Sports betting maths

[GoldenBootIRL]

Assume you have two bets one is 4/6 and the other is 10/1.

You have a 20% edge in both.


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With a 20% edge you should win?:

The 4/6 shot 72.2% of the time?
The 10/1 shot 10.9% of the time?

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What is the expected run of losses in a row that can be expected with the above edge?

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Can you show the maths behind the answer, thanks.

[bztips]

I don't know the answer, but what you're asking about is equivalent (I think) to what's called "average run length" in quality control studies. Google it and you may find an answer (probably in the form of a chart, rather than a formula).

[GoldenBootIRL]

For anyone that comes across the thread and would like to know:

convert the odds the decimal, take the reciprocal to find the percentage, then increase the percentage by your edge.

(1 / (4/6)+1) )*1.2 = 0.72 = 72%

(1 / (10/1)+1) )*1.2 = 0.109 = 10.9%




The more bets you make, the greater your chance of seeing consecutive winners and losers will be.

but, you can calculate the expected maximum losing run in a given number of bets.

the formula is,

Log(Bets)/ -Log(1-win_percentage)

so for a 1000 bets and a win percentage of 72%, you would have,

Log(1000) / -Log(1-0.72) = 5.42

that is, you would expect a maximum of around 5 losing bets in a row.

[GoldenBootIRL]

convert the odds the decimal, take the reciprocal to find the percentage, then increase the percentage by your edge.

(1 / (4/6)+1) )*1.2 = 0.72 = 72%

(1 / (10/1)+1) )*1.2 = 0.109 = 10.9%




The more bets you make, the greater your chance of seeing consecutive winners and losers will be.

but, you can calculate the expected maximum losing run in a given number of bets.

the formula is,

Log(Bets)/ -Log(1-win_percentage)

so for a 1000 bets and a win percentage of 72%, you would have,

Log(1000) / -Log(1-0.72) = 5.42

that is, you would expect a maximum of around 5 losing bets in a row.

Thanks.