How often do +15 OOTM streaks happen? [Archive]

se2schul

09-30-2005, 01:08 PM

[ QUOTE ]
I know it's random, but...

[/ QUOTE ]

Not to nitpick, but but misuse of the word random really bugs me. This certainly isn't random.

Results from previous SNGs are very likely to affect your play in later SNGs. If your play were "constant", then it would be random, but I doubt your play will be constant.

An example is if you start a losing by doing nothing wrong (random) and notice that you have just gone 8 OOTM. You start thinking that this is the start of a loosing streak and begin playing weak-tight and not making bubble pushes that you'd ordinarily make. Now it ceases to be random as you are changing your play and increasing your odds of subsequent OOTMs.

If you could play the same regardless of your bankroll, your current streak, the fight you had with your GF, the dickhead at the table that you want to spite-call, etc, then it would be random and you could precisely work it out with the math the others have shown you.

pineapple888

09-30-2005, 01:14 PM

[ QUOTE ]
well, if u are a 40% itm player, then .6^15=.00047, so .047%or 1/2000

but there are other factors like tilt, etc...

[/ QUOTE ]

This doesn't really answer the question anyway.

In fact, there is no precise formula to answer the OPs question. You have to assume a certain ITM and run an enormous simulation (millions of trials), then find how often these streaks occur.

From similar problems I've seen in the past, it will happen much more often than most people's intuition would say.

pineapple888

09-30-2005, 01:19 PM

LOL I've gone 8 in a row OOTM dozens of times. I don't change my play a bit. So please don't generalize.

se2schul

09-30-2005, 01:48 PM

[ QUOTE ]
LOL I've gone 8 in a row OOTM dozens of times. I don't change my play a bit. So please don't generalize.

[/ QUOTE ]

Well, I'm referring to most people, not all people. I'm sure that there are even lots of people who believe their play doesn't change during an OOTM streak, when in fact it does (I'm not implying this is you).

Anyway, my post was directed at the OP. Someone who asks a question like he did is much more likely to change his style of play than a veteran SNG player during an OOTM thus prolonging the streak. Besides, the point of my post wasn't about people changing their play, it was about
the misuse of the word random which bugs me more than nails on a chalkboard.

Nicholasp27

09-30-2005, 01:57 PM

[ QUOTE ]
[ QUOTE ]
well, if u are a 40% itm player, then .6^15=.00047, so .047%or 1/2000

but there are other factors like tilt, etc...

[/ QUOTE ]

This doesn't really answer the question anyway.

In fact, there is no precise formula to answer the OPs question. You have to assume a certain ITM and run an enormous simulation (millions of trials), then find how often these streaks occur.

From similar problems I've seen in the past, it will happen much more often than most people's intuition would say.

[/ QUOTE ]

no, there is a precise formula, which is P(X)^n

now, u can argue that tilt, etc may change this value...but if u truly have 60% probability of finishing ootm in any given game, then .6^15 is correct

in practice, u are prolly more than 60% likely to finish ootm in the tourney following losing 13 in a row...but theoretically, it's .6^15

Slim Pickens

09-30-2005, 01:58 PM

[ QUOTE ]
[ QUOTE ]
well, if u are a 40% itm player, then .6^15=.00047, so .047%or 1/2000

but there are other factors like tilt, etc...

[/ QUOTE ]

This doesn't really answer the question anyway.

[/ QUOTE ]

Really? I thought it did (in a purely statistical sense anyway), except for being (1-ITM)^(N-1) because the streak is 14 OOTM after any given OOTM.

Freudian

09-30-2005, 02:00 PM

I never had one in 4k+ SnGs. My worst is 13. Of course that 13 could have easily slipped to 15.

SCfuji

09-30-2005, 02:03 PM

clayton

i can hit about 10 ootm everyday i play.

pineapple888

09-30-2005, 02:05 PM

Dudes, carefully read the OP's question.

Of course there is an easy formula for the chance at any given moment of losing the next 15 in a row.

That's not what he's asking.

Nicholasp27

09-30-2005, 02:09 PM

yes it is...the chances of it happening at any given moment tell you the odds of it occuring, ie 800-1, 2000-1...

therefore, in the long run, u can expect one every 800 tourneys or every 2000 tourneys...

just like u can expect to roll a 3 from a single die every 6 rolls...

TruFloridaGator

09-30-2005, 02:14 PM

Good post...

I haven't hit 15, but some of my longer ones have been affected by a bit of tilt and other outside influences, so it's certainly true for some players.

Slim Pickens

09-30-2005, 02:14 PM

[ QUOTE ]
Dudes, carefully read the OP's question.

Of course there is an easy formula for the chance at any given moment of losing the next 15 in a row.

That's not what he's asking.

[/ QUOTE ]

As long as the OP is asking about any decent sample size, much larger than 15, you take your 1/1000 or whatever you get from a given ITM and X^N is your answer. I know that. Don't nit me, son.

Nicholasp27

09-30-2005, 02:18 PM

if your play is unaffected by previous tourney results and u are a 38% itm'er, then you should expect 'one 15+ ootm streak per 1000 sng's'

if that number is 40%, then it's every 2000 sngs

polley

09-30-2005, 02:27 PM

[ QUOTE ]

no, there is a precise formula, which is P(X)^n

now, u can argue that tilt, etc may change this value...but if u truly have 60% probability of finishing ootm in any given game, then .6^15 is correct

in practice, u are prolly more than 60% likely to finish ootm in the tourney following losing 13 in a row...but theoretically, it's .6^15

[/ QUOTE ]

I don't post often but I'm a big lurker. Anyway, the statement that you theoretically have .6^14 chance to finish out of the money after 13 straight losses is wrong. The outcome of one trial shouldn't be affected by the outcome of the previous one. You still have a 60 percent chance of finishing OOTM.

--Coconut Creek

pineapple888

09-30-2005, 02:29 PM

OK, OK, you're right. I got myself confused thinking this was harder than it is.

My apologies.

Nicholasp27

09-30-2005, 02:33 PM

[ QUOTE ]
[ QUOTE ]

no, there is a precise formula, which is P(X)^n

now, u can argue that tilt, etc may change this value...but if u truly have 60% probability of finishing ootm in any given game, then .6^15 is correct

in practice, u are prolly more than 60% likely to finish ootm in the tourney following losing 13 in a row...but theoretically, it's .6^15

[/ QUOTE ]

I don't post often but I'm a big lurker. Anyway, the statement that you theoretically have .6^14 chance to finish out of the money after 13 straight losses is wrong. The outcome of one trial shouldn't be affected by the outcome of the previous one. You still have a 60 percent chance of finishing OOTM.

--Coconut Creek

[/ QUOTE ]

no, the .6^15 is the probability of finishing 15 in a row ootm

u still have a 60% chance of finishing ootm after 13 straight losses...i never said u didn't...of course the outcome of one trial isn't affected by previous outcomes...

altho i did say that, in real world applications, ur prolly not exactly 60%, as you may have tilt factor there...this isn't rolling a fair die or spinning a ball on a fair roulette wheel or dealing a card from a fair deck...

fnord_too

09-30-2005, 02:40 PM

meh, I typed a big reply and you deleted your post so I got an error trying to post it.

Here's a fun thing to bet on that is related to the OP's question (but more in line with what I think you were thinking in your replies):

I will bet at even money that you cannot roll double sixes on a fair pair of dice given N rolls. For what values of N am I getting the best of it on this bet?

pineapple888

09-30-2005, 02:48 PM

[ QUOTE ]
meh, I typed a big reply and you deleted your post so I got an error trying to post it.

Here's a fun thing to bet on that is related to the OP's question (but more in line with what I think you were thinking in your replies):

I will bet at even money that you cannot roll double sixes on a fair pair of dice given N rolls. For what values of N am I getting the best of it on this bet?

[/ QUOTE ]

Exactly. This is a much more complicated problem.

I told people to carefully read the OP's question when I hadn't done so myself. Classic donk move. At least I didn't have any chips on the line. /images/graemlins/grin.gif

fnord_too

09-30-2005, 02:54 PM

[ QUOTE ]
[ QUOTE ]
meh, I typed a big reply and you deleted your post so I got an error trying to post it.

Here's a fun thing to bet on that is related to the OP's question (but more in line with what I think you were thinking in your replies):

I will bet at even money that you cannot roll double sixes on a fair pair of dice given N rolls. For what values of N am I getting the best of it on this bet?

[/ QUOTE ]

Exactly. This is a much more complicated problem.

I told people to carefully read the OP's question when I hadn't done so myself. Classic donk move. At least I didn't have any chips on the line. /images/graemlins/grin.gif

[/ QUOTE ]

It's actually pretty easy to solve in closed form (no need for simulation) but the answer is very non intuitive, and is a great way to employ Barnum's directive.

mmbt0ne

09-30-2005, 02:56 PM

<blockquote>En réponse à:<hr />
<blockquote>En réponse à:<hr />
I know it's random, but...

[/ QUOTE ]

Not to nitpick, but but misuse of the word random really bugs me. This certainly isn't random.

Results from previous SNGs are very likely to affect your play in later SNGs. If your play were "constant", then it would be random, but I doubt your play will be constant.

An example is if you start a losing by doing nothing wrong (random) and notice that you have just gone 8 OOTM. You start thinking that this is the start of a loosing streak and begin playing weak-tight and not making bubble pushes that you'd ordinarily make. Now it ceases to be random as you are changing your play and increasing your odds of subsequent OOTMs.

If you could play the same regardless of your bankroll, your current streak, the fight you had with your GF, the dickhead at the table that you want to spite-call, etc, then it would be random and you could precisely work it out with the math the others have shown you.

[/ QUOTE ]

I really like this post. This is why I hate it when people start to use statistics for long term poker outcomes. Almost all of the time, you are not going to get accurate results.

Poker isn't a random game. There is nothing in the Normal distribution that people keep using that can account for the many subtle aspects of poker. Stop trying to measure your winrates, and confidence intervals so closely.

If you focused your energy on poker theory and tilt control you would be much much better off than if you found out that your real ROI whould be 16.8473% (+- 3sigma of course) so that when you dip to 9 you start getting pissed at your play.

The long run is longer than most of us will ever experience. Ride the waves of variance, and play your best. Let luck sort out the rest.

Nicholasp27

09-30-2005, 03:07 PM

[ QUOTE ]
The long run is longer than most of us will ever experience.

[/ QUOTE ]

i'd bet money that NONE of us will ever experience the long run

mmbt0ne

09-30-2005, 03:10 PM

<blockquote>En réponse à:<hr />
<blockquote>En réponse à:<hr />
The long run is longer than most of us will ever experience.

[/ QUOTE ]

i'd bet money that NONE of us will ever experience the long run

[/ QUOTE ]

That's a bad bet, because the only way someone ends up paying will be you paying if they do. Do you see why? /images/graemlins/tongue.gif

09-30-2005, 03:24 PM

If you have a large enough sample size to get a very accurate ITM percentage, then the inconsistency in the game will not affect your chances in the least because it's already incorporated into your ITM percentage. Say you played plastered one day a week, that still doesn't change anything because it's already in the negative side of your ITM ratio. Basically, in the long run, the inconsistency in your game will still even out because of the consistency in your inconsistency =P.

Edit: Never mind, that can't be right. If there's higher variance, then the chance of you having a streak will be that much higher. Granted that 15 OOTM is extremely tough to do, higher variance will give higher chances.

Nicholasp27

09-30-2005, 04:04 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The long run is longer than most of us will ever experience.

[/ QUOTE ]

i'd bet money that NONE of us will ever experience the long run

[/ QUOTE ]

That's a bad bet, because the only way someone ends up paying will be you paying if they do. Do you see why? /images/graemlins/tongue.gif

[/ QUOTE ]

It's impossible for me to lose this bet.
Do you see why?

pineapple888

09-30-2005, 04:57 PM

[ QUOTE ]

It's actually pretty easy to solve in closed form (no need for simulation) but the answer is very non intuitive, and is a great way to employ Barnum's directive.

[/ QUOTE ]

At N=25, you become a dog, if you are betting against the roller hitting 66. (35/36)^25 is just less than 50%. Try calculating that value in a bar to win a bet.

But wouldn't some people guess 18 and some guess 36? I'd want to ask the person (or let a group have a debate) and choose which side to bet based on their answer.

mmbt0ne

09-30-2005, 05:01 PM

<blockquote>En réponse à:<hr />
<blockquote>En réponse à:<hr />
<blockquote>En réponse à:<hr />
<blockquote>En réponse à:<hr />
The long run is longer than most of us will ever experience.

[/ QUOTE ]

i'd bet money that NONE of us will ever experience the long run

[/ QUOTE ]

That's a bad bet, because the only way someone ends up paying will be you paying if they do. Do you see why? /images/graemlins/tongue.gif

[/ QUOTE ]

It's impossible for me to win this bet.
Do you see why?

[/ QUOTE ]

FYP

Wait, would it maybe be possible to both never lose and never win a bet? If this is the case, I might need to become the escrow for some such wager!

pineapple888

09-30-2005, 05:04 PM

A better prop would be "I bet I can hit a 7 in five rolls. It usually takes six rolls, but I feel lucky today. I can do it in five." You're a 60-40 favorite, and it's over fast, and you might be able to repeat it a bunch of times.

MelchyBeau

09-30-2005, 05:07 PM

many of you are ignoring Standard Deviation. You can't think of this as a dice roll. Think about ring games, people go through long breakeven or losing streaks when they are winners.

Melch

pergesu

09-30-2005, 05:18 PM

I don't see why there's any arguments here. The correct answer has been posted a bunch of times, by Slim, citanul, etc.

This is basic math guys.

[ QUOTE ]
don't nit me, son

[/ QUOTE ]
fcukin pro

Nicholasp27

09-30-2005, 07:36 PM

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
The long run is longer than most of us will ever experience.

[/ QUOTE ]

i'd bet money that NONE of us will ever experience the long run

[/ QUOTE ]

That's a bad bet, because the only way someone ends up paying will be you paying if they do. Do you see why? /images/graemlins/tongue.gif

[/ QUOTE ]

It's impossible for me to win this bet.
Do you see why?

[/ QUOTE ]

FYP

Wait, would it maybe be possible to both never lose and never win a bet? If this is the case, I might need to become the escrow for some such wager!

[/ QUOTE ]

it's impossible for me to lose because u can never reach the 'long run'...it's infinity

Nicholasp27

09-30-2005, 07:37 PM

std deviation has nothing to do with this question

u can use std deviation to say that u are 95% confident that u'll hit a -15 streak in the next X tourneys...but that wasn't the question

09-30-2005, 10:59 PM

Most of the answers here don't really answer the question. That simplistic figure double counts your OOTM, so say if you finish, OOTM 16 times in a row, it would count as 2 samples of finish OOTM 15 times in a row. They only answer the simple question of for any given moment what is the chance of going OOTM in the next 15 tournaments. So anyway, firstly you must define your OOTM.

If your OOTM is 60%. Then the % chance of finishing OOTM 15 in a row in a set of N games is thus given by 100*(1-(1-0.6^15)^(N-15)) where N > 15.
So over 1000 games you have a 37% chance of going OOTM 15 times in a row. Over 10000 games you have a 99% chance of going OOTM 15 times in a row.

Nicholasp27

10-01-2005, 01:37 AM

u are doing the odds that someone will get a 15 ootm streak in the next X tourneys

10-01-2005, 01:47 AM

[ QUOTE ]

In fact, there is no precise formula to answer the OPs question. You have to assume a certain ITM and run an enormous simulation (millions of trials), then find how often these streaks occur.

[/ QUOTE ]

This is very dumb.

Here's the formula, assuming 40% ITM:

OOTM = 60%
OOTM all of one specific set of 15 = .6^15*100 = ~.047%
Not OOTM in that set = ~99.953%

Now, in a set of N games where N > 15, the number of sets of 15 is N - 14. Thus, the odds of having at least one streak of 15 OOTM in a row in a set of N games is just:

(1 - .9953^(N-14))*100.

So, for 1000 SnGs, the odds of one streak is 37%. For 2k, it's 99.99%.

10-01-2005, 02:17 AM

Hehe Noah, we're both almost right but we both made some mistakes in our formulas.

(1-.99953^(N-14))*100 should be right for ITM of 0.4

And 99.99% chance of 15 OOTM would be more like 20k.

But yeah, learnt this neat formula off you last time, so thanks.

Slim Pickens

10-01-2005, 02:01 PM

Don't mind me. I was a little jumpy yesterday morning anyway. If I remember correctly, a similar thread came up about two months ago where we figured out that nobody really cares enough to get the exact answer.

10-03-2005, 03:35 AM

The expected number of streaks of length S or more in N tourneys is

[(N-S)*ITM+1]*(1-ITM)^S

For S=15 and ITM=30% this is about 1 per 702.
For S=15 and ITM=40% this is about 1 per 5317.

Even for a bot, where the play is consistent, the quality of the opposition faced is autocorrelated. So streaks would be more frequent.

Slim Pickens

10-03-2005, 12:14 PM

OOTM in one tournament: 1-ITM
OOTM in S consecutive: (1-ITM)^S
not OOTM in S consecutive: 1-(1-ITM)^S
number of S-long trials in N tournaments: N-S
probability that NONE of those trials is a S-long OOTM streak: [1-(1-ITM)^S]^(N-S)
probability that NOT NONE of them are S-long OOTM streaks: 1-[1-(1-ITM)^S]^(N-S)

62% for an ITM of 0.37 over 1000 tournaments...

So the answer to the OP's question is the value of N that gives 0.5 using his ITM and S=15. This is also a lower bound because SNG results can autocorrelate (so you just happen to systematically log on at certain times such that your ITM can be split into two distinct values... it averages to your real ITM, but those "harder" times you play can make the streaks come a little more often) but that effect should be small.

10-03-2005, 05:22 PM

I take the original question to be "how often do 15+ length OOTM streaks happen?" The average number of such streaks per tournament is as I calculated above. The average length of "time" (tournaments) from streak to streak is the reciprocal (e.g., 702 above).

You are answering a different question.

[ QUOTE ]

probability that NONE of those trials is a S-long OOTM streak: [1-(1-ITM)^S]^(N-S)

[/ QUOTE ]

This step is wrong. The "trials" are not independent.

Slim Pickens

10-03-2005, 05:34 PM

[ QUOTE ]

[ QUOTE ]

probability that NONE of those trials is a S-long OOTM streak: [1-(1-ITM)^S]^(N-S)

[/ QUOTE ]

This step is wrong. The "trials" are not independent.

[/ QUOTE ]

You are correct, but could you show why whatever it was you did accounts for this?

[(N-S)*ITM+1]*(1-ITM)^S

We have to be answering the same question. There aren't two different questions.

10-03-2005, 11:56 PM

Think about how many streaks of length S or more occur in N independent tournaments.

For each OOTM streak of length S or more, we will count it multiple times if we just say (N-S+1)*(1-ITM)^S. To avoid overcounting, we note that each streak has its own beginning (and ending) tournament.

There is a sequence of ITM/OOTM results, call it ABCDEF....

Consider the possibility that B is the beginning of a streak of S+ OOTM. For this to be true, A must be ITM, B..P must be OOTM, and we don't care what happens with Q onward. So B is the beginning of a S+ streak with probability ITM*(1-ITM)^S. The same is true for C, D, etc.

Result A is different, since it does not have a preceding tourney. For A, the probability it begins a S+ length streak is just (1-ITM)^S.

There are N-S tournaments like B, and 1 like A. Hence [(N-S)*ITM+1]*(1-ITM)^S. A, B, C, etc. being the beginning of streaks are not independent. But when we care about the mean number of streaks, we can still add their probabilities; they don't need to be independent.

The other problem, which you address asks whether, in a sequence of N tournaments, there was no streak of S+ OOTM, and what is the probability of that. This problem is called the consecutive S-out-of-N:F system reliability problem. The solution (for the case where each tournament "failure" has identical probability 1-ITM) is

Q(N;S) = Q(N-1;S) + [1-Q(N-S-1;S)]*ITM*(1-ITM)^S

A web search turned up a calculator for this function at k/n:F system calculator (http://iew3.technion.ac.il/sqconline/reliability.html)