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Question to mathematicians

fab

08 Jan 12 15:46

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Date Joined: 24 Oct 05

| Topic/replies: 33 | Blogger: fab's blog

Hi

A question to mathematicians regarding odds.

I understand how to calculate odds for an accumulator ...
eg 1.2 X 1.4 X 1.7 X 1.3 = 3.71 (or 26.95 %) So I have a 26.95 % chance of winning this acca.

But I tend to bet on much longer shots and I want to know when I am likely to win my next bet.
So if I bet on let's say 4 matches with odds such as 21 150 12 65 how do I calculate the chance (in %age) of winning at least once?

in percentage the odds are 21 (4.76%) 150 (0.67%) 12 (8.33%) 65 (1.54%)

Thanks for your help

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Report Mr Magoo • January 8, 2012 3:59 PM GMT

It's much easier to work out the chance of losing all the bets.

If individual winning chances are: 4.76%, 0.67%, 8.33% and 1.54%
Then the individual losing chances are: 95.24%, 99.33%, 91.67% and 98.46%

So: chance of losing them all is .9524 x .9933 x .9167 x .9846 = .8539, or 85.39%
Which means the chance of winning at least one is 100 - 85.39 = 14.61%

Report fab • January 8, 2012 11:42 PM GMT

Thanks MR Magoo, that's very clear.

One more question...
What would be the probability of all bets succeeding?

Is it 0.0476 x 0.0067 x 0.0833 x 0.0154 = 0.000000409 (about 1 in 250 million?)

Thanks in advance!

Report Lori • January 9, 2012 12:30 AM GMT

Making it too hard on yourself on the last one... just go back to multiplying the numbers together like you did for the first acca!

Your number is right, although you've forgotten it's a percentage so it's 1 in 2.5 million (ish)

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