I was trying to do a roulett technique, without the right amount of money, and I just lost 120$.I knew I did not have enough money to do the strat. But did it any way.

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[–][deleted] 2 insightful - 1 fun2 insightful - 0 fun3 insightful - 1 fun -  (18 children)

I know this makes sense in your head. I’m trying to politely tell you that you are wrong. I am trying to politely guide you to the correct solution. I am not trying to be rude or smart.

You could say gambling is stupid. The method I’m discussing about gambling is stupid. The house still has the edge. It’s relies on chance. Etc.

All of that would be correct.

These are two different statistical equations.

One equation is what are the odds for red, for a a single coin flip. That’s 50/50.

Now what are the odds for red on 10 coin flips. Remember, it’s not what are the odds for red per coin flip, in 10 coin flips. The question is what are the odds you will flip red, on 10 coin flips.

The odds tend towards 50/50 on thoes ten coin flips.

We’re not talking about the odds per flip. We’re talking about the odds for one color on ten flips.

Now turn it into 100 flip. Or 1000 flips.

You do not have the same odds, on 1000 flips, for them all to be red. You have the same odds on each flip that it will come up red. But if you had to pick, prior to flipping, you would be roughly retarded, to say they will all be red.

The odds of that happening are much smaller than winning the lotto.

It’s that same thing with flipping a coin 10 times. It’s unlikely it will be red every time.

[–]Dragonerne 3 insightful - 1 fun3 insightful - 0 fun4 insightful - 1 fun -  (17 children)

I will make it simple for you
Red, red
Red, black
Black, red
Black, black

These 4 are all equally likely to happen. It doesn't matter if you pick red, red or black, red. It will be just as likely as any other.

Conditional statistics can be difficult to understand but these are independent coin flips, not conditional. Every single output is equally likely, you cannot improve your chances by not voting red.

It is just as stupid to vote red every time as it is to vote red or black randomly. This might seem surprising to you but it is factual.

[–][deleted] 3 insightful - 1 fun3 insightful - 0 fun4 insightful - 1 fun -  (16 children)

You really don’t get it.

[–]TitsAndWhiskey 2 insightful - 1 fun2 insightful - 0 fun3 insightful - 1 fun -  (15 children)

Seems like maybe you’re the one not getting it.

[–][deleted] 2 insightful - 1 fun2 insightful - 0 fun3 insightful - 1 fun -  (14 children)

They are both correct, but in different ways.

If you place a new bet after every roll without considering previous outcomes, each roll will obviously have an equal 50/50 chance. This continues to remain equal infinitely no matter how many reds or blacks are seen, there will always be a 50/50 chance per bet.

But, the odds of seeing 10 reds in succession are 1 in (2.06) to the power of 10 = 1 in 1376. So it becomes more probable that the next roll will be a black as the odds of seeing another red increases. If you ask what is more probable, seeing 6 reds, or 3 reds and 3 blacks, the latter would be a more probable outcome.

[–]TitsAndWhiskey 2 insightful - 1 fun2 insightful - 0 fun3 insightful - 1 fun -  (2 children)

Yeah, that’s the meaningless part. It’s hokum. Spurious. Not real.

Every roll is a 50/50, regardless of prior outcomes.

The statistics you’re citing are better understood as your personal chances of losing X times in a row, regardless of color.

Those odds grow exponentially in a 50/50 game just as you’ve cited.

Let’s say you hit a black streak. It’s landed on black 3x in a row. You’ve been betting on red. You keep betting on red, but on the next roll it hits black again. Sure of your strategy, you bet on red again. It hits on red. Success! The strategy worked!

But not really. That win was the same 50/50 chance as your other losses.

If you had bet black, red, black, red, black, you would have won $15, lost $15, won $30, lost $15, then lost $30. For a total loss of $15.

In the first example, black-black-black-black-red, you’d be down $15, down $30, down $60, down $120, then up $240, for a total gain of $15.

Whether you run the color strategy or not, your total gain/loss in a 50/50 game will always be plus or minus your initial bet.

It is extremely unlikely for you to win or lose more than that even after 5 turns. On a long enough timeline, that unlikelihood becomes a statistical impossibility.

Try it yourself with a coin and a piece of paper. Flip it 10x for one strategy where you’re following the streaks, 10x doing the OPPOSITE of that, and 10x just going back and forth.

They’ll all be approximately the same. Do it 100x and I guarantee they will all work out to +/- $15

[–][deleted] 2 insightful - 1 fun2 insightful - 0 fun3 insightful - 1 fun -  (1 child)

Im not in disagreement. Like I said, there is a 50/50 chance with every turn taken.

The calculation of multiplying odds as repeat reds/black outcomes is only relevant when attempting to use the strategy of foreseeing an inevitable streak of losses by sticking to one's guns and doubling the stakes which was the main topic of the discussion. I'm not saying it will work.

[–]TitsAndWhiskey 1 insightful - 1 fun1 insightful - 0 fun2 insightful - 1 fun -  (0 children)

I’m just saying the idea of odds around color streaks is meaningless. This is purely a betting strategy dressed up in some mumbo-jumbo statistical bullshit.

I just tested both ways on a free online roulette site with the expected results. I might have to download a roulette app now. Need a new time waster lol

[–]Dragonerne 2 insightful - 1 fun2 insightful - 0 fun3 insightful - 1 fun -  (10 children)

No, you have it wrong.

Seeing 3 reds and then 3 blacks is not more likely than seeing 6 reds in a row.

No matter the order you choose, the likelihood will be the same. You know this because of the phrasing, 3 reds and 3 blacks, not in a row, but as an unordered set. But nowhere did I say as a set. I always did an order to the set.

This is important because this is what he is doing with his strategy. He is choosing an ordered set. A fixed "red, black, red, red, black, black" or whatever. Not a random 3 blacks and 3 reds in any order.

The specific one you choose will always be an ordered one and will always have the same probability of happening as 6 reds in a row.

[–][deleted] 4 insightful - 1 fun4 insightful - 0 fun5 insightful - 1 fun -  (9 children)

Probability of consecutive spins:

1 1 in 2.06

2 1 in 4.24

3 1 in 8.74

4 1 in 18.00

5 1 in 37.09

6 1 in 76.41

7 1 in 157.42

8 1 in 324.29

9 1 in 668.04

10 1 in 1376.17

The likelihood of getting 6 in a row is 1 in 76, so if you pull 6 colours out of a bag 76 times, probability rules you are likely to pull 6 reds about one time.

[–]Dragonerne 1 insightful - 1 fun1 insightful - 0 fun2 insightful - 1 fun -  (8 children)

Red, red, red
has the same probability as
red, red, black

You are mistaken

[–][deleted] 1 insightful - 1 fun1 insightful - 0 fun2 insightful - 1 fun -  (7 children)

If that were true, the odds of every football match at a bookies would be 1/1. But we all know that isn't true, odds are calculated accumulatively. There is a difference between turn based chance and stacked odds.

[–]Dragonerne 1 insightful - 1 fun1 insightful - 0 fun2 insightful - 1 fun -  (6 children)

Throw a coin 3 times.
See if it is more likely to get red, then red, then black or if it more likely to get red, then red, then red.

They are equally likely. You can test it yourself.