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For 3 cases, i particularly referred here and probably the best answer but when I take n=4 i.e. no of times dice can rolled, for I get expected off as 4.94 and for n=5, i get 5.12 that means for n=6, logically it means payoff i expect should be 6, and for any case above 6, for example here n=1000, the fair value which i should pay is 6 as that will be expected payoff. Am i right here, not missing something?

pkg7724
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2 Answers2

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The expected payoff will never be exactly $6$, but will approach it in the limit of infinite rolls.

For $4$ and $5$ dice the exact expected payoffs are $\frac{89}{18}$ and $\frac{277}{54}$ respectively. Since $\frac{277}{54}>5$, the expected payoff for $n$ dice from $n\ge6$, $E_n$, is computed from $E_{n-1}$ as $$E_n=\frac66+\frac16E_{n-1}=1+\frac56E_{n-1}$$ So $E_6=\frac{1709}{324}$, for example. It is easy to show by induction that $E_n<6$ for all $n\ge1$, and that the sequence $\{E_n\}$ is strictly increasing, so $\lim_{n\to\infty}E_n=6$.

An intuitive reason for the expected payoff never being $6$ is that there is a nonzero probability of not getting a $6$ at all in any finite number of rolls, so sometimes you will have to settle for less than that.

Parcly Taxel
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  • Thankyou. Apparently i made a rookie mistake while saying the payoff would be 6. Another ques i should have added there, since the payoff is increasing concavely, for a case like n=1000, the player should continue to roll right till he get highest payoff, no matter probability of the same can be exhausting the total 1000 times still not getting 6, right? – pkg7724 Apr 24 '20 at 14:23
  • @pkg7724 Yes, like that. – Parcly Taxel Apr 24 '20 at 14:23
  • @Parcly Taxel No! You would not toss a 1000 times if your 999th toss is a 5. That’s what I explained in my answer. Because at 999th toss your expected payoff is just the same problem for n=2. So you’re happy to let go with a 5 as well. As 5 is larger than the expected pay off for n=2. – Vishaal Sudarsan Apr 25 '20 at 06:58
  • @VishaalSudarsan My reply was only for the first $995$ tosses. For the last $5$ tosses, as you said, we have to tread a little more carefully. – Parcly Taxel Apr 25 '20 at 08:07
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Adding the Parcly’s answer

In the given strategy you might not even toss the coin n times and choose to settle with something less than 6. That is, say you’re in your $(n-2)^{th}$ toss according to the strategy you would quit if $(n-2)^{th}$ toss is a 5. Since you have only 2 more tosses left and your expected pay off given this situation is less than 5, so even though you had $n$ tosses to begin with you’re happy to let go off with a 5 now given there are only 2 more tosses left.

Hence your expected payoff would always be less than 6 for a given finite number of tosses.

Vishaal Sudarsan
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