# Probability calcuation

Working on a side project … Been asked to write a service for a mobile app to calculate some betting odds.

If the odds prior to the start of a soccer game for the number of passes a player will make is eighteen passes. That means the over 18 passes has approximately 50% probability while the under 8 passes has 50% probability (minus the bookie spread).

So you can say the player will average .2 passes per min over 90 minute game. What’s the best way to calculate the odds, let’s say immediately after 5 minutes of the game, the player already takes 3 passes (averaging .6).

Will bayers theorem work? What other ways can you sort of make an algorithm for calculating the odds as things change?

Just looking for linear interpretation, not all the actual play mechanics in real games.

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125mph wrote:
Working on a side project … Been asked to write a service for a mobile app to calculate some betting odds.

If the odds prior to the start of a soccer game for the number of passes a player will make is eighteen passes. That means the over 18 passes has approximately 50% probability while the under 8 passes has 50% probability (minus the bookie spread).

No it doesn’t.

You’re confusing the mean number of passes with the median number of passes.

125mph wrote:
So you can say the player will average .2 passes per min over 90 minute game.

Yes.

125mph wrote:
So you can say the player will average .2 passes per min over 90 minute game. What’s the best way to calculate the odds, let’s say immediately after 5 minutes of the game, the player already takes 3 passes (averaging .6).

Will bayers theorem work? What other ways can you sort of make an algorithm for calculating the odds as things change?

Just looking for linear interpretation, not all the actual play mechanics in real games.

It’s not a Bayes Theorem problem.

The expected number of passes for the remainder of the game (assuming a uniform rate of passing throughout the game, and that the two extra passes in the first 5 minutes don’t portend an increase in his passing frequency) would be:

0.2 passes / minute × 85 minutes = 17 passes

The expected number of passes for the entire game would therefore be 20 (= 3 + 17).

In general, the number of passes expected for an entire game (assuming a uniform rate of passing throughout the game) is:

# of passes in the first t minutes + 0.2 (90 − t)

Simplify the complicated side; don't complify the simplicated side.

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This is helpful! So you’re assuming that the 0.2 rate will remain, giving a linear interpolation and the earlier event was an anomoly?

How can I set that to a probability. So if the payout at the start of the game is 1:1 for 18 over or 18 under, and immediately they land 3 passes in 5 minutes, therefor the expected pass is now 20.. so the probability is > 50% that an over 18 will win, but can I get closer and say its 66.7%.. what calculation would that be, if even possible?