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50/50 Betting Math

Xour

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I understand that if you bet $1 1000 times, you will end up with $0 profit, with 500 losses and 500 wins.

Can someone explain to me what happens when you throw in the variance of bet sizes?

For example... 219 of the bets are $1, 486 are $3, and 295 are $5.

Assuming 50/50 odds
 
I'm not a gambler a friend of mine is confident that sizing makes a difference.

I told him the three bet sizes would just average out is this correct?
 
It makes zero difference. Your expected outcome is still 50% of what you wagered.

Math: (219)(1)(.5) + (486)(3$)(.5) + (295)(5$)(.5) = 1576$ which is half the money you started off with.

Source: Took math in elementary school which is above average for the Runescape community so I know where the frustration comes from.
Your equation is autistic so you obviously didn't pay attention lol
 
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Are you stupid? My equation is the basic probability of expectations equation. What's wrong about it? Care to explain?

The events of probability are independent of one another which is why you can add them like that. Betting 1$ won't effect your chances of winning while betting 5$ and vice versa.

You might want to check the number of chromosomes your friend has.

Because you didn't factor in that you'll win half of them as well.

But let's move on from the first example and focus on ten bets of different value.
 
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Variance is still higher on bigger bets, meaning with bigger bets your actual value will differ more from the expected value than on a smaller bet

Your bet size is the expected value of the bet if it's 50/50 odds - so different bet sizes don't matter for the EV but it does for the variance
 
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@Xour you're right, I read the question wrong. You would end with what you started with no matter the variability of your bets. If you start with 1000$ and bet 1$, 5$, 10$ or any other value adding to 1000$ the chances are that you'll end up with 1000$ in the end.
Yeah I figured it would average in the first example that's what I told him.

But how do you calculate the variance in ten or one hundred different sized bets? That's a better way to put this I guess. @Searchy
 
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You're not guaranteed to end up with +/-0, it's just the statistic probability. The more bets you make the closer you get to evening out. i.e. if you made a billion $1 bets you would likely be pretty close to 50%.

In your scenario you bet $3152 and wind up with a similar amount of money due to the 50% odds
 
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With 50/50 odds your expectation value (the average result if you any sequence of bets thousands of times) is $0.

Changing the bet size only reduces the total number of bets (if you are betting with a fixed cap). This pushes you to the left on the graph above.

Making fewer bets makes your net position larger randomly in either the negative or positive direction.

If you wanted to get technical about it, look at the cumulative binomial distribution for p=0.5. Blue line below.

300px-Binomial_distribution_cdf.svg.png


As the total number of bets (N) approaches about 20 your chance of deviating from the expectation value ($0) is virtually 0%
 
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Still zero EV gained and in an infinite number of events you'll have the same amount of money as when you started

Bet size does not matter, although I'll add that larger stakes is more likely to deplete your bankroll faster than if you stayed at 1$ over and over but yeah
 
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