The probability is 0.67 that the favorite in a horse race will finish in the money( first, second and third place) in the next 5 races, what is the probability that the first one finishes in the money between 2 and 4 times inclusive
Answers
ll simplify this problem a bit to make it a little easier. Instead of using that probability as how often the horse finishes in the money, let’s assume that it’s the odds of them finishing first. This does not change much, as it’s just looking at things from another angle, but it’ll be easier for me to explain.
This makes the question, in this new form, “What is the probability that the favourite wins 4, 3 or 2 of the next five races?”
To work this out, we need to know the probability that the favourite would win at least 2 of the next 5 races, and then subtract the probability of them winning all 5 of them.
As the win chance is 0.67, this means the chance not to win is 0.33. The question does not make clear if the odds are exactly 0.67, or if it’s meant to be two thirds - if the latter, we can use the fractions 1/31/3 and 2/32/3 to represent the losing and winning percentages. If the former, we would need to use 33/10033/100 and 67/10067/100 instead. We will call the losing percentage ll and the winning percentage ww .
The probability of them losing any four races (chosen at random) is l4l4 . It doesn’t matter if they win or lose the remaining race, as that would only mean they won one race at most.
In the case where l=1/3l=1/3 , this value, fa(4)=(14)/(34)=1/81fa(4)=(14)/(34)=1/81
In the case where l=33/100l=33/100 , this value, fb(4)=(334)/(1004)=1,185,921/100,000,000fb(4)=(334)/(1004)=1,185,921/100,000,000
The probability of them winning all five races is w5w5
In the case where l=1/3l=1/3 , this value, ta(5)=(25)/(35)=32/243ta(5)=(25)/(35)=32/243
In the case where l=33/100l=33/100 , this value, tb(5)=(675)/(1005)=1,350,125,107/10,000,000,000tb(5)=(675)/(1005)=1,350,125,107/10,000,000,000
Finally, the probability of winning at least 2 but at most 4 is 1−(f(4)+t(5))1−(f(4)+t(5))
In the case where l=1/3l=1/3 , this becomes 1−((1/81)+(32/243))=(243/243)−(35/243)=208/2431−((1/81)+(32/243))=(243/243)−(35/243)=208/243 , which is roughly equal to 85.6%
In the case where l=33/100l=33/100 , this becomes 1−((1,185,921/100,000,000)+(1,350,125,107/10,000,000,000))=(10,000,000,000/10,000,000,000)−(1,468,717,207/10,000,000,000)=8,531,282,793/10,000,000,0001−((1,185,921/100,000,000)+(1,350,125,107/10,000,000,000))=(10,000,000,000/10,000,000,000)−(1,468,717,207/10,000,000,000)=8,531,282,793/10,000,000,000 , which is roughly equal to 85.3%
As you can see, not knowing if it’s meant to be 0.66 recurring or a flat 0.67 to win (or, in the case of the original question, to be in the money) can lead to an error of about 0.3% compared to the other answer for the final result, and that error is big enough and the answers positioned in such a way that if you were to give it as a whole percentage, one would give you 85% and the other would give you 86%