The gamblers fallacy | Bitcointalksearch.org

a1choi

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Quote from: dolphin2014 on March 13, 2014, 01:55:31 PM

Quote from: goose20 on March 06, 2014, 07:33:26 AM

The gamblers fallacy is that you can find a 'system' to win at gambling.

So true. You can find a system that might increase your chances of winning, but there is no system that can guarantee you will win every single time.

i disagree. for dice, there is no way to increase your chance of winning. Your chances are set at the expected value of the house edge, which is negative. What you can do is skew the potential results in such a way so that you can win more at the cost of losing even more. This can happen while still holding the same overall chance of winning. see my above post for a graphical representation

dolphin2014

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Quote from: goose20 on March 06, 2014, 07:33:26 AM

The gamblers fallacy is that you can find a 'system' to win at gambling.

So true. You can find a system that might increase your chances of winning, but there is no system that can guarantee you will win every single time.

Qwiner

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I don't know if probability is really @ work when betting. The games seemed to become rigged, as U increase your games no.

The more U play,
1. the actual results seems to be the opposite of what U bet (eg. when U bet the result to be >50%, the actual result = <50%, & vice versa) it seems that the gaming system know your betting patterns, so U always ended up losing all your bank roll; +

2. your winning rate (compared to losing rate) will be decreasing over the LR

FUR11

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Quote from: monsterer on March 07, 2014, 04:47:45 AM

Quote from: FUR11 on March 07, 2014, 04:09:37 AM

In terms of odds-to-lose?

So, for example, the odds of losing 10 consecutive 50% in a row is the same as the odds of losing one 99.90234375%?

I guess I need ask about more than just raw probabilities, otherwise it is trivial as you've just demonstrated.

So, I'd like to know:

I place a martingale doubling series bet, starting at size X. It takes Y number of doublings before a win is generated. What single bet size and odds is that equivalent to?

There ought to be a nice graph in there somewhere

The problem is, it's a non-definitive answer. If you are betting at 50%, for example, it's 50% that you'll generate a win after one roll, 75% that you'll generate a win within two rolls, ect, but technically you could go on forever without winning a roll (thus then the gamblers fallacy).

a1choi

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All bets are equal to all other bets independently, as their EV is the same. But a chain of bets, plus differing bet sizes will product different possible outcomes if you take a non-infinite time scale. EV is still the same, but skew is different.

For more analysis, please go here http://wizardofodds.com/gambling/betting-systems/

Take a look at the Flat-betting outcomes versus Martingale system.

Martingale can be more alluring for the Gambler, as it provides chance for larger wins, but at a price of much larger losses. This is why martingale bettors believe in the martingale method. That is until they hit bad luck and lose their shirts.

But thats the thrill of gambling. Good luck!

monsterer

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Quote from: FUR11 on March 07, 2014, 04:09:37 AM

In terms of odds-to-lose?

So, for example, the odds of losing 10 consecutive 50% in a row is the same as the odds of losing one 99.90234375%?

I guess I need ask about more than just raw probabilities, otherwise it is trivial as you've just demonstrated.

So, I'd like to know:

I place a martingale doubling series bet, starting at size X. It takes Y number of doublings before a win is generated. What single bet size and odds is that equivalent to?

There ought to be a nice graph in there somewhere

FUR11

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Quote from: monsterer on March 07, 2014, 04:05:58 AM

Quote from: FUR11 on March 07, 2014, 03:31:55 AM

What do you mean a mathematical analysis? Do you mean something like "what are the odds that you'll go X rolls without a win"?

No, I'd like to see a proof which says that a chain of margingale bets X, is exactly equivalent to a one time bet of Y.

In terms of odds-to-lose?

So, for example, the odds of losing 10 consecutive 50% in a row is the same as the odds of losing one 99.90234375%?

monsterer

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Quote from: FUR11 on March 07, 2014, 03:31:55 AM

What do you mean a mathematical analysis? Do you mean something like "what are the odds that you'll go X rolls without a win"?

No, I'd like to see a proof which says that a chain of margingale bets X, is exactly equivalent to a one time bet of Y.

brian123321

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Quote from: monsterer on March 07, 2014, 03:18:32 AM

Has anyone seen a mathematical analysis of a martingale chain?

I'd be interested to see what single bet odds are equivalent to a given martingale chain of bets. This might help address the addiction to margingale that a lot of gamblers suffer with.

Cheers, Paul.

Martingale Strategies are never a good thing. You will always go broke in the long run and most gambling sites will have a limit on what you can bet which you might hit quickly.

I have messed around with them in the past and they can seem like a great strategy if you get on a nice run out of the gate but in the long run they will cripple/break you. I dabbled with a martingale strategy about a year ago at the beginning of last baseball season when i first started dabbling in MLB. I got off to a killer start but a month or two later i went on a bad streak, which will happen, and absolutely crippled myself.

Its a tough but valuable lesson to learn. Listen to others when they say its a recipe for disaster. They have been through it and are trying to warn you.

FUR11

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Quote from: monsterer on March 07, 2014, 03:18:32 AM

Has anyone seen a mathematical analysis of a martingale chain?

I'd be interested to see what single bet odds are equivalent to a given martingale chain of bets. This might help address the addiction to margingale that a lot of gamblers suffer with.

Cheers, Paul.

What do you mean a mathematical analysis? Do you mean something like "what are the odds that you'll go X rolls without a win"?

monsterer

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Has anyone seen a mathematical analysis of a martingale chain?

I'd be interested to see what single bet odds are equivalent to a given martingale chain of bets. This might help address the addiction to margingale that a lot of gamblers suffer with.

Cheers, Paul.

FUR11

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Quote from: Kyraishi on March 06, 2014, 12:04:39 PM

The winning strategy is to not gamble.

That's probably not the most popular opinion in this forum.

Kyraishi

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The winning strategy is to not gamble.

spixel

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Quote from: b!z on March 06, 2014, 10:08:31 AM

I have noticed that some people believe they have a "winning strategy". This is simply not possible. If you are having gaming problems, please read http://bitcoinreviewer.com/safe-responsible-gambling/ for more information.

There's no such thing as a winning strategy, but there can be ones which are better than others. For example, going all in @ 0.01% chance is not a good strategy.

BitCoinPokerBro

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Quote from: monsterer on March 06, 2014, 06:10:25 AM

The gamblers fallacy is this:

Scenario: Dice game, 50/50 random chance, bet on <50 wins

"I've rolled >50 four times in a row now, probability of that happening is (1/2)*(1/2)*(1/2)*(1/2) = 1/16 (or 0.0625), the chances of rolling 5 times <50 are therefore 1/32, so I'm likely to win this time."

Seems logical enough? The reality is this:

The probability of <50 on any given roll is 1/2. It doesn't matter that you've rolled 4 times < 50, probability doesn't have a memory. To see why this is the case, consider this:

What's the probability of rolling 3 times < 50 and 1 time > 50? Guess what? (1/2)*(1/2)*(1/2)*(1/2) = 1/16.

What about 2 times > 50 and 2 times < 50, (1/2)*(1/2)*(1/2)*(1/2) = 1/16. The same probability.

It doesn't matter that you've rolled 4 times < 50, the probability of the next roll being > or < is still 50/50.

In fact, the probability of any combination of rolls occurring only depends on the number of rolls, not the previous outcomes.

Hope that helps some of you

Cheers, Paul.

http://wizardofodds.com/ask-the-wizard/betting-systems/gamblers-fallacy/

b!z

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I have noticed that some people believe they have a "winning strategy". This is simply not possible. If you are having gaming problems, please read http://bitcoinreviewer.com/safe-responsible-gambling/ for more information.

Light

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Quote from: goose20 on March 06, 2014, 07:33:26 AM

The gamblers fallacy is that you can find a 'system' to win at gambling.

Well at dice betting at least. Sports betting may and can be profitable as it is truly impossible for anyone to actually quantify the probability of a sports event exactly (or even extremely close) and hence these variations in lines can create opportunities for bettors.

goose20

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The gamblers fallacy is that you can find a 'system' to win at gambling.

monsterer

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Quote from: spixel on March 06, 2014, 07:15:00 AM

Yes, but alot of people think 'gamblers fallacy' means that
<50, <50, <50, <50
is equally as likely as this sequence:
<50, <50, <50, <50, <50, <50, <50, <50, <50

But it's not, because each roll gives you another chance to win.

They do? I was under the impression that the fallacy is primarily concerned with the history of previous bets having an influence.

I guess the takeaway is that rolling <50 on any go is always 50/50, no matter what the previous bets have been.

spixel

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Quote from: monsterer on March 06, 2014, 07:05:41 AM

Agreed.

Importantly, and in laymen's terms:

This sequence:
<50, <50, <50, <50
is equally as likely as this sequence:
<50, >50, <50, >50
and this one:
<50, <50, >50, >50
and this one:
<50, >50, >50, >50
...and so on

Yes, but alot of people think 'gamblers fallacy' means that
<50, <50, <50, <50
is equally as likely as this sequence:
<50, <50, <50, <50, <50, <50, <50, <50, <50

But it's not, because each roll gives you another chance to win.

monsterer

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Quote from: Light on March 06, 2014, 07:01:29 AM

In layman's terms with the use of math, basically the past has no bearing or influence on your future results (for completely unrelated events ie dice rolls). So don't believe that just because something is so unlikely that it cannot happen when you lose 28 times in a row and try for a 29th.

That's really the reason I stopped dice betting, the more you play the more you accentuate the house edge against you, whereas in something like sports betting/poker you have greater control over your results and hence there is a correlation between your skill and your earnings. In effect you can improve your probability of winning in those, whereas in dice you cannot.

Agreed.

Importantly, and in laymen's terms:

This sequence:
<50, <50, <50, <50
is equally as likely as this sequence:
<50, >50, <50, >50
and this one:
<50, <50, >50, >50
and this one:
<50, >50, >50, >50
...and so on

Light

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In layman's terms with the use of math, basically the past has no bearing or influence on your future results (for completely unrelated events ie dice rolls). So don't believe that just because something is so unlikely that it cannot happen when you lose 28 times in a row and try for a 29th.

That's really the reason I stopped dice betting, the more you play the more you accentuate the house edge against you, whereas in something like sports betting/poker you have greater control over your results and hence there is a correlation between your skill and your earnings. In effect you can improve your probability of winning in those, whereas in dice you cannot.

monsterer

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Quote from: spixel on March 06, 2014, 06:37:54 AM

Rolling <50 x amounts of times in a row still has its own probability, true a previous roll doesn't affect the next but this is obvious.

Rolling <50 1000 times in a row has the exact same probability of rolling any combination of < or > 1000 times in a row.

The question shouldn't be: "Would would rather bet $1,000 on rolling <50 once or <50 ten times in a row?", but instead:

"Would would rather bet $1,000 on rolling <50 once or any combination of under or over ten times in a row?"

<50 once has a probability = 1/2.
10 rolls (< or >) in a row has a probability = (1/2)^10 = 1/1024.

spixel

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Rolling <50 x amounts of times in a row still has its own probability, true a previous roll doesn't affect the next but this is obvious.

Would would rather bet $1,000 on rolling <50 once or <50 ten times in a row?

monsterer

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The gamblers fallacy is this:

Scenario: Dice game, 50/50 random chance, bet on <50 wins

"I've rolled >50 four times in a row now, probability of that happening is (1/2)*(1/2)*(1/2)*(1/2) = 1/16 (or 0.0625), the chances of rolling 5 times <50 are therefore 1/32, so I'm likely to win this time."

Seems logical enough? The reality is this:

The probability of <50 on any given roll is 1/2. It doesn't matter that you've rolled 4 times < 50, probability doesn't have a memory. To see why this is the case, consider this:

What's the probability of rolling 3 times < 50 and 1 time > 50? Guess what? (1/2)*(1/2)*(1/2)*(1/2) = 1/16.

What about 2 times > 50 and 2 times < 50, (1/2)*(1/2)*(1/2)*(1/2) = 1/16. The same probability.

It doesn't matter that you've rolled 4 times < 50, the probability of the next roll being > or < is still 50/50.

In fact, the probability of any combination of rolls occurring only depends on the number of rolls, not the previous outcomes.

Hope that helps some of you

Cheers, Paul.

Topic: The gamblers fallacy (Read 1835 times)