GAMBLERS RUIN

[teddy]

GIVEN A 90% HIT RATE OVER 1000 ROLLS HOW MANY TIMES WILL I LOSE CONSECUTIVE BETS. LONGEST LOSING STREAK? 15 IS MY GUESS.

[Dave Schwartz]

Gambler's ruin addresses the chances of losing a particular stake (of units) given a hit rate and payoff.

It is not the tool for determining consecutive losers.

[teddy]

SOMEONE ON HERE ADDRESSED THE chance of consecutive losses even with a high hit rate of 90%. I cant remember his name but he posted his system picks alot and had a website. From what I remember its far higher than one would expect.

[TrifectaMike]

GIVEN A 90% HIT RATE OVER 1000 ROLLS HOW MANY TIMES WILL I LOSE CONSECUTIVE BETS. LONGEST LOSING STREAK? 15 IS MY GUESS.

You can calculate a estimation of consecutive losses over N samples from the following:

Consecutive Losses = Ln(N)/-Ln(1-P)

where
N = number runs in a sequence (1000)
P = probability of a win (.9)
Ln = the natural log (base e)

Using your example:

Consecutive Losses = Ln(1000)/-Ln(1-.9) = 3

Mike (Dr Beav)

[Stillriledup]

GIVEN A 90% HIT RATE OVER 1000 ROLLS HOW MANY TIMES WILL I LOSE CONSECUTIVE BETS. LONGEST LOSING STREAK? 15 IS MY GUESS.

In one series of 1,000 rolls, its unlikely you'll lose 15 in a row.

[teddy]

I have to think thats pretty wrong...has to be at least 6. Its just 27 to one to lose 3 in a row isnt it.

[TrifectaMike]

I have to think thats pretty wrong...has to be at least 6. Its just 27 to one to lose 3 in a row isnt it.

No, 3 is correct.

The probability of losing 3 in a row independent of sequence is .001

If you increase the sequence from 1000 to 2000 the number of consecutive losses will increase. If you decrease the win % and keep the 1000 sample sequence, the number of consecutive losses will increase.

However, for a sequence of a 1000 runs and a win% of 90, consecutive losses of 3 is what is expected.

Mike (Dr Beave)

[Cholly]

SBR Forum has a streak calculator at
http://www.sbrforum.com/betting-tools/streak-calculator/

Its result for a 1,000 sample run (w/ a 90% hit-rate) gives a 60% chance for losing 3 in a row, and a 10% chance of losing 4 in a row.

[TrifectaMike]

SBR Forum has a streak calculator at
http://www.sbrforum.com/betting-tools/streak-calculator/

Its result for a 1,000 sample run (w/ a 90% hit-rate) gives a 60% chance for losing 3 in a row, and a 10% chance of losing 4 in a row.

The above as derived from the calculator are correct. However, the expected consecutive losses are 3.

The probability of the 3 consecutive losses ocurring over the sequence (1000) is .60. That means even though 3 is the longest losing streak, the probability of the 3 losses being observed over the sequence is .60 or 60%....40% of the time 3 consecutive losses will not be observed.

As I said 3 is the expected consecutive losses.

Mike (Dr. Beav)

[Cholly]

The above as derived from the calculator are correct. However, the expected consecutive losses are 3.

The probability of the 3 consecutive losses ocurring over the sequence (1000) is .60. That means even though 3 is the longest losing streak, the probability of the 3 losses being observed over the sequence is .60 or 60%....40% of the time 3 consecutive losses will not be observed.

As I said 3 is the expected consecutive losses.

Mike (Dr. Beav)

Don’t mean to quibble, but the term “expect” is somewhat imprecise. Does it mean you expect with 100% certainty, or do you merely mean that it is the most likely result. And “3” is not the longest losing streak; there is a 10% chance in that 1000-sample sequence that one will experience 4 losses in a row.

While I’m here, I’d like send out a thank-you to the person on a previous thread who recommended the book “Black Swan” by Nassim Taleb…fascinating read! Teddy, and anybody else who may be proceeding counting on a 90% hit rate of any event, should read that book if they’re considering investing serious money. He gives the example of a turkey who had a 1,000 "samples" of empirical evidence indicating that the advance of a bi-ped was a good thing because it indicated food was coming. But despite this mass of evidence, the advance of a bi-ped on November 23rd was disastrous, to say the least.

[098poi]

He must not have been aware of the third Thursday in November angle.

[teddy]

Ok, my model suggests it returns .98 on the dollar for the last 1000 races. My avg rebate is 8% if not betting twinspires tracks. While on some of my show bets there will be no rebate because of neg pool, still many will pay off. So if the chances of losing 5 in a row over the next 1000 bets is probably under 1.0%. I am safe to assume if my bank has 2k in it. And I bet 100 to show on each race, we will be safe to assume that in the end I should be between 4 to 6% return. Should produce $12 to $18 a day on the show bets and at least the same on place and win bets. Still this is very hard to do because of low play.. only 3 bets a day on avg.

[098poi]

He must not have been aware of the third Thursday in November angle.
Fourth Thursday!!

[Dave Schwartz]

But despite this mass of evidence, the advance of a bi-ped on November 23rd was disastrous, to say the least.

Amazing! They spend an entire year lulling them into a sense of false security.

:lol:

[TrifectaMike]

Don’t mean to quibble, but the term “expect” is somewhat imprecise. Does it mean you expect with 100% certainty, or do you merely mean that it is the most likely result. And “3” is not the longest losing streak; there is a 10% chance in that 1000-sample sequence that one will experience 4 losses in a row.

Expectation is not imprecise.

Let me try another way. How often will one experience a sequence of 3 losses given a win% of 90?

Run = 1/(1-p)^3 = 1/.1^3 which equals 1000 races.

That is one can expect a sequence of 3 losses for every 1000 races.

Here's another way...Hope I don't lose anyone

The probability of three losses is .1*.1*.1 = .001

Consecutive losses = log(,001) =3 where log is to base .1

It is as precise as the the win % given.

Mike (Dr Beav)

[teddy]

ok so what area the chances of losing 5 bets not in a row but within 30 plays when the total number of plays is 1000. Exaple win win win loss loss win loss win win loss loss win win win win loss.. ect.....90% win rate still.

[TrifectaMike]

ok so what area the chances of losing 5 bets not in a row but within 30 plays when the total number of plays is 1000. Exaple win win win loss loss win loss win win loss loss win win win win loss.. ect.....90% win rate still.

Teddy, I've copied a post from another thread, which explains and answers your question.

Mike (Dr, Beave)


The majority of the posts relate to Binomial experiments and the Binomial distribution. So, let's first define them.

A Binomial distribution is obtained from a probability experiment called a binomial experiment. The experiment must satisfy these conditions:

1. Each trial can have only two outcomes or outcomes that can reduced to two outcomes. The outcomes are usually considered as a success or failure (win loss).

2. There is a fixed number of trials.

3. The outcomes of each trial are independent of each other.

4. The probability of success must remain the same for each trial.

You will notice that order of the success or failure is not included in the above conditions.

Let's perform a binomial experiment.

If we play three races and want to know the probability of winning two races and losing one race and we have a hit rate of 40%. This can be determined as:

P(winning 2 races and losing 1 race) = P(winning 2 races)*P(losing 1 race)

P(winning 2 races and losing 1 race) = (.4)(.4)(.6) = 9.6%

The probability is only correct if the win/loss is order specific (win-win-loss).

Now, if order is not specific, we would also have win-loss-win and loss-win-win. In this case the probability is 9.6% + 9.6% + 9.6% or 28.8%.

This leads to the binomial theorem: P(x=k) = n!/k!(n-k)! *p^k(1-p)^(n-k)

where
K= successes
n= number of trials
p= probability of success
!= factorial

The main point is that the binomial theorem is independent of order.

Mike (Dr Beav)

[teddy]

This leads to the binomial theorem: P(x=k) = n!/k!(n-k)! *p^k(1-p)^(n-k)

can someone run this for
90% probability of winning
1000 trials.


where
K= successes
n= number of trials
p= probability of success
!= factorial


Not sure what some of this is like factorial.

[teddy]

You cant really know your min bankroll till you figure out what can go wrong like 10 losses out of 30... bad runs

[TrifectaMike]

You cant really know your min bankroll till you figure out what can go wrong like 10 losses out of 30... bad runs

These are exact Binomial calculations (no estimates)

P: exactly 25 wining bets out of 30 bets = .102 (10.2%)
P: 25 or fewer winning bets out of 30 bets = .175 (17.5%)
P: 25 or more winning bets out of 30 bets = .927 (92.7%)

Mean = 27 winning bets
Standard deviation = 1.64 winning bets

Teddy just be certain of your 90% win rate.

Mike (Dr Beav)

[teddy]

See h ow decieveing 90% is. There is a 10.% chance of losing 5 of 30 bets. So you can figure there is a pretty good chance of losing 7 of 30 at some point. I think anything above 1% is a good chance. When gambling.

[TrifectaMike]

See h ow decieveing 90% is. There is a 10.% chance of losing 5 of 30 bets. So you can figure there is a pretty good chance of losing 7 of 30 at some point. I think anything above 1% is a good chance. When gambling.


Yes, there is a high statistical chance of losing 5 of 30 bets. Things are not always as good as they first appear. But this is much different than the initial question of consecutive losing bets in a sequence.

Here are few more probabilities from the same distribution.

P: exactly 23 wining bets out of 30 bets = .018 (1.8%)
P: exactly 22 wining bets out of 30 bets = .006 (0.6%)

Mike(Dr Beav)

[teddy]

90% is as deceiving as the chances of having 3 losses n a row being 1% or 7 of 23 losses in no order being 1.8%. Im pretty dead certain onthe 90% show bets. Still even if you are only betting 100.00 you need to be starting with a bank of 1k to be almost certain to survive till you can be far enough ahead that the chances of busting are infinite. With a 4% return that might be quite a few races if you start with under $500 as a bank.

[highnote]

GIVEN A 90% HIT RATE OVER 1000 ROLLS HOW MANY TIMES WILL I LOSE CONSECUTIVE BETS. LONGEST LOSING STREAK? 15 IS MY GUESS.


In theory, your longest losing streak could be 1000. It's not likely, but it's statistically possible.

The way you phrased your post seems to be confusing to people.

The way you phrased your sentences, you actually asked two questions:

1. How many times will I lose consecutive bets? Consecutive means two in a row.

2. [what is the] LONGEST LOSING STREAK [I can expect]?

Maybe you're asking what is the probability of losing 15 bets in a row? Or how likely is it that I would lose 15 in a row?

Maybe using Standard Deviation would help make this clearer?

How many consecutive loses are to be expected within 1 Standard Deviation? ... within 2 Standard Deviations?... within 3 Standard Deviations?

:confused:

[TrifectaMike]

I
Maybe using Standard Deviation would help make this clearer?

How many consecutive loses are to be expected within 1 Standard Deviation? ... within 2 Standard Deviations?... within 3 Standard Deviations?

:confused:

Can you explain how you would do this? I'm interested.

Mike (Dr Beav)

[Capper Al]

Thanks for the Streak Calculator. It is probably scientifically exact. For the railbirds, who care less about being scientific, squaring your hit ratio is probably close enough to figure your ruin and needed bank roll to support your habit.

Here's how to do it. The average capper hits 1 out of 4. Square 4 and you have an estimate of how much a minimum bank roll should be to support the game, 16 times the bet size. So if this railbird wins 1 out of 4 and plays $2.00 win bets, his bank roll should be at least $32.00.

[TrifectaMike]

Thanks for the Streak Calculator. It is probably scientifically exact. For the railbirds, who care less about being scientific, squaring your hit ratio is probably close enough to figure your ruin and needed bank roll to support your habit.

Here's how to do it. The average capper hits 1 out of 4. Square 4 and you have an estimate of how much a minimum bank roll should be to support the game, 16 times the bet size. So if this railbird wins 1 out of 4 and plays $2.00 win bets, his bank roll should be at least $32.00.

I love shortcuts....especially when they work.

Let's see...a 90% hit rate

Here's how to do it. The average capper hits 1 out of 1.1. Square 1.1 and you have an estimate of how much a minimum bank roll should be to support the game, 1.2 times the bet size. So if this railbird wins 1 out of 1.1 and plays $2.00 win bets, his bank roll should be at least $2.20.

Is that correct? I only need a $2.20 bankroll?

Mike (Dr. Beav)

[thaskalos]

Thanks for the Streak Calculator. It is probably scientifically exact. For the railbirds, who care less about being scientific, squaring your hit ratio is probably close enough to figure your ruin and needed bank roll to support your habit.

Here's how to do it. The average capper hits 1 out of 4. Square 4 and you have an estimate of how much a minimum bank roll should be to support the game, 16 times the bet size. So if this railbird wins 1 out of 4 and plays $2.00 win bets, his bank roll should be at least $32.00.

Is 16 betting units the recommended bankroll for ONE DAY?

If so...then I agree.

If not...then I vehemently disagree. IMO...a win bettor cashing 25% of his tickets needs a bankroll of 100 bets in order to support his habit...assuming he is a winning player!

If he is a losing player...then NO bankroll is enough, no matter how large it is.