Does martingale really works?

[CoinCidental]

It's a pity that you can't think outside the box. You multiply the two chances, so they are already related, being of the same kind, right? Therefore they are not truly independent of each other since each of them is still dependent on something else common for them. You can't multiply a number with something which is not a number. This is likely not a very good example, but still...

As to me, your example is not relevant to successive rolls

Pick one card from each of two decks then. It doesn't change anything. What's the chance you get a queen from the first and a heart from the 2nd?

Now, of course, the "heart, then red" events *are* independent, and so 1/8 is the chance of the first card being a heart and the 2nd being red.

Why do you think that multiplying the probabilities of two independent events causes the events to stop being independent?

I don't argue with your examples, I know what you say perfectly well myself. But you give examples which are not the same as successive rolls. If you pick a queen of hearts from a deck, will the probability of picking another queen of hearts from the same deck (provided there is only one queen of hearts in the deck) be of the same kind as the probability of picking next a king of spades?

the probabilty of picking a 2nd queen of hearts from the deck is 0 irrespective of everything else
(magic tricks excluded)

[deisik]

It's a pity that you can't think outside the box. You multiply the two chances, so they are already related, being of the same kind, right? Therefore they are not truly independent of each other since each of them is still dependent on something else common for them. You can't multiply a number with something which is not a number. This is likely not a very good example, but still...

As to me, your example is not relevant to successive rolls

Pick one card from each of two decks then. It doesn't change anything. What's the chance you get a queen from the first and a heart from the 2nd?

Now, of course, the "heart, then red" events *are* independent, and so 1/8 is the chance of the first card being a heart and the 2nd being red.

Why do you think that multiplying the probabilities of two independent events causes the events to stop being independent?

I don't argue with your examples, I know what you say perfectly well myself. But you give examples which are not the same as successive rolls. If you pick a queen of hearts from a deck, will the probability of picking another queen of hearts from the same deck (provided there is only one queen of hearts in the deck) be of the same kind as the probability of picking next a king of spades?

the probabilty of picking a 2nd queen of hearts from the deck is 0 irrespective of everything else
(magic tricks excluded)

As you might have noticed (in fact, actually failed to notice), I didn't ask what the probability of picking a second queen of hearts was. I ask if this probability is of the same kind as the probability of picking next any other card but the queen of hearts. Can you say for sure that these probabilities are alike, or, in a more broad sense, are there different types of probabilities?

I think these are different types of probabilities, and this can be proven

[CoinCidental]

It's a pity that you can't think outside the box. You multiply the two chances, so they are already related, being of the same kind, right? Therefore they are not truly independent of each other since each of them is still dependent on something else common for them. You can't multiply a number with something which is not a number. This is likely not a very good example, but still...

As to me, your example is not relevant to successive rolls

Pick one card from each of two decks then. It doesn't change anything. What's the chance you get a queen from the first and a heart from the 2nd?

Now, of course, the "heart, then red" events *are* independent, and so 1/8 is the chance of the first card being a heart and the 2nd being red.

Why do you think that multiplying the probabilities of two independent events causes the events to stop being independent?

I don't argue with your examples, I know what you say perfectly well myself. But you give examples which are not the same as successive rolls. If you pick a queen of hearts from a deck, will the probability of picking another queen of hearts from the same deck (provided there is only one queen of hearts in the deck) be of the same kind as the probability of picking next a king of spades?

the probabilty of picking a 2nd queen of hearts from the deck is 0 irrespective of everything else
(magic tricks excluded)

As you might have noticed (in fact, actually failed to notice), I didn't ask what the probability of picking a second queen of hearts was. I ask if this probability is of the same kind as the probability of picking next any other card but the queen of hearts. Can you say for sure that these probabilities are alike, or, in a more broad sense, are there different types of probabilities?

I think these are different types of probabilities, and this can be proven

i think youve lost the plot already .......its not improbable to pick a 2nd queen of hearts ,its IMPOSSIBLE

yes ,these are different probabilities since any other card is still probable to some degree if its still in the deck ........

[deisik]

I don't argue with your examples, I know what you say perfectly well myself. But you give examples which are not the same as successive rolls. If you pick a queen of hearts from a deck, will the probability of picking another queen of hearts from the same deck (provided there is only one queen of hearts in the deck) be of the same kind as the probability of picking next a king of spades?

the probabilty of picking a 2nd queen of hearts from the deck is 0 irrespective of everything else
(magic tricks excluded)

As you might have noticed (in fact, actually failed to notice), I didn't ask what the probability of picking a second queen of hearts was. I ask if this probability is of the same kind as the probability of picking next any other card but the queen of hearts. Can you say for sure that these probabilities are alike, or, in a more broad sense, are there different types of probabilities?

I think these are different types of probabilities, and this can be proven

i think youve lost the plot already .......its not improbable to pick a 2nd queen of hearts ,its IMPOSSIBLE

yes ,these are different probabilities since any other card is still probable to some degree if its still in the deck ........

My goal was not to show that something is impossible, and in that it differs from something which is possible. My example reveals how a previous choice affects the future probability of an event. And more specifically, if we have 20 losing bets in a row, then the probability of the next 20 losing bets will be the same, right? But in this case these probabilities are different from the probability of 40 losses in a row, which seems paradoxical. Thus, in a sense, previous outcomes affect the probability of future rolls...

[futureofbitcoin]

Why are there so many pages? This question is easy to answer.

Martingale works if 2 conditions are met:

1. You have an infinite balance

2. There is no maximum bet

My goal was not to show that something is impossible, and in that it differs from something which is possible. My example reveals how a previous choice affects the future probability of an event. And more specifically, if we have 20 losing bets in a row, then the probability of the next 20 losing bets will be the same, right? But in this case these probabilities are different from the probability of 40 losses in a row, which seems paradoxical. Thus, in a sense, previous outcomes affect the probability of future rolls...

It's not paradoxical. The solution is that because the amount of information known is different.

For example, I ask you "what's the area of a square?"

Well, that question is unsolvable. Then I ask you "What's the area of a square with each side being 2?" Well, the answer is simple, it's 4. Is that paradoxical? Somehow the first time I ask you for the area it was unsolvable, and then suddenly it became solvable? No, because the amount of information given is different.

I would specifically argue that the reason we drive to work is because, although driving to work is essentially a -ev bet (life expectancy not withstanding), we do it because our goal is to make a profit, not to "beat the house."  Hence, I think that is more often why people gamble, i.e. to "profit," and that's it.
 

Driving to work is not necessarily a -ev bet. If, for example, by not working, you will starve to death after a week with 100% certainty, then the MARGINAL EV is positive, because you necessarily have to play the game, you can only pick the best option.

And then there are people who derive pleasure from working. If P = Pleasure, 1/X = the probability of dying in a car crash while driving to work, and N = the negative effect of dying, if:

XP > N, then it's positive EV, no?

And then there are the people who neither HAVE to work, nor gain a positive EV from working. These, are people I would call "stupid"

[deisik]

My goal was not to show that something is impossible, and in that it differs from something which is possible. My example reveals how a previous choice affects the future probability of an event. And more specifically, if we have 20 losing bets in a row, then the probability of the next 20 losing bets will be the same, right? But in this case these probabilities are different from the probability of 40 losses in a row, which seems paradoxical. Thus, in a sense, previous outcomes affect the probability of future rolls...

It's not paradoxical. The solution is that because the amount of information known is different.

For example, I ask you "what's the area of a square?"

Well, that question is unsolvable. Then I ask you "What's the area of a square with each side being 2?" Well, the answer is simple, it's 4. Is that paradoxical? Somehow the first time I ask you for the area it was unsolvable, and then suddenly it became solvable? No, because the amount of information given is different.

I see your point but it is not strictly the same. We know the probability of each loss in a row of 40 successive losses (so we know the area of a square beforehand), but with each roll made the probabilities for rolls yet to be made change, right? Therefore, it turns out that the outcome of each previous roll changes the probability of the next roll...

I don't find a fallacy in this logic, do you?

[adaseb]

...


Again, it depends on what you mean by "odds in your favour".

If you mean "more likely to win than to lose" then yes.

...

That was what I mean

I of course agree with you that the stakes must be considered.

The only reason I made the post was because it seemed to me as though people were treating "profit" and "doubling money" as equivalent.  I felt it wasn't totally irrelevant or meaningless to point out that one's goal when betting is an important consideration.

Practical scenario:  Rent is due tomorrow and you owe BTC1.01 and you have BTC1.00 only.  If you are late on your payment, you will face a 10% fee and will then owe BTC1.10.

To avoid the 10% fee, you could gamble your BTC1.00 with somewhere between a 98-99% chance of profiting/winning BTC0.01 so that you can pay on-time and in full.

Would you take that bet to make your rent payment on time?  I'm guessing there would be mixed opinions depending on who you ask.

In this case, the bettor's goal isn't to double his money.  His goal is to make a minute profit.

Honestly with my luck, I would probably be the unlucky one and hit the 1% losing edge. Rather just do it in partial ammounts. Or just get a loan from someone at BTCTalk

LOL

[Bardman]

My goal was not to show that something is impossible, and in that it differs from something which is possible. My example reveals how a previous choice affects the future probability of an event. And more specifically, if we have 20 losing bets in a row, then the probability of the next 20 losing bets will be the same, right? But in this case these probabilities are different from the probability of 40 losses in a row, which seems paradoxical. Thus, in a sense, previous outcomes affect the probability of future rolls...

It's not paradoxical. The solution is that because the amount of information known is different.

For example, I ask you "what's the area of a square?"

Well, that question is unsolvable. Then I ask you "What's the area of a square with each side being 2?" Well, the answer is simple, it's 4. Is that paradoxical? Somehow the first time I ask you for the area it was unsolvable, and then suddenly it became solvable? No, because the amount of information given is different.

I see your point but it is not strictly the same. We know the probability of each loss in a 40 successive losses, but with each roll made the probabilities for rolls yet to be made change, right? Therefore, it turns out that the outcome of each previous roll changes the probability for the next roll...

Look, lets say we have a 50% chance of red/black and we play 4 times, what are the chances of getting 4 blacks in a row? 6.25% what are the chances of getting 4 reds in a row? 6.25%, what are the chances of getting 1 red 1 black 1 red 1 black ? 6.25% and so on

Here are all the possibilietes :

RED RED RED RED  -  1
RED RED RED BLACK  - 2
RED RED BLACK RED  -  3
RED RED BLACK BLACK  - 4
RED BLACK BLACK RED  - 5
RED BLACK BLACK BLACK  - 6
RED BLACK RED BLACK  - 7
RED BLACK RED RED  - 8
BLACK BLACK BLACK BLACK  - 9
BLACK BLACK BLACK RED  -  10
BLACK BLACK RED RED     -  11
BLACK BLACK RED BLACK    - 12
BLACK RED BLACK BLACK    - 13
BLACK RED BLACK RED    - 14
BLACK RED RED BLACK     - 15
BLACK RED RED RED   - 16


16 Possible outcomes , guess whats 6.25% x 16 ? = 100%

[futureofbitcoin]

I see your point but it is not strictly the same. We know the probability of each loss in a 40 successive losses, but with each roll made the probabilities for rolls yet to be made change, right? Therefore, it turns out that the outcome of each previous roll changes the probability for the next roll...

I don't find fallacy in this logic, do you?

No, it doesn't change. The probability distributions of 40 rolls, all unknown and the probability distribution of 40 rolls, with X amount known is different.

It's like the probability of there being rain within a 100km radius of point A in the Amazon is X

But the probability of there being rain within a 100km radius of point B is not necessarily X, they're different.

This is, of course ignoring the possibility of things such as chaos theory, or dark matter or some form of physics that is still completely hidden from humanity affecting future events.

[CoinCidental]

I don't argue with your examples, I know what you say perfectly well myself. But you give examples which are not the same as successive rolls. If you pick a queen of hearts from a deck, will the probability of picking another queen of hearts from the same deck (provided there is only one queen of hearts in the deck) be of the same kind as the probability of picking next a king of spades?

the probabilty of picking a 2nd queen of hearts from the deck is 0 irrespective of everything else
(magic tricks excluded)

As you might have noticed (in fact, actually failed to notice), I didn't ask what the probability of picking a second queen of hearts was. I ask if this probability is of the same kind as the probability of picking next any other card but the queen of hearts. Can you say for sure that these probabilities are alike, or, in a more broad sense, are there different types of probabilities?

I think these are different types of probabilities, and this can be proven

i think youve lost the plot already .......its not improbable to pick a 2nd queen of hearts ,its IMPOSSIBLE

yes ,these are different probabilities since any other card is still probable to some degree if its still in the deck ........

Thus, in a sense, previous outcomes affect the probability of future rolls...

but they dont ....

lets make this easy to understand

flip a coin 100 times ,last 20 flips were heads
getting ready to flip it for 101,what do you predict the odds will be  for heads again ?
its still 50/50 no matter how many times you play

[deisik]

I see your point but it is not strictly the same. We know the probability of each loss in a 40 successive losses, but with each roll made the probabilities for rolls yet to be made change, right? Therefore, it turns out that the outcome of each previous roll changes the probability for the next roll...

Look, lets say we have a 50% chance of red/black and we play 4 times, what are the chances of getting 4 blacks in a row? 6.25% what are the chances of getting 4 reds in a row? 6.25%, what are the chances of getting 1 red 1 black 1 red 1 black ? 6.25% and so on

Here are all the possibilietes :

RED RED RED RED  -  1
RED RED RED BLACK  - 2
RED RED BLACK RED  -  3
RED RED BLACK BLACK  - 4
RED BLACK BLACK RED  - 5
RED BLACK BLACK BLACK  - 6
RED BLACK RED BLACK  - 7
RED BLACK RED RED  - 8
BLACK BLACK BLACK BLACK  - 9
BLACK BLACK BLACK RED  -  10
BLACK BLACK RED RED     -  11
BLACK BLACK RED BLACK    - 12
BLACK RED BLACK BLACK    - 13
BLACK RED BLACK RED    - 14
BLACK RED RED BLACK     - 15
BLACK RED RED RED   - 16

16 Possible outcomes , guess whats 6.25% x 16 ? = 100%

I understand what you mean, but here is my example. We have a 50% chance of red/black, and we play just 2 times, the probability of two reds in a row before we start betting is 0.5 x 0.5 = 0.25. All is good so far. We make a bet and get a red, what is the probability of having two reds in a row now, after we have got a red? Right, 0.5...

[Bardman]

I see your point but it is not strictly the same. We know the probability of each loss in a 40 successive losses, but with each roll made the probabilities for rolls yet to be made change, right? Therefore, it turns out that the outcome of each previous roll changes the probability for the next roll...

Look, lets say we have a 50% chance of red/black and we play 4 times, what are the chances of getting 4 blacks in a row? 6.25% what are the chances of getting 4 reds in a row? 6.25%, what are the chances of getting 1 red 1 black 1 red 1 black ? 6.25% and so on

Here are all the possibilietes :

RED RED RED RED  -  1
RED RED RED BLACK  - 2
RED RED BLACK RED  -  3
RED RED BLACK BLACK  - 4
RED BLACK BLACK RED  - 5
RED BLACK BLACK BLACK  - 6
RED BLACK RED BLACK  - 7
RED BLACK RED RED  - 8
BLACK BLACK BLACK BLACK  - 9
BLACK BLACK BLACK RED  -  10
BLACK BLACK RED RED     -  11
BLACK BLACK RED BLACK    - 12
BLACK RED BLACK BLACK    - 13
BLACK RED BLACK RED    - 14
BLACK RED RED BLACK     - 15
BLACK RED RED RED   - 16

16 Possible outcomes , guess whats 6.25% x 16 ? = 100%

I understand what you mean, but here is my example. We have a 50% chance of red/black, and we play just 2 times, the probability of two reds in a row before we start betting is 0.5 x 0.5 = 0.25. All is good so far. We make a bet and get a red, what is the probability of having two reds in a row now, after we have got a red? Right, 0.5...

First of all 2 in a row is 25% not 0.25 

So we make 1 bet

And we have 50% of that bet to be RED or BLACK

It is RED

What are our chances now of getting another RED ? 50%

in 2 plays we have this:

RED RED
RED BLACK
BLACK RED
BLACK BLACK

You said its RED what we got in the first bet so:

RED RED

RED BLACK

as you can see you have 50% chances

[futureofbitcoin]

Uh, 0.25 = 25%.... and you are just agreeing with him LOL.

@deisik:

You're comparing apples to oranges. The previous roll doesn't affect the outcome of the next roll. It affects the distribution of BOTH rolls. It's different.

In 40 rolls, every individual roll affects ITSELF, and only itself. But because it is a part of the probability distribution, it changes the whole distribution, but it doesn't change the probabilities of other rolls.


Think of it like this:

The average height of a family with a father, a mother, a brother and a sister is 1.50m tall.

Then the brother grew 10cm, and the sister grew 6cm. Now the average height of the family is 1.54m tall.

As you can see, the height of the brother affects his own height and no one else's. Same with the sister. It affects the total distribution, but not the other independent variables.

[deisik]

I understand what you mean, but here is my example. We have a 50% chance of red/black, and we play just 2 times, the probability of two reds in a row before we start betting is 0.5 x 0.5 = 0.25. All is good so far. We make a bet and get a red, what is the probability of having two reds in a row now, after we have got a red? Right, 0.5...

First of all 2 in a row is 25% not 0.25 

So we make 1 bet

And we have 50% of that bet to be RED or BLACK

It is RED

What are our chances now of getting another RED ? 50%

in 2 plays we have this:

RED RED
RED BLACK
BLACK RED
BLACK BLACK

You said its RED what we got in the first bet so:

RED RED

RED BLACK

as you can see you have 50% chances

You seem to have not read my post. And I meant unit fractions for probabilities, not percentages...

[Bardman]

I understand what you mean, but here is my example. We have a 50% chance of red/black, and we play just 2 times, the probability of two reds in a row before we start betting is 0.5 x 0.5 = 0.25. All is good so far. We make a bet and get a red, what is the probability of having two reds in a row now, after we have got a red? Right, 0.5...

First of all 2 in a row is 25% not 0.25 

So we make 1 bet

And we have 50% of that bet to be RED or BLACK

It is RED

What are our chances now of getting another RED ? 50%

in 2 plays we have this:

RED RED
RED BLACK
BLACK RED
BLACK BLACK

You said its RED what we got in the first bet so:

RED RED

RED BLACK

as you can see you have 50% chances

You seem to have not read my post. And I meant unit fractions for probabilities, not percentages...

What did you not understand? Or were you just making it clear that its always 50%?

[deisik]

You're comparing apples to oranges. The previous roll doesn't affect the outcome of the next roll. It affects the distribution of BOTH rolls. It's different.

I know all of this. I don't say that the previous rolls directly affect the outcome of the next rolls. They don't, and practice just confirms that. Whenever I said anything of the kind, I always emphasized "in a sense". I just want to look at the probabilities from another angle, maybe, to find a weak point (and ruin a casino)...

[tspacepilot]

Earlier in the thread I left a comment about my own particular conundrum on this topic and I'm honestly a little surprised that no one followed up with me given the smart statistical heads that hand around this forum.  Here's my issue, we know that, as has been said above, in these sorts of games each trial is independent.  However, I think we also know that increasingly long streaks are increasingly unlikely.  How do we resolve these two (both valid, I think) intuitions?

[Bardman]

Earlier in the thread I left a comment about my own particular conundrum on this topic and I'm honestly a little surprised that no one followed up with me given the smart statistical heads that hand around this forum.  Here's my issue, we know that, as has been said above, in these sorts of games each trial is independent.  However, I think we also know that increasingly long streaks are increasingly unlikely.  How do we resolve these two (both valid, I think) intuitions?

i explained that here:

Look, lets say we have a 50% chance of red/black and we play 4 times, what are the chances of getting 4 blacks in a row? 6.25% what are the chances of getting 4 reds in a row? 6.25%, what are the chances of getting 1 red 1 black 1 red 1 black ? 6.25% and so on

Here are all the possibilietes :

RED RED RED RED  -  1
RED RED RED BLACK  - 2
RED RED BLACK RED  -  3
RED RED BLACK BLACK  - 4
RED BLACK BLACK RED  - 5
RED BLACK BLACK BLACK  - 6
RED BLACK RED BLACK  - 7
RED BLACK RED RED  - 8
BLACK BLACK BLACK BLACK  - 9
BLACK BLACK BLACK RED  -  10
BLACK BLACK RED RED     -  11
BLACK BLACK RED BLACK    - 12
BLACK RED BLACK BLACK    - 13
BLACK RED BLACK RED    - 14
BLACK RED RED BLACK     - 15
BLACK RED RED RED   - 16

16 Possible outcomes , guess whats 6.25% x 16 ? = 100%


Even tho getting 4 blacks in a row is unlikely (6.25%) getting 2 reds 2 blacks is unlikely aswell (6.25%) so every combination is equally unlikely

[tspacepilot]

Earlier in the thread I left a comment about my own particular conundrum on this topic and I'm honestly a little surprised that no one followed up with me given the smart statistical heads that hand around this forum.  Here's my issue, we know that, as has been said above, in these sorts of games each trial is independent.  However, I think we also know that increasingly long streaks are increasingly unlikely.  How do we resolve these two (both valid, I think) intuitions?

i explained that here:

Look, lets say we have a 50% chance of red/black and we play 4 times, what are the chances of getting 4 blacks in a row? 6.25% what are the chances of getting 4 reds in a row? 6.25%, what are the chances of getting 1 red 1 black 1 red 1 black ? 6.25% and so on

Here are all the possibilietes :

RED RED RED RED  -  1
RED RED RED BLACK  - 2
RED RED BLACK RED  -  3
RED RED BLACK BLACK  - 4
RED BLACK BLACK RED  - 5
RED BLACK BLACK BLACK  - 6
RED BLACK RED BLACK  - 7
RED BLACK RED RED  - 8
BLACK BLACK BLACK BLACK  - 9
BLACK BLACK BLACK RED  -  10
BLACK BLACK RED RED     -  11
BLACK BLACK RED BLACK    - 12
BLACK RED BLACK BLACK    - 13
BLACK RED BLACK RED    - 14
BLACK RED RED BLACK     - 15
BLACK RED RED RED   - 16

16 Possible outcomes , guess whats 6.25% x 16 ? = 100%


Even tho getting 4 blacks in a row is unlikely (6.25%) getting 2 reds 2 blacks is unlikely aswell (6.25%) so every combination is equally unlikely

I think that does help actually.  So, even though my intuition is that 5 blacks in a row is more unlikely than 4 blacks in a row, in fact, any combination of 5 is more unlikely than any combination of 4?  Is that right?  I think I said that very inelegantly but this does help, thanks Bardman. 

[Bardman]

Earlier in the thread I left a comment about my own particular conundrum on this topic and I'm honestly a little surprised that no one followed up with me given the smart statistical heads that hand around this forum.  Here's my issue, we know that, as has been said above, in these sorts of games each trial is independent.  However, I think we also know that increasingly long streaks are increasingly unlikely.  How do we resolve these two (both valid, I think) intuitions?

i explained that here:

Look, lets say we have a 50% chance of red/black and we play 4 times, what are the chances of getting 4 blacks in a row? 6.25% what are the chances of getting 4 reds in a row? 6.25%, what are the chances of getting 1 red 1 black 1 red 1 black ? 6.25% and so on

Here are all the possibilietes :

RED RED RED RED  -  1
RED RED RED BLACK  - 2
RED RED BLACK RED  -  3
RED RED BLACK BLACK  - 4
RED BLACK BLACK RED  - 5
RED BLACK BLACK BLACK  - 6
RED BLACK RED BLACK  - 7
RED BLACK RED RED  - 8
BLACK BLACK BLACK BLACK  - 9
BLACK BLACK BLACK RED  -  10
BLACK BLACK RED RED     -  11
BLACK BLACK RED BLACK    - 12
BLACK RED BLACK BLACK    - 13
BLACK RED BLACK RED    - 14
BLACK RED RED BLACK     - 15
BLACK RED RED RED   - 16

16 Possible outcomes , guess whats 6.25% x 16 ? = 100%


Even tho getting 4 blacks in a row is unlikely (6.25%) getting 2 reds 2 blacks is unlikely aswell (6.25%) so every combination is equally unlikely

I think that does help actually.  So, even though my intuition is that 5 blacks in a row is more unlikely than 4 blacks in a row, in fact, any combination of 5 is more unlikely than any combination of 4?  Is that right?  I think I said that very inelegantly but this does help, thanks Bardman. 

5 blacks in a row is equally unlikely as 4 blacks and 1 red or any other combination, so even tho 5 blacks in a row is unlikely so are the other combinations