probabilities

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middler
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middler

probabilities
« on: December 21, 2015, 06:36:00 PM »

Hey guys,

I'm posting this here as it might be noobish not to know it.

I would like to compare the probability given to odds in 2-outcome vs 3-outcome events.

For example. I know that if an event has two possible results, i can calculate the probability of each one with the next formula (always using european odds): 1/odd*100=%

How would i calculate the probability of an outcome for an event with three possible outcomes?

Thanks in advance!

FORGET IT, I GOT IT, STUPID QUESTION :)
« Last Edit: December 21, 2015, 07:05:00 PM by middler » Logged
middler
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Re: probabilities
« Reply #1 on: December 22, 2015, 10:38:29 AM »

thanks for your reply.

the question was stupid, the probability is always 1/odd. what got me confused for a moment was trying to compare odds from a 2-leg with a 3-leg.
For example, DNB vs moneyline, but i will just do it manually..

thanks anyways
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qbet
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Re: probabilities
« Reply #2 on: December 22, 2015, 11:58:35 AM »

The implied probability isn't always 1/odds. For example this isn't the case for draw no bet or Asian handicaps.
What you are saying holds for bets that are all or nothing,  like 1x2.
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middler
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Re: probabilities
« Reply #3 on: December 22, 2015, 12:36:57 PM »

you are absolutely right. what i wanted is a formula to compare these kinds of bets.

So let's say we have team a and team b.

Moneyline: 2-3-4
DNB: 1,95-1,85

Of course you can calculate the actual probability given to team b in both cases manually, taking into account the amount you get back if the result is a draw. However, i haven't found a way to calculate these kind of 3 way arb and save some time (it's not always moneyline vs dnb, it can be ah round number vs ah non-round number)

thanks for the answer anyways
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middler
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Re: probabilities
« Reply #4 on: December 30, 2015, 09:20:18 AM »

So if we have a basketball game and we bet on moneyline (ot included), probabilities of each outcome are 1/ODD, so if 1/ODD A + 1/ODD B < 1 , you are getting extra value and therefore are able to place an arb.

For example, Team A - 2,1 and Team B - 2,1. 1/2,1 = 0,475. So if you have two outcomes, you get an arb of 5% (0,475 + 0,475 = 0,95).

On a 3-way moneyline bet, such as football, if we had Team A - 2,1, Draw - 2,1, Team B - 2,1, we would have 142,5% total probability, and therefore a negative arb of 42,5%.

But on other kinds of bets, such as AH, or DNB, the real probability (or implied probability as qbet said) is different. So if you have a football DNB with Team A - 2,1 and Team B ML - 2,1, you are not in the same situation as above.
If Team A +0 (DNB) - 2,1 and Team B (Moneyline) - 2,1, the probability of Team A winning is lower and better paid than Team B winning, although the odds are the same. Therefore, the previous formula does not apply to this case.

The same thing happens when you have AH (round number, .25 or .75), but then again it would be a different formula for each case. (I'm not going to write down each example, but you can get for instance AH1 -2, AH2 +1,5, AH2 +2,5).

Actually, the result that we get here is the probability for us, not the bookmaker, since it is the amount we get paid. So this is not even an accurate calculation, as it has to be taken into account the juice of each bookmaker in order to know the real probability that they assign to each possible outcome of an event, so the calculation is much harder than just this.

I am sorry for the bad explanation, but this is not an easy topic for me to explain, and I don't even have it that clear in my head.
I guess I will have to calculate manually depending on the type of bet.
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middler
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middler

Re: probabilities
« Reply #5 on: December 30, 2015, 09:31:03 AM »

I would love to have a formula and I would be happy to share it but it would be different for each case.. even for DNB, you can put them against ML or AH or even some cross market odds...
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raizzak
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Re: probabilities
« Reply #6 on: December 30, 2015, 12:34:35 PM »

If i understand this correct you need the formulas for the asian handicaps in relation with 1x2 odds?
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middler
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Re: probabilities
« Reply #7 on: December 30, 2015, 12:54:43 PM »

No, thanks anyways, I guess I can calculate that.

I just wanted to know if I could find a common formula to calculate possible arbs in cases involving ML and DNB and AH, but I guess there is a different formula for every kind, so it is almost like doing it manually every time.

The aim was to save some time, but it will be hard.

Thanks anyways
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middler
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Re: probabilities
« Reply #8 on: December 30, 2015, 02:37:58 PM »

For AH1(0), X, 2,

Team 1 win: AH1(0) * STAKE 1 = AMOUNT WON

X: STAKE 1 + ((AMOUNT WON-STAKE 1)/X) * X =AMOUNT WON

Team 2 win: 2 * STAKE 3 =AMOUNT WON

Where (AMOUNT WON - STAKE 1)/X = STAKE 2

On the other example, it is similar, just take into account that in that case the draw would be the stake on ah1(0) + the outcome of ah2(+0,5) winning bet. And team b winning would include 2 and ah2 (+0,5).

There are many possible combinations for these types of bets... I think that the best option is to calculate manually...
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whitesnake
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Re: probabilities
« Reply #9 on: January 01, 2016, 07:42:56 AM »

what is the use of doing these calculations if they do not provide a no lose scenario?
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