If there are 39 runners in a race and prizes will be awarded to first, second, and third places only, what is the probability of correctly p

As of late, the probability that I get a correct answer on a probabilty problem has been something less than 1, but I'm still gonna' give this one a try.....

The probabilty of picking the first place finisher correctly is just 1/39. And the probabilty of picking a second place finisher correctly (given that we've picked the first place finsher correctly) is just 1/38. And the probability that we pick the third place finisher correctly is just 1/37.

So ....  (1/39)*(1/38)*(1/37)  = 1/54834

Another way to look at this is count all the sets that are possible when we choose any 3 people from 39. But since order matters within the set, we must use a permutation. Thus, (39 P 3) = 54834. And since only one of these sets is "correct," we have 1/54834 chances of choosing this set.

Assuming everyone has equal probablilty of getting a place and there are no ties I am going with you Chris.  Last time this gor both of us into trouble but hopefully we will fare better this time!

Let P₂  = Probabilty that Chris and Melody answer subsequent probability question correctly

Let P₁ =  Probabilty that Chris and Melody answer prior probability question incorrectly

So we have

[ P₂ l  P₁ ] = [P₁ * P₂] /  [P₁]    →  Somewhere between 0  and 1  ....... (probably)

Nice one Phill -re the 39 runners, two different methods, identical results, can't be bad !

Well..bertie, as you can tell, probability isn't my strong suit.....I am happy when my answer actually agrees with anyone's .....but, when I actually agree with myself, it's truly odd !!!    LOL!!!