# 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

##### Ambassador to the humans
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.

It's probably easier to think about 3 people insteads of 10. 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.