Probability : Voting problem - Game strategy

[mathmari]

Hey!

I am looking the following:

1. There are 1 Million voters. 2000 of them know exactly that they will vote for A, but the other 998000 will decide if whether they will vote for A or B in the voting booth using a coin.
   With which possibility will A get more votes?
   
2. At a quiz we get 20.000 Euro. We have the chance to increase the money tenfold. For this we have to find out from 7 boxes, the one which contains the number '10'.
   We can unpack 2 boxes. However, if we are unlucky and cannot find the '10', the 20.000 euros will go as well lost.
   But can can not participate in the game and take the 20.000 euros home.
   How do we decide when we join the game once and what strategy do we use if we play every year?

I have done the following:

1. Suppose that the voters vote for a if at tossing a coin they get head. So that A wins we have to get such a number of heads that \begin{align*}&2000 + \text{ number of heads }> 998000 - \text{ number of heads } \\ & \Rightarrow 2\text{ number of heads }>998000-2000 \\ & \Rightarrow 2\text{ number of heads }>996000 \\ & \Rightarrow \text{ number of heads }>498000\end{align*}
   
   So, the possibility that will A win is equal to the possibility that we get more that $498000$ times "head", or not? (Wondering)
2. Do we choose one of the two correct boxes out of 7 with probability $\frac{2}{7}$ ? (Wondering)
   

 

[I like Serena]

Hey mathmari! (Smile)

Yep. All correct.
We can calculate 1 using either the binomial distribution, or the approximation with a normal distribution.
For 2 I don't understand which choices we have, to create a strategy. All we seem to be able to do is just open 2 boxes, which is always best. (Thinking)

 

[mathmari]

Yep. All correct.
We can calculate 1 using either the binomial distribution, or the approximation with a normal distribution.

Using the binomial distribution we have the following:

We have that $\mu=\mathbb{E}[X]=p=\frac{1}{2}$ and $\sigma^2=\text{Var}(X_i)=p(1-p)=\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$.

The total number of tossing a coin is $n=998000$.
$X$ is the random variable that describes the number of times that we get "head".
\begin{align*}P(X>498000)&=1-P(X\leq 49800)\\ & =1-P\left (\frac{X-n\mu}{\sqrt{n\cdot \sigma^2}}\leq \frac{498000-n\mu}{\sqrt{n\cdot \sigma^2}}\right ) \\ & = 1-\Phi \left ( \frac{498000-n\mu}{\sqrt{n\cdot \sigma^2}}\right ) \\ & = 1-\Phi \left ( \frac{498000-998000\cdot \frac{1}{2}}{\sqrt{998000\cdot \frac{1}{4}}}\right ) \\ & = 1-\Phi \left ( \frac{498000-499000}{\sqrt{249500}}\right ) \\ & \approx 1-\Phi \left ( \frac{-1000}{499.5}\right )\\ & = 1-\Phi \left ( -2\right ) \\ & = \Phi (2) \\ & = 0,97725\end{align*}

Is everything correct? (Wondering)

 

[mathmari]

For 2 I don't understand which choices we have, to create a strategy. All we seem to be able to do is just open 2 boxes, which is always best. (Thinking)

We can either play that game, i.e. to open 2 boxes, or we don't play that game and leave with the 20000 Euros.

So, we have to calculate the probability that we get the right box, and the money gets increased by tenth, or not?

If this is a small probability it is better not to play the game and get the 20000 Euros because otherwise the probability that we leave with no money is big.

But if the probability that we open the right box is big, we would play that game, or not?

(Wondering)

 

[I like Serena]

Using the binomial distribution we have the following:

Is everything correct? (Wondering)

Looks correct to me.

We can either play that game, i.e. to open 2 boxes, or we don't play that game and leave with the 20000 Euros.

So, we have to calculate the probability that we get the right box, and the money gets increased by tenth, or not?

If this is a small probability it is better not to play the game and get the 20000 Euros because otherwise the probability that we leave with no money is big.

But if the probability that we open the right box is big, we would play that game, or not?

Ah okay.
So if we play and win, we get 20.000 + 10 x 20.0000.
If we play and lose, we get nothing.
And if we don't play, we get 20.000.
Yes? (Wondering)

I think we will have to calculate the expected winnings in each case. (Thinking)

 

[mathmari]

Re: Probability : Voting problem - Game stratergy

So if we play and win, we get 20.000 + 10 x 20.0000.
If we play and lose, we get nothing.
And if we don't play, we get 20.000.

So if we play and win, we get 10 x 20.0000.
If we play and lose, we get nothing.
And if we don't play, we get 20.000.

I think we will have to calculate the expected winnings in each case. (Thinking)

If we play the expected value is $\frac{2}{7}\cdot 200.000+\frac{5}{7}\cdot 0\approx 57000>20.000$, that means we should play, or not? (Wondering)

 

[I like Serena]

Re: Probability : Voting problem - Game stratergy

If we play the expected value is $\frac{2}{7}\cdot 200.000+\frac{5}{7}\cdot 0\approx 57000>20.000$, that means we should play, or not? (Wondering)

Depends on our strategy.
If we play only once, it may be wise to just take the 20000 rather than have a $\frac 57$ chance to end up with nothing.
But if we play every year, gambling will ultimately indeed net us more than we would otherwise have. (Happy)

It's a classical gambling dilemma.
Suppose at every gambling step we can expect to gain more by continuing to play then stopping.
Should we continue?
If we only play the once, we can be almost certain of losing everything if we keep playing.
But if we play often enough, the one time or so that we win the grand prize should compensate all losses. (Nerd)

 

[mathmari]

I understand! Thank you! (Sun)