# Help with how much to allocate to varying probabilities to get specific probability?

• Jan 15th 2012, 06:30 AM
dmbeas12
Help with how much to allocate to varying probabilities to get specific probability?
Problem:
I must spend $1000. I have 3 strategies that have different Probabilities Of Success (POS). Strategy A POS = 70% Strategy B POS = 50% Strategy C POS = 25% How much do I spend on each strategy (must spend total of$1000) to have an overall POS of 65%?

D$• Jan 15th 2012, 08:19 AM Ridley Re: Help with how much to allocate to varying probabilities to get specific probabili You can use the law of total probability and fixate one of the betting strategies to a reasonable number. Then you end up with the percentages spent on each betting strategy like this (you can choose to fixate either a or b): $P(A)=\frac{a}{1000}$, $P(B)=\frac{b}{1000}$ and $P(C)=\frac{1000-a-b}{1000}$. • Jan 15th 2012, 10:58 AM dmbeas12 Re: Help with how much to allocate to varying probabilities to get specific probabili Thanks Ridley, but I'm not sure how to plug in the numbers to fit your formula above, and I don't think I can come up with a reasonable number. Here is what I tried. Can anyone see where I went wrong? I tried to determine the ratio needed between the 3 strategies to equal 65%: .65 = .70x + .5x + .25x .65 = 1.45x x = .45 .70(.45) = .31 .50(.45) = .23 .25(.45) = .11 Ratios: .31/.65 = 48% .23/.65 = 34% .17/.65 = 18% I thought I could then take those ratios and multiple by 1000 to get how much I should apply to each strategy: 1000(.48) = 480 1000(.34) = 340 1000(.18) = 180 However when I multiple those numbers by their respective probability of successes for each strategy, I don't get the 65% total probability of success: 480(.70) = 336 340(.50) = 170 180(.25) = 45 336+170+45= 551 551/1000 = 55% ---- not the 65% I was expecting to get. Where did I go wrong? Thanks so much! D$
• Jan 15th 2012, 11:26 AM
Ridley
Re: Help with how much to allocate to varying probabilities to get specific probabili
If X is success, then the probability of success $P(X)=0.65 = 0.70 \cdot P(A) + 0.50 \cdot P(B) + 0.25 \cdot P(C)$

If you let $b=0$ then you end up with $0.7a + 0.50 \cdot 0 + 0.25 \cdot (1000-a)=650$

Law of total probability - Wikipedia, the free encyclopedia
• Jan 15th 2012, 11:44 AM
Ridley
Re: Help with how much to allocate to varying probabilities to get specific probabili
Quote:

Originally Posted by dmbeas12
.65 = .70x + .5x + .25x

This is where you went wrong. Your solution assumes that you bet the same amount on each betting strategy. The percentages you've posted can be interpreted as conditional probabilities. Given a betting strategy A, B or C, what is the probability X that it success. This is where the law of total probability comes into the picture. If you have a sample space where the probability of all events add up to 1, then you can write the probability for an arbitrary event A as:

$P(A)=P(A|H_1)P(H_1) + P(A|H_2)P(H_2)+ ... + P(A|H_N)P(H_N)$, $P(H_1)+P(H_2)+...+P(H_N)=1$