How many losses in a row are to be EXPECTED?

[BellagioOwner]

Statistical question. Ignore Zero for now. On a fair and random roulette game of 100 spins, while betting on EC (50% Win chance ignoring zero) how many losses in a row on average are expected a player to have?
Since 1/2*1/2*1/2*1/2*1/2*1/2= 1/64 and (1/2)^7= 1/128 (more than 100) does that mean that it is EXPECTED in these 100 spins to have AT LEAST once 6-7 losses in a row?

What happens also on dozens? How many losses in a row on a Dozen bet are to be expected in 100 spins? 

[Firefox]

Depends what you mean by expected.

Is it expected as in the mean/average or expected as in the chance greater than 0.5.

I look at it as like the chance that a number will occur in some number of spins.

For example, the chance that a number  will not occur is 36/37. The chance that it will not occur in 37 spins is

(36/37)^37 = 0.36

And the chance that it will occur is thus 0.64 Although we know the mean number of occurences in any group of 37 is 1. So it's kind of a paradox that the expected number of occurences is 1 but the chance of it happening is considerably less than 1.

I guess this reflects the fact that in the next group of 37, it may appear twice or even thee or more times, so we need to have the chance it appears exactly once less than one to allow for the repeats on other occasions and still end up with an expected value of  1, if that makes any sense.

Similarly, a run of 6, chance 1/2^6 =0.0156, chance of not happening is 0.9844.

Chance of not happening in 64 attempts is (0.9844)^64 =0.365

Chance of happening = 0.635, although the expected number is still once.

Coincidentally, just by chance very similar to the probability of a number in 37 spins.

This explains why a martingale player can think he has a winning system, getting through a couple of sessions or more without a big loss, but the very next session he bemoans his luck and takes two big hits. When in fact it's exactly what probability predicts!

[BellagioOwner]

Similarly, a run of 6, chance 1/2^6 =0.0156, chance of not happening is 0.9844.

Chance of not happening in 64 attempts is (0.9844)^64 =0.365

Chance of happening = 0.635, although the expected number is still once.

I get it yes, so it's like saying that in 64 spins there is a 63.5% chance of having a 6 in a row losses (or wins by the way too) on EC bets

More specifically though on my question, in 100 spins betting on a single dozen, what is the max number of losses that we will more probably see happening? Is it like 14? 20? Is there a way to calculate this? I know that it won't always be like this and that sometimes in 100 spins I will see longer or shorter max losses streaks than the calculated above.

[Firefox]

Assuming the probability of a loss is 25/37 including zero, then using a similar method to above, I calculate:

Prob of run of 6 = 0.9999
Prob of run of 7 = 0.9987
Prob of run of 8 = 0.9881
Prob of run of 9 = 0.9489
Prob of run of 10= 0.8426
Prob of run of 11 = 0.7405
Prob of run of 12 = 0.6000
Prob of run of 13 = 0.4577
Prob of run of 14 = 0.3369

So you have less than 50% chance of 13 losses in a row but greater than 50% chance of 12 losses in a row.

A run of 12 losses is a 1 in 110 event.

Perhaps someone else would like to check my figures, in case I made a mistake!

[Joe]

More specifically though on my question, in 100 spins betting on a single dozen, what is the max number of losses that we will more probably see happening?

Nothing wrong with Firefox's answer, but you seem to be talking about the mode (the most frequent outcome). I wrote a little simulation of 10,000 sessions each consisting of 100 spins and recorded the longest losing run for each session. Here are the results :

Code Select Expand

Frequency distribution for mlr, obs 1-10000
number of bins = 29, mean = 9.8844, sd = 3.08214

       interval          midpt   frequency    rel.     cum.

           < 4.5000    4.0000        22      0.22%    0.22%
    4.5000 - 5.5000    5.0000       191      1.91%    2.13%
    5.5000 - 6.5000    6.0000       711      7.11%    9.24% **
    6.5000 - 7.5000    7.0000      1215     12.15%   21.39% ****
    7.5000 - 8.5000    8.0000      1574     15.74%   37.13% *****
    8.5000 - 9.5000    9.0000      1558     15.58%   52.71% *****
    9.5000 - 10.500    10.000      1314     13.14%   65.85% ****
    10.500 - 11.500    11.000      1009     10.09%   75.94% ***
    11.500 - 12.500    12.000       727      7.27%   83.21% **
    12.500 - 13.500    13.000       509      5.09%   88.30% *
    13.500 - 14.500    14.000       377      3.77%   92.07% *
    14.500 - 15.500    15.000       236      2.36%   94.43%
    15.500 - 16.500    16.000       189      1.89%   96.32%
    16.500 - 17.500    17.000       126      1.26%   97.58%
    17.500 - 18.500    18.000        86      0.86%   98.44%
    18.500 - 19.500    19.000        35      0.35%   98.79%
    19.500 - 20.500    20.000        47      0.47%   99.26%
    20.500 - 21.500    21.000        29      0.29%   99.55%
    21.500 - 22.500    22.000        14      0.14%   99.69%
    22.500 - 23.500    23.000         9      0.09%   99.78%
    23.500 - 24.500    24.000        11      0.11%   99.89%
    24.500 - 25.500    25.000         2      0.02%   99.91%
    25.500 - 26.500    26.000         1      0.01%   99.92%
    26.500 - 27.500    27.000         3      0.03%   99.95%
    27.500 - 28.500    28.000         0      0.00%   99.95%
    28.500 - 29.500    29.000         0      0.00%   99.95%
    29.500 - 30.500    30.000         2      0.02%   99.97%
    30.500 - 31.500    31.000         2      0.02%   99.99%
          >= 31.500    32.000         1      0.01%  100.00%

From the frequency column you can see that the losing run which occurred the most often was of length 8. Naturally this isn't an exact answer because I'm not using an infinite sample size, but I reckon it's good enough. It's between 8 and 9.

[BellagioOwner]

thank you guys. these all help a lot. I get a clear picture

[Firefox]

It makes sense that 8 or 9 would be the most likely longest as 6 or 7 are almost certain to ocurr either by themselves or part of a longer run.

[zhone]

For a single-dozen betting, at worst it could be 38 losses in a row. Analysis based on 40,000 random numbers generated by random.org

[Gitano]

∞ ?

[Taotie]

Nothing wrong with Firefox's answer, but you seem to be talking about the mode (the most frequent outcome). I wrote a little simulation of 10,000 sessions each consisting of 100 spins and recorded the longest losing run for each session. Here are the results :




From the frequency column you can see that the losing run which occurred the most often was of length 8. Naturally this isn't an exact answer because I'm not using an infinite sample size, but I reckon it's good enough. It's between 8 and 9.

If you are trying to survive the above bell curve entirely, it doesn't matter where you park your progression along the way because the expense combined with the frequency of the very next step of the bell curve will consume all profits gained from the previous steps survived.

Even without the expense of the next step, just the frequency combined with the accumulated loss will eventually consume all profits gained from the previous steps survived.

It's plain to see.

[Firefox]

If you are trying to survive the above bell curve entirely, it doesn't matter where you park your progression along the way because the expense combined with the frequency of the very next step of the bell curve will consume all profits gained from the previous steps survived.

Even without the expense of the next step, just the frequency combined with the accumulated loss will eventually consume all profits gained from the previous steps survived.

It's plain to see.

This is unfortunately true of any system using a negative progression, on any type of chance.

The table limits are set, such that reasonably common events eg 1 in 100/200/,400 etc lead to one's ruin, and return of profits to the house, in a timely fashion. With some nice interest for them after you expose massive progression bets to the house edge.

Better to be the table owner 

[ego]

 
Many times during the years I have seen the same mistake over and over again.
The first lesson for anyone who will start gambling should be to understand that all system and staking plans fail.
If someone can do that conclusion from the beginning they will start to ask questions like how to keep the money I earn without giving it all back to the casino. Then someone develops a sense for risk management.

So many times I have seen punters being ahead using progressions and systems and still, they continue playing the same way without taking some counter measuring action. Just such a simple thing to stay longer in the game and keeping the bankroll intact.
The best weapon in any given situation is a regression. Why not start after 1/3 or half the cost and operate with casino money.
You are not only playing the game with the games money you also need to face the bad sequence twice to bust and give all the money back to the casino. Such a simple action can make wounder.

Cheers

[Joe]

The first lesson for anyone who will start gambling should be to understand that all system and staking plans fail.

But then you go on to recommend a staking plan, namely regression. Don't you see the contradiction? And quitting when ahead doesn't work either. Actually, I disagree with your first statement because I'm open to the possibility that some kind of progression can work long term if you can manage to reduce the length of the losing runs. In that case even though you don't have an edge which gives you a flat bet profit, you will nevertheless make a profit (and without ever reaching house limits). But in this case you still need to focus on bet selection, so either way, bet selection is king.

[Firefox]

The trouble with shallow progressions or insurance lines is that they don't recover all losses, although they are likely to stay below limits, you just end up losing a different way. There's nothing special about progression bets, they are simply larger bets subject to the same percentage negative expectation as smaller bets. Just that the size of the expected loss is greater as it's the same percentage but a bigger bet.

But I agree increased accuracy in bet selection is the way to go. No sensible way to do that with even chances, dozens, lines or corners though.

[nottophammer]

But I agree increased accuracy in bet selection is the way to go. No sensible way to do that with even chances, dozens, lines or corners though.

But staring at the wheel all day is