# Best Decision Probability Problem

#### statsdog123

##### New Member
Hi Everyone,

I've been thinking about a probability problem that has been bugging me; I would appreciate any insight.

Problem: Let's say that me and my 9 colleagues each come up with an independent marketing strategy for a product; and let's assume that we are all equally skillful at coming up with a good strategy. What is the probability that I come up with the best strategy? I have 2 "solutions".

1) Comparing all strategies all at once: Since there are 10 possible strategies, I have a 1/10 chance of having the best strategy, or 10%.

2) Comparing strategies one at a time: I have a 50% chance of having a better strategy than colleague A. I have a 25% (50% x 50%) chance of having a better strategy than colleagues A and B. I have a 12.5% (50% x 50% x 50%) chance of having a better strategy than colleagues A, B, and C, so on until I finally conclude that I have a ~2% (0.5^9) chance of having a better strategy than my 9 colleagues A-I.

I'm very sure that solution 1 is the correct solution, but I am having trouble explaining the logical fallacy of solution 2. Could you guys and gals please help provide some statistical explanations? Thank you in advance.

#### Dason

This is a tricky one to think about. The issue is that the conditional probabilities don't actually stay constant. Because of the way you construct the sequence you actually have a larger than 0.5 probability to of getting through any round other than the first round.

So with three people there are six possible ways of assigning "rank" to their strategies. For example in the first possibility we have Person A having the "best" strategy and Person C having the "worst" strategy. In actuality before doing anything we don't know which possibility is reality so if we wanted to know the probability that Person A had the best strategy we could look at the number of possibilities where they are the best divided by the total (2/6) which gives 1/3 like your intuition for your first solution would give. If we want to use your second solution we would really want to figure out P(A beats B and A beats C) and we can reduce this to: P(A beats B)*P(A beats C | A beats B). In three possibilities A beats B so P(A beats B) = 0.5. So your logic is fine up to here. However, now is where it gets tricky since we want P(A beats C | A beats B). So conditional on "A beats B" we only consider the possibilities where A beat B, so the rows with possibilities 1, 2, 4. Looking at the outcomes we see that in 2 out of the 3 of those possibilities A beats C so really P(A beats C | A beats B) = 2/3. So P(A beats both B and C) = (1/2)*(2/3) = 1/3. Somewhat surprisingly these probabilities would remain exactly the same if there was a Person D as well (although the total number of possibilities jumps to 4!=24). And then you might begin to notice a pattern as you find that P(A beats D | A beats B and C) = (3/4).

So you might be able to catch the pattern and in your actual problem the probabilities of interest would boil down to : (1/2)*(2/3)*(3/4)*(4/5)*(5/6)*(6/7)*(7/8)*(8/9)*(9/10) = 1/10.