## Mean value!!!

**Moderators:** gmalivuk, Moderators General, Prelates

### Mean value!!!

Hi!!!!I hope you can help me at the following exercise:

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)???

Thanks in advance!!!

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)???

Thanks in advance!!!

- gmalivuk
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### Re: Mean value!!!

What do you already know about means? What have you done already? Where are you getting stuck?

We are not here to do your homework for you.

We are not here to do your homework for you.

### Re: Mean value!!!

I can't parse this unless "mean value" is to be interpreted as "expected value". At the end of the trials you will still only have one value, so it doesn't make sense to find the mean. One could find the mean of possible values, but I don't think that's a way of solving this (?).

Anyway, I hope the following advice won't count as helping too much: One of the fun things about certain probability questions (and many other math problems) is that if you're totally stuck you can try it out yourself! Play with pennies and a twenty-sided die (or, ha, another coin),~~pick a number that you have to roll better than to win,~~ and see what happens after a bajillion trials. Then see if you can explain it with math. Of course, you can't just assume that your results are what "will" happen, since this is randomness after all. That's why larger numbers of trials help. (If you know how to program it on a computer, then try that!)

Edit: I misread the question as asking for a generalization where the probability of victory is another variable along with c and n.

Anyway, I hope the following advice won't count as helping too much: One of the fun things about certain probability questions (and many other math problems) is that if you're totally stuck you can try it out yourself! Play with pennies and a twenty-sided die (or, ha, another coin),

Edit: I misread the question as asking for a generalization where the probability of victory is another variable along with c and n.

### Re: Mean value!!!

I write you what I have done until now:

n=1: E=(2c+c/2)/2=5c/4=5

n=2: E=(4c+2c+c/4)/4=25c/16=5

n=3: E=(14c+3c/2+c/8)/8=125c/64=5

etc.

So, playing the game n times, the expected value is E=5

Is this right???

n=1: E=(2c+c/2)/2=5c/4=5

^{1}c/2^{2*1}n=2: E=(4c+2c+c/4)/4=25c/16=5

^{2}c/2^{2*2}n=3: E=(14c+3c/2+c/8)/8=125c/64=5

^{3}c/2^{2*3}etc.

So, playing the game n times, the expected value is E=5

^{n}c/2^{2*n}=(5/4)^{n}c.Is this right???

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### Re: Mean value!!!

evinda wrote:I write you what I have done until now:

n=1: E=(2c+c/2)/2=5c/4=5^{1}c/2^{2*1}

n=2: E=(4c+2c+c/4)/4=25c/16=5^{2}c/2^{2*2}

n=3: E=(14c+3c/2+c/8)/8=125c/64=5^{3}c/2^{2*3}

etc.

So, playing the game n times, the expected value is E=5^{n}c/2^{2*n}=(5/4)^{n}c.

Is this right???

It seems right to me. Although, you could prove your result with a bit of work, rather than just noticing it.

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### Re: Mean value!!!

Ok,I while make a try...Thank you very much!!!

### Re: Mean value!!!

Isn't that the common way of thinking about EV? The average result of infinite trials?Lenoxus wrote:I can't parse this unless "mean value" is to be interpreted as "expected value". At the end of the trials you will still only have one value, so it doesn't make sense to find the mean. One could find the mean of possible values, but I don't think that's a way of solving this (?).

I think I actually prefer the term "mean value" because the MEDIAN is usually the value that you would expect. In this case it's 1 (for all n?), I think (Or maybe not?). And in a round of standard russian roulette, the expected value is "live" (compared to the Expected Value, which is a meaningless "one sixth dead").

Please correct me, I have no idea what I'm talking about.

-Adam

### Re: Mean value!!!

evinda wrote:I write you what I have done until now:

n=1: E=(2c+c/2)/2=5c/4=5^{1}c/2^{2*1}

n=2: E=(4c+2c+c/4)/4=25c/16=5^{2}c/2^{2*2}

n=3: E=(14c+3c/2+c/8)/8=125c/64=5^{3}c/2^{2*3}

etc.

So, playing the game n times, the expected value is E=5^{n}c/2^{2*n}=(5/4)^{n}c.

Is this right???

So, since you've got this far, do you want to try proving the expected value you've found? It looks like a problem that opens itself up quite nicely to induction (although maybe not).

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### Re: Mean value!!!

Adam H wrote:I think I actually prefer the term "mean value" because the MEDIAN is usually the value that you would expect. In this case it's 1 (for all n?), I think (Or maybe not?). And in a round of standard russian roulette, the expected value is "live" (compared to the Expected Value, which is a meaningless "one sixth dead").

You really shouldn't 'expect' to live if you're playing a game of Russian Roulette. Not from a maths point of view, but just a common sense one.

**Spoiler:**

"One sixth dead" does have some sensible meaning though - other than just the "Princess Bride"-esque meaning that you are getting from it.

I like to interpret the "one sixth dead" as meaning that you should expect to die one sixth of the time.

### Re: Mean value!!!

To me 'expected result' implies median and 'expected value' implies mean, if I'm trying to parse layman's terms into mathematical ones.

But, in any case, 'one sixth dead' seems to be invoking common parlance in a nonsensical fashion. I'd interpret that in the same fashion as calling someone 'half dead' - ie. left for dead or nearly dead. If someone means it in a mathematical sense and not a layman sense, they should use the well-defined mathematical terms.

But, in any case, 'one sixth dead' seems to be invoking common parlance in a nonsensical fashion. I'd interpret that in the same fashion as calling someone 'half dead' - ie. left for dead or nearly dead. If someone means it in a mathematical sense and not a layman sense, they should use the well-defined mathematical terms.

### Re: Mean value!!!

OK, russian roulette was a bad example because death is not a number (and it's extreme!).

If you have a 90% chance of getting a value of 0, and a 10% chance of getting a value of 10, what value would you expect to get? Definitely not 1.

Phrases mean whatever people think they mean, so I'm not saying we should stop using the phrase Expected Value. It gets the point across fine. But if whoever is in charge of math terminology wanted to change it to Mean Value, they'd have my vote of approval.

If you have a 90% chance of getting a value of 0, and a 10% chance of getting a value of 10, what value would you expect to get? Definitely not 1.

Phrases mean whatever people think they mean, so I'm not saying we should stop using the phrase Expected Value. It gets the point across fine. But if whoever is in charge of math terminology wanted to change it to Mean Value, they'd have my vote of approval.

-Adam

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### Re: Mean value!!!

The problem with that is that mean value already has a well defined meaning in mathematics.

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