Can a recurrence relation predict property fluctuations in a gambling game?

[evinda]

Hello!Could you help me at the exercie below?

Consider a gamble,with the same possibility to win or to lose.If we win,we double our property,but if we lose we halve our property.Let's consider that we begin with an amount c.Which will be the mean value of our property,if we play n times(independent repetitions of the game)?

Is there any recurrence relation for the expectation for the time n?
Thanks in advance! :)

 

[Nono713]

Re: Gamble

What is the expected value for one round of the gamble? How can you extend this to multiple independent rounds? :p

 

[chisigma]

Re: Gamble

Hello!Could you help me at the exercie below?

Consider a gamble,with the same possibility to win or to lose.If we win,we double our property,but if we lose we halve our property.Let's consider that we begin with an amount c.Which will be the mean value of our property,if we play n times(independent repetitions of the game)?

Is there any recurrence relation for the expectation for the time n?
Thanks in advance! :)

Setting $P_{n}$ Your property at the step n and $L_{n}= \ln_{2} P_{n}$ You obtain...

$\displaystyle L_{n}= \ln_{2} P_{0} + \sum_{k=1}^{n} X_{k}$ (1)

... where the $X_{k} = \pm 1$ are independent r.v. and $\displaystyle P\{X_{k}=1\}= P\{X_{k}=-1\} = \frac{1}{2}$. The (1) is a process called random walk and a well known result is that...

$\displaystyle E\{\sum_{k=1}^{n} X_{k}\} = 0$ (2)

In other words the expected value of Your property at the step n is Your original property...

Kind regards

$\chi$ $\sigma$

 

[evinda]

Re: Gamble

Could you explain me why you used the equation \(\displaystyle L_{n}= \ln_{2}Pn\) ? How did you find this?

 

[chisigma]

Re: Gamble

Could you explain me why you used the equation \(\displaystyle L_{n}= \ln_{2}Pn\) ? How did you find this?

If You use $\log_{2} P_{n}$ instead of $P_{n}$ You transform the process in the well known 'random walking' process, one of the most deeply studied...

Kind regards

$\chi$ $\sigma$

 

[evinda]

Re: Gamble

Thank you very much! But...is the random walking process the only way to solve the exercise?I haven't get taught this method yet...

 

[I like Serena]

Re: Gamble

Thank you very much! But...is the random walking process the only way to solve the exercise?I haven't get taught this method yet...

The expectation is that you win as many times as that you loose.
What will your property be if you win say 2 times and loose 2 times in some order?

 

[evinda]

Re: Gamble

If you win 2 times and loose 2 times in some order, your property will be equal to the property you had at the beginning of the game...Right?

 

[I like Serena]

Re: Gamble

If you win 2 times and loose 2 times in some order, your property will be equal to the property you had at the beginning of the game...Right?

Yep!

 

[evinda]

Re: Gamble

A ok...And what can I tell about the general case,playing n times(if it is not clear that I win n/2 times and loose n/2 times) ?

 

[I like Serena]

Re: Gamble

A ok...And what can I tell about the general case,playing n times(if it is not clear that I win n/2 times and loose n/2 times) ?

What will your property be if you win 3 times and lose 1 time?
There is an equal probability that you win 1 time and lose 3 times. What will your property be then?

 

[I like Serena]

Re: Gamble

Let me rephrase that.

Suppose you play 1 time.
Then it's fifty-fifty whether you win or lose.
So the expectation is:
$$E_1 = \frac 1 2 \cdot \frac 1 2 c + \frac 1 2 \cdot 2c = \frac 5 4 c$$

If you play 2 times, your expectation is:
$$E_2 = \frac 1 4 \cdot \frac 1 4 c + \frac 1 2 \cdot c + \frac 1 4 \cdot 4c = \frac {25} {16} c$$

Can you find the expectation if you play 3 times?

 

[Nono713]

Re: Gamble

Let me rephrase that.

Suppose you play 1 time.
Then it's fifty-fifty whether you win or lose.
So the expectation is:
$$E_1 = \frac 1 2 \cdot \frac 1 2 c + \frac 1 2 \cdot 2c = \frac 5 4 c$$

If you play 2 times, your expectation is:
$$E_2 = \frac 1 4 \cdot \frac 1 4 c + \frac 1 2 \cdot c + \frac 1 4 \cdot 4c = \frac {25} {16} c$$

Can you find the expectation if you play 3 times?

Is it just me or does this result contradict chisigma's result? I found the same general formula via the binomial distribution and while I understand now why the expected final property is not the original property (losing does balance out winning, however you still earn more overall by winning than by losing, i.e. if you start with \$1 and you win you earn \$1, but if you start with \$1 and lose you only lose \$0.5) I am still confused as to why we get seemingly contradictory results.​

 

[I like Serena]

Re: Gamble

Is it just me or does this result contradict chisigma's result? I found the same general formula via the binomial distribution and while I understand now why the expected final property is not the original property (losing does balance out winning, however you still earn more overall by winning than by losing, i.e. if you start with \$1 and you win you earn \$1, but if you start with \$1 and lose you only lose \$0.5) I am still confused as to why we get seemingly contradictory results.​

chisigma's result leads to $E(\ln P_n) = \ln c$.
However, $E(\ln P_n) \ne \ln E(P_n)$.

As you said, winning earns you more than losing costs you, which is quite different from the game of roulette.

 

[evinda]

Re: Gamble

Let me rephrase that.

Suppose you play 1 time.
Then it's fifty-fifty whether you win or lose.
So the expectation is:
$$E_1 = \frac 1 2 \cdot \frac 1 2 c + \frac 1 2 \cdot 2c = \frac 5 4 c$$

If you play 2 times, your expectation is:
$$E_2 = \frac 1 4 \cdot \frac 1 4 c + \frac 1 2 \cdot c + \frac 1 4 \cdot 4c = \frac {25} {16} c$$

Can you find the expectation if you play 3 times?

$$E_3 = \frac 1 8 \cdot \frac 1 8 c + \frac 3 8 \cdot \frac c 2 + \frac 3 8 \cdot 2c+ \frac 1 8 \cdot 8c= \frac {125} {64} \cdot c$$
So for playing the gamble n times,is the expected value E_n=(5/4)ⁿc ?

 

[I like Serena]

Re: Gamble

$$E_3 = \frac 1 8 \cdot \frac 1 8 c + \frac 3 8 \cdot \frac c 2 + \frac 3 8 \cdot 2c+ \frac 1 8 \cdot 8c= \frac {125} {64} \cdot c$$
So for playing the gamble n times,is the expected value E_n=(5/4)ⁿc ?

Yes. :)

If you want you can prove it with the binomial theorem.
A full induction proof is probably possible as well.
But you'll just get the same result.

 

[evinda]

Ok!Thank you! ;)