MHB A wheel of fortune is divided into 10 parts, one of them brings the jackpot.

  • Thread starter Thread starter mathmari
  • Start date Start date
  • Tags Tags
    parts Wheel
AI Thread Summary
The discussion revolves around calculating probabilities related to a wheel of fortune divided into ten equal parts, with one part yielding a jackpot. The player spins the wheel 20 times and seeks to determine the probability of winning at least one jackpot, aiming for a 90% confidence level in their results. There is also a focus on the probability of misjudging the wheel if less than twice the jackpot appears in those 20 spins, and how to adjust the significance level to about 6%. Participants discuss the correct application of binomial probability formulas and the implications of the null hypothesis in their calculations. The conversation emphasizes the need for accurate statistical methods to assess the wheel's fairness.
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

A wheel of fortune is divided into $10$ equal sized parts, of which one of them brings the jackpot.

  • A player wants to investigate the regularity of the wheel of fortune. He turns the wheel 20 times. Calculate the probability that he will get at least one main prize if it is a laplace wheel.
  • How often does the player have to spin the wheel of fortune to get at least one main prize with a probability of at least 90%.
  • The player objects to the wheel when less than twice the jackpot appears at 20 times turn. Calculate the probability that the wheel will be misjudged.
  • What does the player have to change in his test so that the significant level is about $6\%$ ?

I have done the following:

  • $$P(X\geq 1)=1-P(X\leq 1)=1-\sum_{i=0}^1\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}$$

    $$ $$
  • \begin{align*}P(X\geq k)\geq 90\% &\Rightarrow 1-P(X\leq k)=90\% \\ & \Rightarrow 1-\sum_{i=0}^k\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}\geq 0.9 \\ & \Rightarrow \sum_{i=0}^k\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}\leq 0.1\end{align*}

Is everything correct so far? (Wondering)

Could you give me a hint for the other two questions? (Wondering)
 
Physics news on Phys.org
mathmari said:
Hey! :o

A wheel of fortune is divided into $10$ equal sized parts, of which one of them brings the jackpot.

  • A player wants to investigate the regularity of the wheel of fortune. He turns the wheel 20 times. Calculate the probability that he will get at least one main prize if it is a laplace wheel.
  • How often does the player have to spin the wheel of fortune to get at least one main prize with a probability of at least 90%.
  • The player objects to the wheel when less than twice the jackpot appears at 20 times turn. Calculate the probability that the wheel will be misjudged.
  • What does the player have to change in his test so that the significant level is about $6\%$ ?

I have done the following:
  • $$P(X\geq 1)=1-P(X\leq 1)=1-\sum_{i=0}^1\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}$$

Hey mathmari! (Smile)

Shouldn't it be:
$$P(X\geq 1)=1-P(\text{NOT }(X\geq 1))=1-P(X<1)$$
(Wondering)
mathmari said:
  • \begin{align*}P(X\geq k)\geq 90\% &\Rightarrow 1-P(X\leq k)=90\% \\ & \Rightarrow 1-\sum_{i=0}^k\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}\geq 0.9 \\ & \Rightarrow \sum_{i=0}^k\binom{10}{i}\left (\frac{1}{10}\right )^{20}\cdot \left (1-\frac{1}{10}\right )^{20-i}\leq 0.1\end{align*}

Is everything correct so far?

You are one off. (Sweating)

Next thing to do is calculate those chances for $i=0$, $i=1$, and so on until we reach a sum of $0.1$. (Thinking)

Or alternatively we can use a tool or table to find the inverse cumulative binomial distribution. (Nerd)

mathmari said:
Could you give me a hint for the other two questions?

Suppose the wheel really is a laplace wheel.
We will call this the null hypothesis ($H_0$).
Then we can calculate the probability that the player will object anyway, can't we?
That is, we can calculate $P(X \le 2\mid H_0)$, can't we? (Wondering)This probability that we misjudge even though the null hypothesis is true, is called the significance or $p$-value.
If just now we found that the probability to misjudge is more than $6\%$, what can we do to change that probability? (Wondering)
 
Back
Top