| imofe |
08-09-2014 02:20 PM |
probable max losses in a row
What would be the correct way to calculate probable max losses in a row at a 45% winning clip over 1000 trials?
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| Robert Fischer |
08-09-2014 05:57 PM |
Quote:
Originally Posted by imofe
What would be the correct way to calculate probable max losses in a row at a 45% winning clip over 1000 trials?
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I'm going to have think more about the subject to really give you any info weighing the pros and cons of drop down formulas and other options, but here is a quick answer:
1000 trials = a maximum of 12 consecutive losses
10000 trials = a maximum of 15 consecutive losses
formula for excel "=ROUND(LN( 1000)/-LN((1- A1)),0)" where A1 is "x" or whatever cell you have the win% and "1000" is the number of trials. |
| imofe |
08-09-2014 06:16 PM |
Thank you Robert. I have an acquaintance at the track who asked me this and I said I would get back to him. He is a big bettor who was recently offered 8% on all his bets. He thinks he can obtain 45% winners at 97 cents on the dollar return. Because he is very aggressive and a rare bird (my dad use to say this) he wants to bet progressive style and wanted to know about losing streaks. I don't know what will happen , but from an entertainment perspective, you can't beat it.
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| Ray2000 |
08-09-2014 08:37 PM |
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| TrifectaMike |
08-09-2014 09:01 PM |
From the past
It is not to difficult to calculate the probability of a losing streak when we start with a loss. Since we start with a loss, it is just the probability of a single loss raised to power of consecutive losses.
So, if q equals the probability of one loss, the probability of k losses simply equals q^k (q raised to the k power).
Example:
q (50% winners) = .5 and k =10 (10 consecutive losses)
q^k = .5^10 is approximately .001
However, let's ask a slightly different question. What is the probability that at least one losing streak of length k will occur somewhere among n independent wagers?
If k is less than n, this probability will be greater than q^k. Why would this be so? Because there now will be many opportunities to start a losing streak (each newly lost wager can be the start of a losing streak).
A good approximation for this probability is
P(>= 1 losing streak of length k in n wagers) is approximately equal to [1 + (n-k)p]q^k, where p is the probability of a win.
As expected this is greater than q^k.
Suppose you make 200 wagers in a month. With a win rate, p, of 50%, the probability of a loss q = 1-p, which is 50%. Most would calculate the probability of a losing streak of 10 losses as q^k = .5^10 = 0.001, or 1 out of 1000.
However, what is really needed is the probability of getting at least one losing streak of length 10 at some point during the 200 wagers. This probability is much higher. It is higher by the factor of [1 + (n-k)p].
[1 + (n-k)p] = [1 +(200 - 10)(.5)] = 96
So, the actual probability of at least one streak of length 10 over 200 wagers is 96 times as large as q^k or (96)(0.001) = .096 (nearly 1 out of 10).
That is quite a difference.
Mike (Dr Beav)
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| Stillriledup |
08-09-2014 10:10 PM |
you gotta eat a LOT of chalk to have 45% winners...but, it can be done if you just pick and choose and stick with prime horses who are 1-2 odds or lower.
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| imofe |
08-09-2014 11:26 PM |
Quote:
Originally Posted by Stillriledup
you gotta eat a LOT of chalk to have 45% winners...but, it can be done if you just pick and choose and stick with prime horses who are 1-2 odds or lower.
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Yep. And I don't think this guy has the patience not to make action bets. Especially if he starts to get some bad breaks, he could go on tilt. |
| davew |
08-10-2014 12:05 AM |
Quote:
Originally Posted by imofe
Thank you Robert. I have an acquaintance at the track who asked me this and I said I would get back to him. He is a big bettor who was recently offered 8% on all his bets. He thinks he can obtain 45% winners at 97 cents on the dollar return. Because he is very aggressive and a rare bird (my dad use to say this) he wants to bet progressive style and wanted to know about losing streaks. I don't know what will happen , but from an entertainment perspective, you can't beat it.
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the parimutual system is designed to take money from people who bet too much into a pool |
| imofe |
08-10-2014 12:49 AM |
True Dave. I guess for me it is very entertaining to watch because I am overly conservative without too many mood swings. This guy is like a roller coaster ride.
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| Stillriledup |
08-10-2014 12:50 AM |
Quote:
Originally Posted by imofe
Yep. And I don't think this guy has the patience not to make action bets. Especially if he starts to get some bad breaks, he could go on tilt.
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The only way you can beat the game, in my opinion, is to bet on horses who look bad on paper but in real life, they're not that bad. Horses who look good on paper are short prices for the most part and horses who look great on paper are shorter prices still yet. The greatest players are able to sniff out the horses that the general public THINK are slugs, but really aren't. |
| TexasDolly |
08-10-2014 07:21 AM |
Quote:
Originally Posted by TrifectaMike
It is not to difficult to calculate the probability of a losing streak when we start with a loss. Since we start with a loss, it is just the probability of a single loss raised to power of consecutive losses.
So, if q equals the probability of one loss, the probability of k losses simply equals q^k (q raised to the k power).
Example:
q (50% winners) = .5 and k =10 (10 consecutive losses)
q^k = .5^10 is approximately .001
However, let's ask a slightly different question. What is the probability that at least one losing streak of length k will occur somewhere among n independent wagers?
If k is less than n, this probability will be greater than q^k. Why would this be so? Because there now will be many opportunities to start a losing streak (each newly lost wager can be the start of a losing streak).
A good approximation for this probability is
P(>= 1 losing streak of length k in n wagers) is approximately equal to [1 + (n-k)p]q^k, where p is the probability of a win.
As expected this is greater than q^k.
Suppose you make 200 wagers in a month. With a win rate, p, of 50%, the probability of a loss q = 1-p, which is 50%. Most would calculate the probability of a losing streak of 10 losses as q^k = .5^10 = 0.001, or 1 out of 1000.
However, what is really needed is the probability of getting at least one losing streak of length 10 at some point during the 200 wagers. This probability is much higher. It is higher by the factor of [1 + (n-k)p].
[1 + (n-k)p] = [1 +(200 - 10)(.5)] = 96
So, the actual probability of at least one streak of length 10 over 200 wagers is 96 times as large as q^k or (96)(0.001) = .096 (nearly 1 out of 10).
That is quite a difference.
Mike (Dr Beav)
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Thanks for the post TM. It was appreciated by me. I am surprised that imofe didn't comment,maybe he missed it. |
| imofe |
08-10-2014 05:06 PM |
Sorry about that. Thank you Mike and Ray. Mike's post and Ray's attachment were both very helpful.
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| imofe |
08-11-2014 01:46 PM |
Okay. This should be interesting. His planned progression based on the probable consecutive losses will be a step up in one unit after three losses to a max of 15 plays ( losses). So max losses would conclude after three 5 unit losses. Looks like he can lose 45 units if 15 in a row go down. He said he will start over after a win even if he has not broke even for that session. He is confident he can do 45% at 97 cents on the dollar ( 8% back). I won't see him again until Friday, so I'll update then.
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| overthehill |
09-01-2014 12:46 PM |
i dont understand why he thinks progression will help him. if anything there is a likelihood that 1. he is in a slump that will be exacerbated by progression, 2. his increased betting size will effect the pools. in my opinion it is very difficult to project 45% winners with confidence under any circumstances. yes it may average out that way over thousands of bets, but its a fallacy to think that therefore there is any certainty that for any race or small group of races that the odds of winning is 45%. horse trainers jockey are not machines and the surface changes and weather effect results as well. i think there have been sequences at new york tracks of a favorite not winning for three days.
the only person i have heard of making money from some sort of progression was a guy i met years ago, who seemed like an astute handicapper though he was an admitted alcoholic. he said he made a living by playing a couple of place parlays a week. he said he would hit 50% of the time and get better than $7 back.
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| thaskalos |
09-01-2014 01:34 PM |
Quote:
Originally Posted by overthehill
i dont understand why he thinks progression will help him. if anything there is a likelihood that 1. he is in a slump that will be exacerbated by progression, 2. his increased betting size will effect the pools. in my opinion it is very difficult to project 45% winners with confidence under any circumstances. yes it may average out that way over thousands of bets, but its a fallacy to think that therefore there is any certainty that for any race or small group of races that the odds of winning is 45%. horse trainers jockey are not machines and the surface changes and weather effect results as well. i think there have been sequences at new york tracks of a favorite not winning for three days.
the only person i have heard of making money from some sort of progression was a guy i met years ago, who seemed like an astute handicapper though he was an admitted alcoholic. he said he made a living by playing a couple of place parlays a week. he said he would hit 50% of the time and get better than $7 back.
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I have a hard enough time with this game when I am sober... |
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