A math question

[soggett]

 

Ok,  I'm not a math genius so i need some help/opinions

Hope I explain it right...
Let's say we have a system that bets on E/C
We wait for a trigger and then bet. We have a 75% chance of winning that bet
If we happen to lose we wait for another trigger and bet again, having 75% chance of winning again.

So if we want to risk 7 units (1-2-4) what chance do we have of winning within 3 tries?

Trigger one - bet 1 unit - 75% of winning - lose
Trigger two - bet 2 units - 75% of winning - lose
Trigger three - bet 4 units - 75% of winning - ?

What are our chances of winning/losing all three bets?
Is it still 75% or is it calculated differently?

[Bayes]

Hi Soggett,

I'm not sure how you get a 75% chance of a win betting one EC for one spin, but anyway for a 75% chance, whatever the bet:

Chance of winning all 3 is 0.75³ = 42%
Chance of losing all 3 is 0.25³ = 1.56%

Roulette probabilities are easy to calculate because spins are independent. That means you can just multiply the probability of a single win by itself X - 1 times, where X is the length of the winning streak. You can find the chance of a bet losing by subtracting the chance of it winning from 100%. This is because either it wins or loses, and the probability of one or the other happening is 100%. So,

Chance of winning + chance of losing = 100%
75% + chance of losing = 100%

Therefore, chance of losing = 100% - 75% = 25%

Then to find the chance of a streak of losses of length X, just use the same rule as above - multiply the probability by itself X - 1 times.

[RouletteExplorer]

I also don t know  how you get a 75% chance of a win betting one EC for one spin.

But because every spin is independent the chances are always the same no matter how many spins have passed

[Chrisbis]

.............because, as I was just discussing with a good friend of mine,  % of wins to loses is Finite, its the DISTRIBUTION of results that we all get confused with, it terms of Bet Chances.is Finit

We are all looking for the distribution of results to go in our favour.

The chance is always the same, at every Betting Point Opportunity.

This is what I (In My Humble Opinion) think is key.

The game can be beaten, that is certain.

Its just a question of ........................
[reveal]timing.[/reveal]

[vladir]

.............because, as I was just discussing with a good friend of mine, opportunity is Finite in the % of wins to loses, its the DISTRIBUTION of results that we all get confused with, it terms of Bet Chances.

We are all looking for the distribution of results to go in our favour.

The chance is always the same, at every Betting Point Opportunity.

This is what I (In My Humble Opinion) think is key.

The game can be beaten, that is certain.

Its just a question of ........................
[reveal]timing.[/reveal]

Only question is... how can we exploit that?

[soggett]

Thank you for your replys

The 75% for E/C was just an example, i didn't know hot to explain it better, sorry

Anyway, thanks guys

P.S.

Are you saying that with 99% of a bet winning it would be 0,99x0,99x0,99= 97% of winning 3 in in a row?
Wow, does that mean that if a system is not 100% it is going to fail?
I don't think any system is 100% chance of winning

Edit,
I just read it again... I think I maybe asked wrong;
I meant: what is the chance of losing a bet in 3 tries if at every try we have a 75% chance of winning?

[mr.ore]

Chance of losing three times in a row is 0.25^3 = 0.015625, that's 1.5625%. It is impossible to get such a hit ratio on EC, if you really HAD it, just flat bet and increase your bet size by Kelly criterion, lol.

[superman]

if you really HAD it, just flat bet and increase your bet size

My thoughts exactly.

The 75% for E/C was just an example, i didn't know hot to explain it better

Which part is the example, 75% ?

[GLC]

Soggett,  What everybody is getting at is an even chance has a 50% chance of winning, not counting in the zero/s.  By saying 75% you're tweaking our brains.

[soggett]

Im sorry for the mixup..

Ok, forget the E/C

If a bet has 75% chance of winning what are the ods of winning in three tries,
I mean:
Bet 1 - lose
Bet 2 - lose.
Bet 3 - win/lose --> here what is our % chance of winning/losing if we lose two times before that and if we know that every bet has 75% chance of winning

Is that explained better?

[Nathanael]

I'm sorry for the mixup..

Ok, forget the E/C

If a bet has 75% chance of winning what are the ods of winning in three tries,
I mean:
Bet 1 - lose
Bet 2 - lose.
Bet 3 - win/lose --> here what is our % chance of winning/losing if we lose two times before that and if we know that every bet has 75% chance of winning

Is that explained better?

100-75 = 25% odds of losing

.25X.25X.25 = .01563 or 1.563 % chance to lose in 3 bets

100-1.563 = 98.437 odds to win in 3 bets.

Beware.  These odds only apply before the 1st placed bet.

Once you lose 1 bet, now the odds change and are for chance to lose in 2 bets.

Once you have lost 2 times, the odds of losing the 3rd bet are 25%.

Nate

[Bayes]

I mean:
Bet 1 - lose
Bet 2 - lose.
Bet 3 - win/lose --> here what is our % chance of winning/losing if we lose two times before that and if we know that every bet has 75% chance of winning

Is that explained better?

Nathanael's post tells you the chance of AT LEAST one win in the 3 spins, but for your specific pattern (which is much less likely), the chance is -

lose ✕ lose ✕ win = 0.25 ✕ 0.25 ✕ 0.75 = 4.69%

[soggett]

Thanks Nathanael, that's what I was thinking, thank you very much

[hanshuckebein]

so if we have a defined sequence of 3 ec like HHL that will make a 0.5 x 0,5 x 0.5 = 0.125 =12.5% chance of losing the sequence.

does this mean that within 100 spins the sequence HHL should show 12,5 times?

I know I'm a real math dummy. and you may laugh if you wish. 

cheers

hans

[Bayes]

Correct Hans. 

You can check this because there are 8 such patterns and 8 × 12.5 = 100