Chit Chat tried their hardest, but could'nt help, so they sent me here....

only one peice of advice, forget all about martingale

15 losses in a row = (2/3)*15 = 0.0023 (4 dp), which is roughly 0.23%

not having 15 losses in a row 1 - 0.23 = 97.7%

which is equivalent to decimal odds of 1.024 (3dp)

YOMOMMA,

That's the easy bit. He wants to know the probability of having 500 bets without having such a losing run.

The maths for this is extremely complex, I think.

Ya it is, I don't know how to do that.

Here's a discussion of the problem, plus a method of approximation.

http://mathforum.org/library/drmath/view/56637.html

I've not come across bernoulli trials before. Bit busy now but I'll study it and come back with an answer later unless someone beats me to it.

Think this workaround may work:- assume 15 consecutive losers in 500 bets is approximately equal to one losing bet in 33 attempts.

Can then use a discrete binomial distribution [rather than the normal approximation to the binomial distribution] in the usual fashion.

Assume: p = 66.67% ^ 15 = 0.23%; q = 99.77%;
        n = 500/15 = 33.33; k = 1

p is defined as the probability of the event [i.e. losing 15 consecutive bets] occurring.

=> Binomial probability using the above parameters: [assuming factorials for 'real numbers' are calculated as per integers, w/o interpolation between the results for: 2/3*33.0 and 1/3*34.0, etc]

=33.33x0.23%^1x99.77%^32.33

=33.33x0.23%x92.94%

=0.070794

=7.08% [approx.]

This was the first, approximate methodology that sprang to mind - would be interested in seeing other solutions.

PS Martingale will land you in the poor house.

Under stochastic calculus, the only possible way a bettor can win using the staking strategy, is perversely, by starting with an infinite bankroll, to mitigate against the exponentially increasing staking required when the inevitable losing run strikes.

As such, risk is increasing exponentially, with no regard to the extent to which your perceived edge supersedes the available market price and liquidity.

HTH

What's wrong with 24pts profit and 20% ROI? Why the need to think about Martingale? Some people don't want to win. 

I'm not sure how genuine these type of threads are. The last one I posted on General Betting about staking, the OP was nowhere to be seen despite receiving several contributions. Threads are started by Moderators just to get some action.

I meant to put, not having 15 losses in a row 1 - 0.0023 = 0.9977
                                                         = 99.77%

No one advises martingale, it's a really poor system and will eventually wipe you out.

thanks for everybodys input, I certainly am not on wind up as suggested....
Anyway, I have managed to find the calculator I was on about on chit chat.

http://www.sbrforum.com/betting-tools/streak-calculator/


if you put in 500 in the first box, 15 in the second, and 0.6666 (66.6%)in the third box

it gives a probability of 31.313%


as stated in the other thread on chit chat, I was looking to have this figure verified, I also do not know ( i presume not) that this figure takes into account the growing bank which allows the losing streak to grow to a run of 16 before you ger halfway through the 'series'..


I have to go out now, but will be interested in any replies when I get back in later.....


thanks again....

I'll verify a figure for you: 0

I repeat what is wrong with 24pts level stakes profit and 20% ROI?

My example above approximates the result for exactly 15 losses.

The blurb on that streak calculator link, applied to your example, is for 15 "or more".

The portion "or more" is the pivotal difference

i.e. 1 - sum(0 loss streak + 1 loss streak + ... + 14 losses, etc)

not sure I quite follow, in fact this is all getting a bit beyond me!!

are you  saying that by your reckoning, there is only a 7.08% chance of reaching 500 bets without hitting a run of 15 losses

from the chit chat sessions, for someone to analyze....


I agree with dunlayin:

You don't want to start a losing run of 15
The chances that the first bet IS the start of a losing run of 15 is 0.00228 ( 1 - POWER(2/3,15)
You want to repeat this feat of avoiding starting a losing run of 15, for a total of 486 times
POWER(0.00228, 486) comes out at 32.9%.

In other words A is 67.1% likely to blow his bank.

Would like another corrobaration though.
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bongo

bongo
08 Mar 13 21:58
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The above is bollocxz
If you're 8 into a losing run then it makes no difference whether the next game is the start of a 15 game losing sequence, it only matters if the game 8 previously was the start of such a sequence. This makes a big difference to the overall probability, and favours A big time.
As an approximation there will be 162 winners in the first 486 games, so what matters is that the very first trial, and the ones after each winner are NOT the start of a 15 run losing sequence.
This is POWER ((1-0.00228), 163) which comes out at around 69%.

In other words A is probably going to make it.

How does dunlayin reconcile such comments with that the fact that each Bernoulli trial is independent??

The probability of each [independent] trial could not be constant otherwise...

I dont know, its all getting beyond me!
Bongo, on the chit chat thread seems to think there is only a 1 in 3 chance of the bank busting during 500 bets, which if true, is a green light to me, but he freely admits you lot may be more qualified to come up with an answer...

Bongos last input to the question...

Hi Ampleforth Roman Catholic College, good thread this - any tips on who wins the big one in Rome next week?

A fuller formula has your % coming it at: =POWER((1-POWER(2/3,15)),163) which is 68.89%.
This would make you a 31.11% chance to empty your wad on or before the 500th trial.

In truth, you would only EXPECT to have 163 successes in 486 trials, 162 successes is slightly more likely than 164, 161 successes more likely than 165 etc, so the above result is only an approximation. This is chit-chat. A proper mathematician on general betting would take account of this, and it would move the odds against you by a few ticks after the decimal point, but I'm not a proper mathematician, don't pay the permutation charge as I lose on specials and most gambling, and like words rhyming with bolloszcz, so am posting here.

Dunlaying said that if we "treat" the trials as independent then we can use a Poisson approximation.Which is the way to do it if there is no calculating device handy.
Your bullshit may baffle others it most certainly will not baffle me.

Ampleworth,

Just carried out a simulation, and the probability comes out as 31.32%, tallying very well with the result from your web calculator (which, I assume, is probably using simulation too).

thats pleasing to know, nice t know there is some experiments to back up the theories

Joined: 12 Feb 11 | Topic/replies: 1,166 | Blogger: dunlaying's blog
Dunlaying said that if we "treat" the trials as independent then we can use a Poisson approximation.Which is the way to do it if there is no calculating device handy.
Your bullshit may baffle others it most certainly will not baffle me.


Whoa there, tiger!

I was merely copying the comment with your username in it - no need to get personal.

Your soln on the chit-chat forum uses a [discrete] poisson distribution function, with the average/"mew" = np; parameter k as the total number of instances required, etc.

It's taken from any decent first year university probability theory course.

Or am I still talking bullshit?...

so this all boils down to a return on investment of 130 percent then right.

given his stated objective, to double his bank every 500 bets, then take the profit and start again.

his average chances are 2/1 off going bust, he will double 2 banks and bust one, on law of averages, ive seen worse EVs.

of course, that presumes I can keep up my 1 win in 3 record. 

One question I do have, is that do the math theories ans simulations take into account the fact that as you get further into the 'series' the odds of hitting that losing run of 15 is constantly diminishing?

For example, the calculator suggests the probability of losing in the first 200 bets is 13.451%

so it seems likely that surviving to 200 bets is quite likely,

then the odds of going bust in the last 300 bets comes out at 19.856%

which looks nice....

and it still hasnt been taken into account that towards the last 1/4 of the series, the 'bust run' would move to 16 due to the increased bank....

You only have to bet 1% of your bank at level stakes to double your
bank(before commission)after 500 bets.

167.67 * 3.6 = 600.(100 profit)

BUT the chances of your strike rate continuing at the current rate is 0.

120 is nowhere near enough games to draw any conclusions.

In the Premership,it's not unusual to get 40+ draws in 120 consecutive
games. I think it happened earlier this season.