50-50 or just abusing the game?

#1
Hi my name is Oscar and I had a question that has been bugging me and some other friends. We cannot wrap our heads around something and we have our own opinions about it. So I will try my best to explain this.

I play a game and there is basically a built in gambling feature. It is 50% win/lose.

Sounds fair.

I came up with a method to abuse(?) this or at least I say so but that isn't important.

So the gambling 'game' is you are against another person and you just fight but there is no skill to it, it is mainly for entertainment purposes that you can watch it.

Both players put up equal amounts of money and whoever wins keeps his money and the opponents

OK so this is my 'method'

Knowing it is 50-50 and I want to make guaranteed profits so what I do is I put up a small amount of money so we will just say 1$. I continue to do this amount until i LOSE.

When I finally lose I put up more money, 10$. If I do lose that 10$ I KEEP gambling that amount until I win.

So to make this easier I will list 3 of the scenarios that basically happen.


***(I do not show me winning the 1$ for that is no harm in this but you can add it in if you would like I just skip to the part where I FINALLY LOSE the 1$ and start from there)***


SCENARIO 1: LOSE 1$ - WIN 10$ (PERFECT SCENARIO)

SCENARIO 2: LOSE 1$ - LOSE 10$ - WIN 10$ (END UP BREAKING EVEN~)

SCENARIO 3: LOSE 1$ - LOSE 10$ - LOSE 10$ - WIN 10$ (DOWN 10$; reset back to 1$ until I lose)


Once again to clarify I do not show me doing the 1$ wins for that doesn't affect anything I just skip to the part where I lose the 1$.

With this method I ALWAYS make profit, but I occasionally go negative but build it right back up and I mean on a daily matter not my overall net.

If I need to clear anything up or add more information I will do so.

THANK YOU for your time and help!
 
#2
You certainly do not ALWAYS make a profit. Your scenario 3 is a loss.

This is a 50-50 game, so any betting strategy will come out even in the long run, with no profit or loss, if you have an infinite amount of money to gamble.

But if you have a finite amount of money to gamble, you will eventually hit a bad streak of luck and lose it all.
 

Dason

Ambassador to the humans
#3
You really do need to account for the parts where you're making $1. Because it's (quite) possible that that never happens - in which cases all of the situations you give as examples you end up with a net loss.

Honestly this just sounds like a less sophisticated, more risky version of the typical example I hear to illustrate the gambler's ruin. The typical betting strategy that we hear about is:
- Start betting with $1 and continue to do this as long as you're winning
- When you lose double your bet to $2 (in this way if you win the $2 you make up the $1 you lost and you win an additional $1).
- If you lose again double your bet to $4 (in this way you make up the $1 you originally lost and the $2 you just lost and you win an additional $1)
- Double you bet every time you lose until you win - Resulting in a net gain of $1. Revert bets back down to $1 and continue the process.

It sounds nice in theory but like Mike Z mentioned you really only have a finite amount of money so all you need is one really bad streak to make you lose everything.

Here's the thing: You're not smart enough to get a positive expected value on a 50-50 game using any type of sophisticated better strategy (in the long term). This isn't to insult your intelligence - I'm just saying that it can't be done and guarantee you that you'll make money. You might make money in the short term but eventually you'll lose it all.
 
#4
That is the beauty of it, I practically have an infinite amount of money that even if I would hit a lose streak, which I do, but lets just say losing 20 in a row which seems statistically impossible or near that still would not clean the bank.

I can add the wins of the 1$s but to me they practically balance out with the loss of the 1$ since it is 50-50 and that's why in scenario 2 I say it a break even point since I may have won 1$ right before I lose it, I just simply didn't add it in to avoid further confusion to my already not perfect explanation.

ie: Scenario 2: win 1$ - win 1$ - lose 1$ - lose 10$ - win 10$ (end up profiting 1$)


So my main question is that is my method of 'gambling' giving me greater than 50% profit (and I don't mean win rate)

^ I think that is the sentence I need to make clear of, that I understand it is a 50% chance to win or lose but my MAIN question is the method that I use is that granting me profits like it ALWAYS has been. AND ONCE AGAIN I have been negative on my daily trials like I will start fresh do 1$ LOSE that and then I would unluckily LOSE like 4 10$'s so I would be in the HOLE for a bit but I EASILY recover out and make my daily profits day by day.

I just cannot think of an equation or an explanation to somehow show proof that to show something that works out so well for me.
 

Dason

Ambassador to the humans
#5
I still don't understand your strategy completely. What happens if you lose $10, lose $10, lose $10, lose$10. You never specify when you'll switch to betting a higher amount. And I don't see the point in raising the bet to $10 if in the end all you end up with is winning $1 overall. It seems that the strategy I mentioned would be much safer if all you want to do is make $1. If you wanted you could up that strategy to start with $10 and then if you lose bet $20, if you lose bet $40, ...., so on until you win at which point you're ahead $10. But one thing you would want to do is figure out exactly how many times you would need to lose in a row to completely lose everything using your gambling scheme.
 
#6
I still don't understand your strategy completely. What happens if you lose $10, lose $10, lose $10, lose$10. You never specify when you'll switch to betting a higher amount. And I don't see the point in raising the bet to $10 if in the end all you end up with is winning $1 overall. It seems that the strategy I mentioned would be much safer if all you want to do is make $1. If you wanted you could up that strategy to start with $10 and then if you lose bet $20, if you lose bet $40, ...., so on until you win at which point you're ahead $10. But one thing you would want to do is figure out exactly how many times you would need to lose in a row to completely lose everything using your gambling scheme.

I try to explain the profitability better, and I know of the 'double-up' method but it was way to risky because the most I have lost was 13 in a row and doubling up 13 in a row is way to risky just to profit 10$ if you do all the math.

Again with the 3 scenarios, NUMBER 1 is where I actually make profit, NUMBER 2 is where I BREAK EVEN, and NUMBER 3 is where I lose some money. So basically, 2/3 of the SCENARIOS are good situations with profiting and breaking even. So hopefully that helps a bit with the explanation.

How I exactly profit is from SCENARIO 1. When I FINALLY LOSE the 1$ I stake 10$ when I do actually win it, I PROFIT 9$. Again, with SCENARIO 2 I break EVEN, and with SCENARIO 3 I LOSE MONEY BUT!!! that is ONLY when I hit a LOSING STREAK.

So I only LOSE money when I HIT A LOSING STREAK. So just that in thought really doesn't make me think that this is a 50% chance.

I am sure by now that you know it is tough for me to explain this so I am very sorry about that.

To reiterate that again:
1. I MAKE MONEY WITH A WIN STREAK OF 1$'s
2. I MAKE MONEY WITH A NORMAL LOSE (1$) AND THEN A WIN right after (10$) *SCEN 1*
3. I BREAK EVEN WITH MYSELF *SCEN 2*
4. I LOSE MONEY WHEN I HIT A LOSE STREAK *SCEN 3*


I ONLY LOSE MONEY WHEN I HIT A LOSING STREAK and we know in a game that is 50% that isn't too often so when I actually do hit a losing streak I recover and profit before I hit another one, so all in all it is just a small hump that I climb over.
 
#8
So your primary and secondary strategies are basically the opposite of each other, correct?

For $1: Play until you lose.
For $10: Play until you win.

The expected value for each part of your strategy is $0. Consider the $1 case.

50% chance you lose $1.
25% chance you break even.
25% chance you make money.

How does this look:

Expected Winnings = (P(L)*-1) + (P(WL)*0) + (P(WWL)*1) + (P(WWWL)*2) + ...

Sum(k=0 to ∞) (k-1)*((1/2)^(k+1)) = 0

It should be clear that the expected value will not change if you bet in $10 increments instead of $1, nor will it change if you play until you win (rather than playing until you lose). Your strategy breaks even.
 

Dason

Ambassador to the humans
#9
And by "breaks even" Janus means that with only a finite amount of money you're almost surely going to lose it all eventually if you keep playing and only use your initial amount. The expected value is 0 but if you play long enough you're going to lose everything.
 

hlsmith

Omega Contributor
#10
I will say it again, gambler's fallacy. If each time it is independent of the last and their is a 50:50 probability you break even if it us repeated to infinity. Can someone write this person a program showing the results for increasing interations, that might help everyone.