0006117   
18  0.0003954   
19  0.0002538   
20  0.0001618   
21  (smaller)

which is not that far away from the constrained case.

We have a box with m blue balls and n red balls ($n \lt m$). We bet r dollars that we will randomly pick a red ball….

In each bet, you either obtain $k\cdot r$ dollars when you pick the red ball, or you lose $r$ dollars if you pick the blue ball. So, using your notation, the expected payout value from a single bet is: $$ P[red]\cdot (k\cdot r) + P[blue]\cdot (-r) = \frac{n\cdot k\cdot r}{m+n} - \frac{m\cdot r}{m+n} = \frac{r}{m+n}(n\cdot k-m) $$

This is the expected payout, i.e. the average payout if you repeat the bet a number of times. For the bet to be profitable to the better, the expected payout must be positive - so you will on average earn money instead of losing it. So we just need to solve an inequality for $k$: $$ \frac{r}{m+n}(n\cdot k-m) > 0 $$ $$ n\cdot k - m > 0 $$ $$ k > \frac{m}{n} $$ We of course assume that $r$ and $n$ are positive.

Best Answer

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[Math] Probability question involving simulations of picking balls from a bag

We have a box with m blue balls and n red balls ($n \lt m$). We bet r dollars that we will randomly pick a red ball….

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