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

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Discussion Overview

The discussion revolves around the mathematical modeling of property fluctuations in a gambling game where the outcomes are binary: doubling the property upon winning and halving it upon losing. Participants explore the expected value of property after multiple independent rounds of the game and the potential recurrence relations that could describe this expectation.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants inquire about the expected value for one round of the gamble and how to extend this to multiple rounds.
  • One participant proposes a recurrence relation involving the logarithm of property values, suggesting that the process can be modeled as a random walk.
  • Another participant questions the validity of using the random walk approach, indicating they have not been taught this method yet.
  • Several participants discuss the implications of winning and losing in various sequences, raising questions about the general case when the number of wins and losses is not equal.
  • One participant calculates the expected values for playing one and two times, leading to a discussion about whether these results contradict earlier claims.
  • Another participant notes that while winning earns more than losing costs, the expected final property is not necessarily the original property, highlighting a potential contradiction in results.
  • Further calculations are presented for three rounds, leading to a proposed formula for the expected value after n rounds.

Areas of Agreement / Disagreement

Participants express differing views on the expected outcomes of the gambling game, with some supporting the random walk model while others present alternative calculations that suggest different expected values. The discussion remains unresolved regarding the consistency of these results.

Contextual Notes

There are limitations in the assumptions made about the independence of rounds and the distribution of wins and losses. The discussion also highlights the complexity of calculating expected values in probabilistic scenarios.

evinda
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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? :confused:
Thanks in advance! :)
 
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Re: Gamble

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

evinda said:
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? :confused:
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

$\chi$ $\sigma$
 
Re: Gamble

Could you explain me why you used the equation $$L_{n}= \ln_{2}Pn$$ ? How did you find this? :confused:
 
Re: Gamble

evinda said:
Could you explain me why you used the equation $$L_{n}= \ln_{2}Pn$$ ? How did you find this? :confused:

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

$\chi$ $\sigma$
 
Re: Gamble

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

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

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?
 
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? :confused:
 
Re: Gamble

evinda said:
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!
 
  • #10
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) ? :confused:
 
  • #11
Re: Gamble

evinda said:
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) ? :confused:

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?
 
  • #12
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?
 
  • #13
Re: Gamble

I like Serena said:
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.​
 
  • #14
Re: Gamble

Bacterius said:
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.
 
  • #15
Re: Gamble

I like Serena said:
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 En=(5/4)nc ?
 
Last edited:
  • #16
Re: Gamble

evinda said:
$$E_3 = \frac 1 8 \cdot \frac 1 8 c + \frac 3 8 \cdot \frac c 2 + \frac 3 8 \cdot 2c+ \frac 1 8 \cdot 8c= \frac {125} {64} \cdot c$$
So for playing the gamble n times,is the expected value En=(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.
 
  • #17
Ok!Thank you! ;)
 

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