True odds vs Bookmaker odds?

#1
Hello, I'm new to the forum and very excited to be here. However I have a question that brought me here. i attach a picture to show more clearly what I'm looking for and what I already tried. So pretty much we have 7 matches with odds.

Odd % probability

1.16 86.21
1.17 85.71
1.31 76.47
1.36 73.33
1.37 72.97
1.59 62.96
1.74 57.45


Now, based on 7 possible outcomes ( I can get 0 right and 7 wrong until 7 right and 0 wrong) what should be the true odds and/or the odds that the bookie should set up?.Is it the one on the right or on the left side in the picture, or maybe something else?. On the right it is just the multiplication, so to hit 2 we have ( 1.16*1.17), for 3 we have (1.16*1.17*1.31) and so on and so forth. It seems true but on the left I tried to convert the odds into %, and pretty much averaged them and came up with 73%. Then I took 7 and took 73% of it, and came with 5 as being the 50/50 margin, so in odds case 2 ( or 1.95 with bookie vig). This seems more realistic however the 5.39 is way of 9.15, and shouldn't they be equal?

Thanks in advance for all responses

Cheers
 

hlsmith

Omega Contributor
#2
You absolutely lost me on what you are trying to do and what you have done. Can you just describe what the game or event is, that might be clearer.
 
#3
Hello Hlsmith,

Okay I will make it short, so we have 7 matches and they have % of winning as follow.

86.21%
85.71%
76.47%
73.33%
72.97%
62.96%
57.45%

The main question is what is the probability that I will win all 7 matches? (We obviously take into account that if there is 86.21% of winning the game, there is 13.79% of losing) what about just 6, 5,4,3,2,1, and finally what is the probability that I will not hit any game.
 

hlsmith

Omega Contributor
#4
That seems much better!

So can I say this is a slot machine with seven spinning wheels and you have an independent chance of getting cherries (the winning image) on the first wheel of 86.21,..., last wheel 57.45?

And you just want to know the chance of getting all of the combinations or at least just how to calculate this (since there are lots of combos; e.g., no cherries,...,all cherries)?
 
#5
Yes, the problem is that each spin has a different chance of getting the cherries, so sometimes the spin of 86.21 comes up with 57.45 ( coming with 2) but there can be 86.21 with 62.96) resulting in 2 cherries as well but different combinations. Is there a way to calculate the average I guess all all combinations of 2's, 3's, 4's etc?

Thanks
 

Dason

Ambassador to the humans
#6
So what you're trying to do is figure out the answer to questions of the form "What is the probability of getting 3 successes on these 7 wheels"?

There isn't a much better way to do this than just saying "what are all the possible ways to get 3 successes (and 4 failures) on the 7 wheels" and then for each of those outcomes figuring out the probability of that exact outcome, and then summing up all of those probabilities. Luckily for you I did just that

Code:
 outcome   probability
       0 0.00005268149
       1 0.00126427723
       2 0.01239987939
       3 0.06438993968
       4 0.19141762732
       5 0.32611856632
       6 0.29499639590
       7 0.10936063266
outcome represents how many "successes" there were and probability represents the corresponding probability of getting that many successes.
 
#7
Thank You Dason, but from the table it implies that 7 successes, meaning hitting all 7 is 0.10%, is higher probable than hitting just 3 succsses and 4 incorrect as the rpobability is 0.064%?
 

Dason

Ambassador to the humans
#8
That is correct. The probability of getting 7 successes is just the product of the 7 success probabilities which in this case ends up being about 10.9%

I should note that the probability column in my table is literally probabilities (so they will always be numbers between 0 and 1). If you want the percents then multiply by 100.
 
#9
Got it, but how did You sum up the probabilities for instance all possible combinations? Can You provide one example on how did You do it on 6 successes ( so it is easier as it is only 2 combinations).

I ask this because I love to complicate things in probabilities/statistics questions and this probably would help in figuring out the probabilities for things like "chances that I will hit 3 or higher, including 3 etc).

Once again, many thanks
 

Dason

Ambassador to the humans
#10
Got it, but how did You sum up the probabilities for instance all possible combinations? Can You provide one example on how did You do it on 6 successes ( so it is easier as it is only 2 combinations).

I ask this because I love to complicate things in probabilities/statistics questions and this probably would help in figuring out the probabilities for things like "chances that I will hit 3 or higher, including 3 etc).

Once again, many thanks
The table provided is all you need to answer that question actually.

And the way I got those probabilities was just like I said. Look at all possible ways to get 3 successes out of 7 and figure out the probability for each of those outcomes, then add up all of the probabilities you calculated. I didn't do any of it 'by hand' - I just told the computer to do it for me.
 

Dason

Ambassador to the humans
#11
Here is the R code I used for anybody interested

Code:
dat <- read.table(textConnection("
1.16    86.21
1.17	85.71
1.31	76.47
1.36	73.33
1.37	72.97
1.59	62.96
1.74	57.45"), col.names = c("odds", "prob"))

probs <- dat$prob/100


outcomes <- do.call(expand.grid, replicate(7, c(T,F), simplify = FALSE))

processOutcome <- function(row){
    prod(c(probs[row], (1-probs[!row])))
}

count <- rowSums(outcomes)
rowprob <- apply(outcomes, 1, processOutcome)
ans <- tapply(rowprob, count, sum)

tab <- data.frame(outcome = names(ans), probability = ans)

print(tab, row.names = FALSE)
 
#12
I figured that not by hand, but I try to understand it and I think the second question( hopefully the last one) will answer it. So based on the table that You provided, then probability of hitting 3 or more is around 98.7%? 10.93+29.49+32.61+19.14+06.43)
 
#14
Which program You guys use that these codes will work and how to put them in in order to see results? I'm new to this so I downloaded Rgui and tried to put Dason's code but it doesn't produce anything.
 

Dason

Ambassador to the humans
#15
I used R. RGui is the default way to interact with R on windows. You didn't say what error you got so it's impossible to help debug that.
 
#16
Is it possible for You to provide a link which R version etc You downloaded, and step by step how You put that code in. In addition, I guess it is possible to change the outcomes within the code, so instead of 7 outcomes, I can change it to 10, 15 etc just by replacing 7's with desired number in that code right? Finally, what is the exact mathematical name for example that I provided? I know it's not binomial distribution, and I even heard of poison distribution ( LOL). Sorry about all those questions, just getting started with probability and statistics.

Cheers

PS. I was thinking of doing that in excel, but can't figure it out either.