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You are playing a fair die game with 'n' die rolls. You have decided on the "fair" price of the game. So, to take part in this game you need to pay upfront this value. Then you get to roll a die, n times. If you ever role a die more than or equal to this value you instantly cash out, and the game ends. If not, you keep rolling and take whatever is the nth die roll value. For ex:

3 Die Roll Game. You decided the fair value is 3 (this is not the actual fair price, just an illustrative example.) A person rolls a 2 in his first attempt. It is less than 3, he continues. Again rolls a 2, again continues. In his 3rd and last attempt he rolls a 1. He has to cash out now with a value of 1, and hence a net loss of 2 because his 3 rolls are done. (Note, at equality when the die's roll is 3 in this case, he cashes out)

What is the fair value of a game with n rolls, such that the expected profit should be zero.

P.S. This is not the simple choose your maximum out of n die rolls game. I have tried working it out, but I am missing it. Can anyone give me the basic logic that I need to write a code for this. I am looking for a cleaner proof than enumerate all the possibilities. I was thinking on the lines of Recursion.

Asaf Karagila
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user300817
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  • When you "cash out" how much do you win? Is it possible to win more than your original bet? – Kaynex Dec 24 '15 at 23:41
  • If I have understood correctly, then the "cutoff" is always effectively an integer, regardless of the expected value $E$, yes? That is, if $E=2.5$ then the cutoff is $3$. But then you need only compute the payout for integer cutoffs until you get an expected payout in the proper range (assuming some value $E$ exists of course. I don't think that is obvious). – lulu Dec 24 '15 at 23:46
  • Are you assuming the number of sides of the dice is 6, or is that also parameterized? – DanielV Dec 25 '15 at 00:12
  • Are you required to stop if you are making a profit? Say I only pay $2$ for a two roll game. If I get a $3$ on the first roll, I would like to roll again. Are you allowed to quit if you are at a loss? If I paid $4.5$ for a two roll game and get $4$ on the first roll, I am better off quitting than rolling again. I think the right question to ask is what is the fair price for an $n$ roll game where the player can stop at any time and collect the value of the last roll. – Ross Millikan Dec 25 '15 at 00:12

2 Answers2

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For $n=1$ the fair price is $3.5$, because that is the expectation of the single roll. As you increase the number of rolls, you should stop whenever the current roll exceeds the expected value of the following series. For $n=2$, you will stop if you get $4,5,6$ on the first roll, and roll again otherwise. The value is then $\frac 12 \cdot 5 + \frac 12\cdot 3.5=4.25$ Then if $n=3$ you should only stop on the first roll with a $5,6$, so the value is $\frac 13 \cdot 5.5 + \frac 23 \cdot 4.25=\frac {14}3$

Ross Millikan
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  • It's not clear how you are getting (1/2) * 5 for the 2 roll case. Why 5? The expected value of the 2nd roll is not (4+5+6)/3 = 5, right? Otherwise, the expected value of the first roll would be 2:: (1+2+3)/3. – VISQL Dec 01 '22 at 14:46
  • @VISQL: The expected value of the second roll assuming you keep it is $5$ because you keep $4,5,6$. You keep it with probability $1\frac 12$. The other term comes from the $\frac 12$ chance you don't keep it and roll again. – Ross Millikan Dec 01 '22 at 15:10
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For reasonably large $n$ I think it is clear that the answer you want, call it $E$, is between $5$ and $6$.

Assuming this, we see that the effective cutoff is $6$. That is, you stop if you ever throw a $6$, otherwise you get the last value.

The probability of reaching the last value is $(\frac 56)^{n-1}$. The probability that you get a $6$ earlier than the last value is $1-(\frac 56)^{n-1}$. Thus the expected payout of this game is $$E_n=(1+2+3+4+5+6)*\frac 16*(\frac 56)^{n-1}+6*(1-(\frac 56)^{n-1})$$

This is the answer you seek, provided $5≤E_n≤6$ . For $n=11$, say, this comes to about $5.125$, which is satisfactory. For $n=10$ the value is less than $5$ (barely) so you'll need to try a lower cutoff.

lulu
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