blonde poker forum

Or perhaps a fifteen year old with a casio...

I do an accumulator on the American Football every Sunday. Five teams at prices usually between 1/2 and 2/1 and it works out to something between 16/1 and 25/1 as a one unit accumulator (in other words, for all five to win).

I am trying to work out what the odds are that I win before the season ends (let's say 16 weeks of the regular season; the final week tends to be hard to predict as some teams rest key players during the game or fringe players get a name out).

Now, for the sake of argument, let's assume I am above average at working out who might beat whom. The actual chance of an average individual accumulator bet I place coming in on any given Sunday is, let's say, 16/1.

Am I right in saying, if I want to work out what the chances are I bink one of these before the curtain comes down, that I use a Poisson Distribution?

Probability of a win on any given week is 1/17

I get 16 goes at it, so 16/17 = 0.9412

Then, to work out the probability of any number of wins, the best way is to work out what no wins is, so I plug that into the Poisson formula with mu = 0 and that pops out at a 40% chance, which means I have a 60% chance of getting a return.

Have I worked this out correctly?

Or is it simpler than that?

(http://www.thatsagoal.com/wp-content/uploads/2014/12/Both-Teams-to-Score-coupon1.jpg)

Thanks in advance. Been a while since I did anything like this, but I was discussing accumulator bets the other day and my mind went a-thinkin'.

Isn't the chance always going to be the same as the previous week's result won't affect this week's?

Doesn't a binomial distribution fit better?

Poison tells you 'time between events' but binomial gives probability of getting k successes given n trials and probability p for each test. Classic example is a coin toss.

Is the accumulate not a really biased coin?

Then sum successes k = 1 to k = 15?

Note: maths fish.

That's what I was trying to figure out as much as anything. If they're not related week to week, you don't need a Poisson Distribution?


Same question, here.

It's been 12 years since I even touched this stuff.

Binomial is right i think

The formula (again i think) is

Probability (0 wins) = f (0) = 0.06^0(1-0.06)^16-0=0.37157429083

Probability of something happening to the power of the amount of times you want it to happen multiplied by one minus the probability of something happening to the power of the number of test's minus the amount of times you want it to happne. (wiki explains it better)

Inverse of this is 63% chance give or take of 1 or more wins. Similar to what you got in Poisson.

https://en.wikipedia.org/wiki/Binomial_distribution (https://en.wikipedia.org/wiki/Binomial_distribution)

When tikay finds out you've been doing accas.......

I suggest you ask Marky.

"asking for a friend".

(http://cosmo.cdn.assets.cougar.bauer-media.com.au/s3/digital-cougar-assets/Cosmo/2012/08/13/13118/H_ED_Common_Cosmo_Online__Site_Content_Lifestyle_Feel_Good_Advice_2012_August_Don't_apologise_100812_apologise_MAIN.jpg?Image=%2Fs3%2Fdigital-cougar-assets%2FCosmo%2F2012%2F08%2F13%2F13118%2FH_ED_Common_Cosmo_Online__Site_Content_Lifestyle_Feel_Good_Advice_2012_August_Don%27t_apologise_100812_apologise_MAIN.jpg&Width=650&AllowUpSizing=False&Mode=Crop)

But Babe, I was just multiplying my edge, I swear...

Please note I am working from your assumptions and not from realistic ones.

Chance of win = 1/17.

Therfore chance of no win =16/17

Chance of no win over 16 weeks = (16/17)^16 = 0.379

Therefore chance of win over 16 weeks = .621 or 62.1%

FWIW You have to assume you are way above average to get from accumulated odds between 16/1 and 25/1 and chance of a win = 1/17.  I'd suggest that it is close to impossibly good at predictions and way above doesn't really cut it either.

Thanks, all.

Doobs, that's fine. I was just using a set of numbers to make the example easier. It's the method of working it out I wanted to clarify.

The insanity of my hedge is greatly exaggerated.

Much obliged.