Expected Value of a Dice Game with an Optimal Minimum Divisor
Summary
The document analyzes a game where two fair dice are summed for an initial score, then the lower result of a second pair is chosen as the divisor. It explains why dividing the expected initial score by the expected divisor does not give the expected final score: in general, the expected reciprocal of a random variable is not the reciprocal of its expectation.
The initial sum has expectation 7, and the minimum of the second pair has a stated probability mass function. Taking the probability-weighted average of the reciprocal of that minimum gives 397/720. Since the two rolls are independent, the expected ratio factors into the expected initial score multiplied by this expected reciprocal. This calculation assumes fair dice and independent rolls; it addresses expected payoff only and does not discuss alternative rules, risk, or strategies beyond choosing the smaller die result.
Key ideas
- The expected value of a ratio is not generally the ratio of expected values.
- For independent variables, the expected ratio factors into the numerator's expectation times the denominator's reciprocal expectation.
- The divisor is the minimum of two fair-die results, so its probability distribution is needed.
- Compute the reciprocal expectation by weighting each reciprocal outcome by its probability.
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Full text
# Sum Over Min Die
# Sum Over Min Die
You have been tasked to find the expected value of a die game in which you are rolling 2 dice at a time. Your first roll of the dice will all be summed up and will be your starting score. Your next, and final, roll of the 2 dice will be your divisor, in which you are going to pick the optimal die out of the two to divide your total by, which will be your final score. Calculate the expected value of playing this game optimally
MY approach - I first got the expected sum on 1st roll which is 7. Then I calculated the expected value of min(d1,d2) on 2nd roll and by evaluating that i got the value 91/36.....(I have verified it)
now to get the max expected return I did - Max Return = Expected Initial score/Expected number you choose on 2nd roll Max return = 7/(91/36) = 36/13 but it is wrong ?? Can anyone correct me where i went wrong and suggest the correct approach
## Answer by Kermittfrog (score 5)
https://quant.stackexchange.com/a/80022
This is an application of the ratio distribution $f(Z)$, $Z=X/Y$.
You are looking for $\mathrm{E}\left(X/Y\right)$. Here, $X$ and $Y$ are independent, thus:
$$ \mathrm{E}(X/Y)=\mathrm{E}(X)\mathrm{E}(1/Y) $$
Note that $\mathrm{E}(1/Y)\neq 1/E(Y)$!
$E(X)=7$ as you have shown already. As $Y$ is the minimum of the points of a throw of two fair dice, its probability mass function is
$$ f(Y= k)=\frac{13-2k}{36} $$
The expectation $\mathrm{E}(1/Y)=\sum_i \frac{1}{i}f(Y=i)$ is
$$ \mathrm{E}(1/Y)=\frac{1}{1}f(1)+\frac{1}{2}f(2)+\cdots+1/6f(6)=\frac{397}{720} $$
Hence, $E(Z)=7\times \frac{397}{720}$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.