Hot answers tagged

80

This is a classical problem. Without changing the problem, we can let the digits on the dice be $0, \ldots, 5$ instead of $1, \ldots, 6$ to make our notation easier. Now we make two polynomials: $$ P(x) = \sum_{i=0}^5 p_ix^i,\qquad Q(x) = \sum_{i=0}^5q_ix^i. $$ Now we can succinctly phrase your condition on $p_i, q_i$: it is satisfied if and only if $$ P(x)Q(...


53

Let's assume that this is possible. We can derive a contradiction from this assumption. The probability of rolling a total of $2$ must be $1/11$, and the probability of rolling a total of $12$ must also be $1/11$, so $$ \begin{align} p_1 q_1 &= 1/11,\ \text{and}\\ p_6 q_6 &= 1/11. \end{align} $$ The probability of rolling a total of $7$ must also be $...


15

The probability generating function of the dice are $P(x) = \sum_{i=1}^6 p_i x^i$ and $Q(x) = \sum_{i=1}^6 q_i x^i$. The probability generating function for their sum is $R(x) = P(x)Q(x)$. You want all possible sums $2, \ldots, 12$ to have the same probability, or equivalently you want $R(x) = \frac{1}{11} x^2 (1+x+\cdots+x^{10})$. Hence an equivalent way to ...


7

To see why the probability is $\frac78,$ let's slightly rephrase the question: let's say instead of simulating the flips until we get a success (let's say heads) let's just flip the coin three times straight away and then look at the results. A set of three flips is successful if and only if there is at least one heads in the three flips, so the probability ...


7

$$E\left[X^{2}\right]=E\left[\left(\sum_{i=1}^{100}X_{i}\right)^{2}\right] \ne E\left[\sum_{i=1}^{100}X_{i}^{2}\right]$$


6

$$X=\sum_{i=1}^{100}X_i$$ $$X^2=\left(\sum_{i=1}^{100}X_i \right)^2= \sum_{i=1}^{100}X_i^2 + 2 \sum_{i < j}X_iX_j$$ You have left out $2 \sum_{i < j}X_iX_j$. Also the second method is not correct. Sum of $100$ variables is not the same as $100$ times a single variable.


6

Your calculation is incorrect because $2^{n}\cdot \left(\frac 56 \right)^n$ is the expected profit of rolling $n$ times and quitting. You should not subtract the $\frac 162^{n}$ because the loss of that is already reflected. Planning to roll $n+1$ times is always better than planning to roll $n$ times. If you plan to roll $n$ times your expected return is $...


4

In your last calculation, you seem to be using a "rule" that if $\lim_{n\to\infty} B(n) = 1$ and $\lim_{n\to\infty} E(n) = \infty$, then $\lim_{n\to\infty} B(n)^{E(n)} = 1$. This is an incorrect rule (and this example actually shows why). This is actually an example of the "$1^\infty$ indeterminate form", which is called an indeterminate ...


3

In a single roll, the probability of rolling $N$ is $p_N^{\,}=\frac{6-|N-7|}{36}$. For example with $N=7$ this is $\frac{6-|7-7|}{36}=\frac{6-0}{36}=\frac{1}{6}$. The probability of rolling a number which is not $N$ is $1-p_N^{\,}$. For example with $N=7$ this is $1-\frac{1}{6}=\frac56$. The probability of $N$ times rolling a number which is not $N$ is $\...


3

Let take a simpler system. There are 2 outcomes for 2 coins $\{C_1,C_2\}$ - $\{1,2\}$ with probabilities $p_1,p_2$ for $C_1$ and $q_1,q_2$ for $C_2$. Given condition: probability after 2 coin throws of the sum being $\{2,3,4\}$ is the same. Question: Can you find some $p_1,p_2,q_1,q_2$ that satisfies the condition? In this the given condition is $p_1q_1 = ...


2

Let $E_1$ be the expected number of trials to see $k$ in a row of any face, and let $E$ be the expected number of trials to see $k$ sixes in a row. First, wait until you see $k$ of anything. That is expected to take $E_1$ trials. At the end of that, two things are possible. Either you got lucky and saw $k$ sixes in a row (probability $\frac 16$) or you ...


2

The combinatorial argument is that you throw until you get $k$ consecutive of any type; if that is of the number you want then stop, and if not then start again. So if $A$ is the expected number of throws until you get $k$ consecutive of any type and $B$ is the expected number of throws until you get $k$ consecutive of the type you want then $$B=A +\frac1n ...


2

If $X_k$ is the $k$-th roll of the dice then all $X_k$ are independent, just note that what happen in a roll doesn't determine what happen in another. You are mixing there the end of the game, what can be seen as a stopping time, and the $X_k$, what are just i.i.d. r.v. Indeed $\{N=n\}=\{(X_1,\ldots ,X_{n-1})\in A^{n-1}, X_n\in B\}$, for $A:=\{4,5,6\}$ and $...


2

Your code is calculating the average number of rolls where you successfully cash out but it doesn't account for the fact that the prize doubles at each step. The expected value does and so taking more risks for the much larger reward is optimal at each step.


2

Say $1, 2, 3$ are considered success and $4, 5, 6$ are considered failure. So we have equal probability of success or failure in each roll of the dice. As we have up to $3$ rolls, the probability of success is, $P(S) = 0.5 + 0.5 \times 0.5 + 0.5^2 \times 0.5 = 0.875$ First term is the probability of success in one roll, second term is the probability of ...


1

I am supposed to discuss your work, before presenting my answer. Unfortunately, while I can solve the problem, my formal knowledge of Probability is inadequate to examine your work. On the re-roll, Player 2's expectation is as follows: $(1/2)$ the time, he will roll $> 10.$ If he does so, his average gain is $(15.5)$. $(1/20)$ of the time, his roll will ...


1

This is hard to do along the lines you propose because winning $1$ in some round does not increase your expectation by $1$, as that $1$ is not certain. However, it is easy to compute the expected value directly. The probability of winning exactly $n$ for $n≥1$ is $\left(\frac 12\right)^n\times \frac 13$ so the expected value of the game is $$\sum_{n=1}^{\...


1

Let n be the number of 6-sided dice thrown and $$k=1,...,6$$. X= "Greatest thrown number". What are the probabilities $P[X=k]$ for every k=1,...,6? let us consider the events: $A_k=${getting a number less than k in the dice} and $X_i$ the result of $i_{th}$ dice $$P[X=k]=P[X\leq k]-P[X \leq k-1]$$ $$P[X=k]=P[\bigcap (X_i \leq k)]-P[\bigcap (X_i \...


1

I wrote an R-script that should simulate your experiment. I used 1 million die tosses. Hopefully I understood it correctly. I have included the code below. Here is the histogram I got from the simulation. It is slightly skewed which makes sense. 6 and 8 occur with the same probability, but since we look through more tosses when $N$ is 8, it has a higher ...


1

If you want to generate numbers 1 through 12 with uniform probability, it is possible by re-labeling the faces of fair dice. One die has the faces labeled 1 through 6. The other has the faces labeled -2, 0, 2, 4, 6, 8. If the total is not in the 1 through 12 range, roll again. This is an acceptance-rejection technique. Also, there is a theorem in ...


1

To make your approach more formal, let $N$ be the number of rolls until the game stops, and $X_i$ the value of the $i$th roll ($i=1,2....$). The total payoff $P$ is $\sum\limits_{i=1}^\infty X_i$. Given $N$: $$E(P\vert N)=E(\sum\limits_{i=1}^\infty$X_i\vert N)=\sum\limits_{i=1}^N E(X_i\vert N)=5\cdot (N-1)+2=5N-3$$ since the expected value of each roll that ...


1

The result is correct, but it's hard to justify along the lines you proposed. As an alternative method: Consider the possible outcomes of the first roll, and note that, if the game does not end, then it restarts (and, of course, will have the same expectation, $E$, going forward). We see that $$E=\frac 16 \times \left(1+2+3+(4+E)+(5+E)+(6+E)\right)\implies ...


1

While $P = NX$ is a valid definition, $N$ and $X$ are not independent, so you cannot say that $\mathbb E(P) = \mathbb E(N)\mathbb E(X)$, which is generally false for dependent variables. Additionally, I do not think that $\mathbb E(X) = 3.5$ -- but to get the right answer for your question, I think you'll want to take a different approach regardless.


1

One good way is to try to convert this to a more standard linear recursion. Letting $$a_n=6(a_{n-1}+1)=6a_{n-1}+6$$ We rewrite this as $$6=a_n-6a_{n-1}=a_{n-1}-6a_{n-2}$$ Whence $$a_n=6a_{n-1}+a_{n-1}-6a_{n-2}=7a_{n-1}-6a_{n-2}$$ and this can now be solved by standard means, yielding $$\boxed{a_n=\frac 65\times \left(6^n-1\right)}$$


1

You are computing the probability that there are no $6$s in the first $x-1$ rolls, but they want to compute the probability that faces $1,2,3,4,5$ each appear at least once in the first $x-1$ rolls.


Only top voted, non community-wiki answers of a minimum length are eligible