quick odds question

nearly 67/1

Would aybody care to share how this is worked out? Thanks.

5 matches played by each team any two from 5 = 10

so at 25/1 that's 26x26 divided by 10 = 67.6 that's 66.6/1

That obviously is taking one specific team to draw twice, if the bet was any of the six teams drawing twice then that becomes more problematic.

Thanks pump.

If you mean twice and no more than twice then your answer is , I think, as follows. Assuming independence.

10* (1/26)^2*(25/26)^3 which comes to around 76/1.

I will give it some more thought though.

For 2 or more there are a few more calculations and my trusty Casio is displaying "low battery" so no can do.

12 minutes to log in today!

Clearly there will be 0,1,2,3,4 or 5 draws in your 5 selected matches. By the laws of probability;
P(0)+P(1)+P(2)+P(3)+P(4)+P(5)=1
then
P(2)+P(3)+P(4)+P(5)= 1 - {P(0)+P(1)}
then using Pascal/Binomial
P(0)= 1*(25/26)^5
P(1)= 5*(1/26)(25/26)^4

Using your calculator you should be able to do that easily enough. One of mine is without battery and the other went under my heel half an hour ago.....it kept giving me the answer as a fraction.


One objection raised against the 76/1 is that a double gives a return of 675/1. But you will have staked ten units to get it and so your odds of reward are 288/5, considerably less than 76/1.

I hope I am correct but do not guarantee to be so.

Sorry 333/5 not 288/5 .

333/5 is 66.6/1 which is what I said earlier.

Using binomial probabilities (n=5, p=1/26), the probability of getting exactly 2 draws is 0.013150834 or almost exactly 75/1. The probability of 2 draws or more is 0.013687 or 72/1.