multiple selections doubles staking help

[Werdnatrebor]

hello all, I would like to back 2 selections in a race in each of four races,how many bets would this be? could somebody please also tell me what the formula is for calculating winning doubles for multiple selections in multiple races.

[cavello]

Re: multiple selections doubles staking help Number of double = ((number of races x number of races -1)/2.) x (number of selections x number of selections) Therefore (4x3)/2x(2x2)=24 bets. If 3 races then (3x2)x(2x2)=12 bets and obviously for 2 races 4 bets. Gets a bit messy when you want 4 in one race and 2 in another etc.

[Werdnatrebor]

Re: multiple selections doubles staking help many thanks cavelloman:)

[slapdash]

Re: multiple selections doubles staking help Another method that works with different numbers of selections in different races is to count the total number of doubles that would be possible if all the horses were in different races, and then subtract the number of "illegal" doubles. The number of doubles from N selections is N*(N-1)/2, so if you had, say, 3 races with 4 selections, 5 selections and 6 selections, the total number of selections is 15, so there are 105 =15*14/2 doubles including illegal ones. Race one gives 6 = 4*/2 illegal doubles. Race two gives 10 = 5*4/2 illegal doubles. Race three gives 15 = 6*5/2 illegal doubles. So the total number of legal doubles is 105 - 6 - 10 - 15 = 84. Applied to your "2 selections in each of 4 races", each race gives one illegal double, and there are 8*7/2 = 28 doubles in total, so 28-4 = 24 legal doubles.

[Werdnatrebor]

Re: multiple selections doubles staking help thanks slapdash, most grateful:)