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Old Feb 03, 2008, 10:08pm   #1
kmay06
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Default An interesting odds question

So a little background story: I was out with some friends, and one of them needed some money and was a couple bucks short. For arguments sake, we'll say he needed $16 and only had $15. Now, I could have just gave him a buck, but instead I said ok, I'll write down a number, either 1, 2 or 3. You pick a number, and if you are right I'll give you the buck, if you are wrong you give me a buck. Now obviously it is a bad deal for him, and having played poker he knew that. So he was like why don't we flip a coin? So I said ok, how about this. We do the number thing, giving you only a 33% chance to win. But if you lose, we'll do the same deal, except for exactly the amount you need. I.e. you need one dollar, you lose the first, you need two dollars, you lose that one, now you need 4 dollars, you lose that one you need 7 dollars. It'll look like this:

15, one short, bet $1. Lose
14, two short, bet $2. Lose
12, four short, bet $4. Lose
8, 8 short, bet $8.

And I jokingly told him, ok so while you only have a 33% chance to win any given time, if you win only once out of any of these you can stop, and since you have four chances to win, you logically should win this at least once.

Now, I said this at somewhat less than mental capacity, and as the night went on I kept thinking about it. Really, I can't find my original flaw in logic with which I was trying to trick him. It seems that it would be a winning proposition for him. But how? Because for any one time he is only getting 33% and if we look at any one seperatly it is a bad deal. But when we look at them all together, he should win at least one mathematically, and be a dollar up from what he was.

Help me forum, less I have to become inebriated again and find out my line of logic.
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Old Feb 03, 2008, 10:38pm   #2
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you cannot add the 33% winning chances. so 33% times 4 is not 132%, meaning he doesn't win always.

Instead the chances of him not winning a single "flip" is 66%^4. That means he loses 4 times in a row an astounding 19% of the time.

so now 19% of the time he loses $15 while 71% he wins a buck only. If he takes on the gambol he does get his missing dollar 71% of the time, but it's a big -EV game for him of around -$2.30.

also this could be a cool read:

http://casinogambling.about.com/od/m...martingale.htm
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Old Feb 04, 2008, 1:21am   #3
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Originally Posted by BubbleBoy View Post
you cannot add the 33% winning chances. so 33% times 4 is not 132%, meaning he doesn't win always.

Instead the chances of him not winning a single "flip" is 66%^4. That means he loses 4 times in a row an astounding 19% of the time.

so now 19% of the time he loses $15 while 71% he wins a buck only. If he takes on the gambol he does get his missing dollar 71% of the time, but it's a big -EV game for him of around -$2.30.

also this could be a cool read:

http://casinogambling.about.com/od/m...martingale.htm

Ahh, ok. Ya, it was a bit harder to remember odds being less than full mental capacity. Thanks, I was wondering where my mind was.
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Old Feb 04, 2008, 8:58am   #4
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Lol that article is gold. I went to the casino once with a friend and we tried the martingale system. I think we went in with about $100 and came out with around $350. Needless to say we thought the system was fool proof and "how the casinos could stay in business if everyone used this system?" Also needless to say that we went back and tried it again with less success. But I am still up $150 from trying the martingale system.
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Old Feb 05, 2008, 3:38am   #5
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It wouldn't be a mistake to say that the martingale system is no more profitable then regular betting would it? Since your chances for any one roll is the same?
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Old Feb 09, 2008, 11:08pm   #6
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EDIT... oops should have read the link before posting, it's covered there.


The martingale system works except that the casinos put a limit on the amount you can bet on the games.

If you had an endless bankroll and there was no limit on the size of your bet, you would always win. You might have to bet 1 million to win 1 dollar but you would eventually win.

I tried to come up with a solution for roulette based on progressive bets on varying odds but.... it didn't work very well as I was not willing to put out $100+ on a single spin.
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Last edited by Hawkwynd; Feb 09, 2008 at 11:11pm.
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Old Feb 25, 2008, 4:23am   #7
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Bump. Can't sleep, so... what would stop a billionaire using the gale system effectively?

For simplicity's sake let's say the bet's $2 on European roulette (a single green slot for the house).

- The houses edge is 2.702...% : this we get from ((1/37)*100)
- The chances of our man's bet coming off each time is: 48.64864865%
- This equates to 1 time in 2.055555556
- He will lose 1 time in 1.947368421

- The chances of him losing 25 times in a row are 1 in 17,226,739.
- But if he does lose 25 times in a row he will need to gamble $33,554,432

- The chances of him losing 30 times in a row are 1 in 482,440,427
- But if he does lose 30 times in a row he will need to gamble $1,073,741824 - might have sweaty palms at this point.


I know the returns are marginal compared to our billionaire's poker roll but in practice would this not work given:

- The virtually unlimited supply of money for the exercise
- The creation of an infinite 'short-run' given the above assumption

Is this not a lovely lil hypothetical idea for a robin hood venture for mr gates. Or is my maths very very wrong making this a blonde moment that's almost as bad as slowrolling another ptipper in a railed game?
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Old Feb 25, 2008, 4:26am   #8
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Quote:
Originally Posted by Hawkwynd View Post
EDIT... oops should have read the link before posting, it's covered there.


The martingale system works except that the casinos put a limit on the amount you can bet on the games.
Ignore above post, I'm the 3rd person to post on it . Gnight
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