1
$\begingroup$

I got a pretty simple problem but I'm not sure about my solution.

Game $A$: We roll a fair die 4 times. If we get the "6" at least one time, we win.

Game $B$: We roll a fair die 8 times. If we get the "6" at least two times, we win.

Which game is more advantageous for us ?

I calculated the first moments which are $\frac{4}{6}$ for game $A$ and $\frac{8}{6}$ for game $B$.

While the expected Value of $B$ is higher than $A$, the probability for $P(X\geq1)$ in game $A$ equals 0.5177 and for game $B$, $P(X\geq2)=0.3953$. This shows that game $B$ is worse than game $A$.

Which solution is right ?

And another question with regard to this problem:

If you multiply the number of trials (here 4 × 2) and the number of minimum successes (here 1 × 2) by a factor $c$ (here 2) why doesn't the probability equal the same number multiplied by $c$ ?

$\endgroup$
14
  • 1
    $\begingroup$ What is the payoff for each outcome? $\endgroup$ – MSIS Nov 20 '19 at 20:10
  • $\begingroup$ There is no payoff it just asks which game is more advantageous $\endgroup$ – Anil K. Nov 20 '19 at 20:11
  • 1
    $\begingroup$ But what notion of advantage do you then use? Just to see which one you're more likely to win? $\endgroup$ – MSIS Nov 20 '19 at 20:13
  • $\begingroup$ Yes, the game with the higher probability of winning, which seems to be A $\endgroup$ – Anil K. Nov 20 '19 at 20:17
  • 1
    $\begingroup$ @whuber I agree, it does not make a lot of sense. However the post does contain a format like "While the expected Value of B is higher than A, the probability ....". So, by using the word 'while' the OP explains which are the two different points of view. The expected value on the one hand and the probability on the other hand. But it is indeed not very clear. It is not made clear why the expected value - of the number of dice with a "6" - matters (so the post sort of indicates that the problem lies in that particulas aspect, although not very clear how exactly). $\endgroup$ – Sextus Empiricus Nov 21 '19 at 16:29
3
$\begingroup$

The number $X$ of 6's in 4 trials has $X \sim \mathsf{Binom}(n=4,p=1/6)$ and he number $Y$ of 6's in 8 trials has $Y \sim \mathsf{Binom}(n=8,p=1/6).$

In R, $P(X \ge 1) = 1 - P(X = 0) = 0.5177.$

1-dbinom(0, 4, 1/6)
[1] 0.5177469

By contrast, $P(Y \ge 2) = 1 - P(X \le 1) = 0.3953,$ smaller than above.

1 - pbinom(1, 8, 1/6)
[1] 0.3953231

By simulation of a million games of each type:

set.seed(1120)
x = replicate(10^6,  sum(sample(1:6, 4, rep=T)==6))
mean(x >= 1)
[1] 0.517721     # aprx 0.5177

y = replicate(10^6,  sum(sample(1:6, 8, rep=T)==6))
mean(y >= 2)
[1] 0.395072      # aprx 0.3953 +/- 0.001
2*sd(y >= 2)/1000
[1] 0.0009777328  # 95% margin of simulation error

In each figure below, the histogram summarizes simulated values and the (centers of) red circles show exact binomial probabilities.

enter image description here

par(mfrow=c(1,2))
 hist(x, prob=T, br = (-1:4)+.5, col="skyblue2")
  points(0:4, dbinom(0:4, 4, 1/6), col="red")
 hist(y, prob=T, br = (-1:8)+.5, col="skyblue2")
  points(0:8, dbinom(0:8, 8, 1/6), col="red")
par(mfrow=c(1,1))

Bottom line: There are only $625$ chances in $1296$ to lose the first game and $1\,015\,625$ chances in $1\,679\,616$ to lose the second.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.