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You play a game using a standard six-sided die. You start with 0 points. Before every roll, you decide whether you want to continue the game or end it and keep your points. After each roll, if you rolled 6, then you lose everything and the game ends. Otherwise, add the score from the die to your total points and continue/stop the game. When should one stop playing this game?

Let $X$ represents the number that comes up on the die.

Therefore the game continues until $X<6$,

So, $P(X=6)=nCr p^r q^{n-r}$ where $r=1$

$$ \dfrac{1}{6}=nC_1 \times \dfrac{1}{6} \times \biggr (\dfrac{5}{6}\biggr )^{n-1}$$

$$1=n \times \biggr(\dfrac{5}{6}\biggr)^{n-1}$$

$$\boxed {n = 1}$$

Am I wrong?


marked as duplicate by MJD, drhab probability May 30 '18 at 9:27

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  • 2
    $\begingroup$ The game continues “as long as” $X<6$ is a better syntax $\endgroup$ – John Cataldo May 30 '18 at 9:11
  • $\begingroup$ You could also consider looking at the probability distribution of the random sum for some overkill. $\endgroup$ – Tony Hellmuth May 30 '18 at 9:29
  • $\begingroup$ However, does my answer seem correct? $\endgroup$ – Busi May 30 '18 at 9:41
  • $\begingroup$ "$X$ represents the number that comes up on the die" But there might be several throws with a die, leading to several numbers. Should it not be a sequence $X_1,X_2,\dots$? $\endgroup$ – drhab May 30 '18 at 10:05
  • 1
    $\begingroup$ @Busi Well, you are the one that introduced $n$ ... so you should know what you meant by $n$ by using $n$ ... and what do you mean y 'both conditions ... given that you only give one? $\endgroup$ – Bram28 May 30 '18 at 17:21

Suppose that you gained $n$ points.

If you decide to go on that situation then the expectation is $$\frac16(n+1)+\cdots+\frac16(n+5)+\frac16\cdot0=\frac56n+\frac52$$

So you could decide to stop if $\frac56n+\frac52<n$ or equivalently $n>15$.

  • $\begingroup$ Does my answer seem correct? $\endgroup$ – Busi May 30 '18 at 9:40
  • $\begingroup$ I cannot discern what you are trying to answer. Is it your conclusion that one should stop after playing exactly $n=1$ game? $\endgroup$ – drhab May 30 '18 at 9:53
  • $\begingroup$ maybe it is. However, I found the answer as $n = 1$. Otherwise, you can comment on the way I found the answer. $\endgroup$ – Busi May 30 '18 at 10:00

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