Let's say I'm playing a game of dice with another 5 people. The game goes like this: each of us has 5 dice (30 dice total) and we roll them, and I can only see what I've rolled (my 5 dice). The purpose of the game is to guess how many 6's have been rolled, and 1 counts as 6, too, so there's a $1/3$ chance for a die to yield 6. I thought about two strategies and I can't tell, mathematically, which one is better. Let's say I have 3 6's (and/or 1's).
-Strategy 1: there are 30 dice total so the probability is that $1/3$ of all the dice to be 6's, and that means that I should guess 10 6's total. This strategy is basically ignoring my own results and assuming that the overall probability is unchanged by what I've rolled.
-Strategy 2: I subtract my 5 dice from the total, and the probability of getting a 6 in the remaining 25 dice is about 8.3. I then add my 3 6's and I get 11.3 total, so I should guess 11 6's. From how I see it, this strategy doesn't treat my own dice "probabilistically" as I already know the results.
I get confused by the fact that the rolls themselves don't influence each other, so I should treat them as if I rolled 30 dice, selected 5 dice randomly (my dice) and it happened to be 3 6's amongst them, even though if I rolled 5 dice individually, the probability of getting 3 6's is less than $1/3$. So, by this thinking, I should guess 10 6's, no matter what I roll. But then, my roll shouldn't influence the other's rolls, so even if I got 3 6's, on average, they should still have about 8.3 6's, so my guess should be influenced by what I've rolled, and guess 11. As the number of dice grows larger, the differences get bigger (for 300 dice, if I roll 30 6's, the first strategy says I should guess 100, and the second, 113, so it's a 13 dice difference)
My questions are:
- Is there a mathematically correct answer, for whatever number of total dice? If not, how does the total influence the answer?
- Should it matter to my guess if I rolled a real 6 or a 1?
- Should my number of dice influence the preferred strategy (as in, if I have 3 dice and the total is 20, is another strategy better than in the case presented above)?
The correct answer should be to guess $\frac{25}{3}$ plus your total number of "sixes." Noting that this isn't an integer, round to the nearest, so $8$ plus your total.
More generally, you should guess $\frac{1}{3}$ times the number of dice that you didn't throw plus the amount of sixes (or ones) that you did throw and rounding to the nearest integer.
This is again a trivial application of linearity of expected value and the binomial distribution along with basic properties of expected value.
"Fake" sixes versus natural sixes doesn't matter.