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I am thinking of a variation of the dice game, where

one has the option to throw a die unlimited number of times.The first throw is free and every next throw costs 1 dollar. One will earn the face value of the die and has the option to stop after each throw and walk away with the money earned. The earnings are not additive. What is the expected payoff of this game?

If I calculate the expected value as

for 2 rolls as  (1/6)(3+4+5+6)  +   (1/3) (3.5-1) = 3.83$ 
for 3 rolls as  (1/6)(3+4+5+6) +   (1/3) (3.83-1) = 3.94$
for 4 rolls as  (1/6)(3+4+5+6) +   (1/3) (3.94-1) = 3.98$
for 5 rolls as  (1/6)(3+4+5+6) +   (1/3) (3.98-1) = 3.99$

it asymptotically tends to 4

Is this approach correct?
Shoudn't the player go bankrupt after 7 rolls (negative expectation)?

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  • $\begingroup$ I don't follow your method here. Can you explain your reasoning? $\endgroup$ Oct 27, 2016 at 7:36
  • $\begingroup$ @ Matthew Conroy let's take 2 rolls. The expectation of a single trow is 3.5, but it costs 1, therefore the expectation is 2.5. Thus, if one gets 1 or 2 in the first try then one keeps rolling because the expectation is higher (2.5). If 3,4,5, or 6 comes one stops and takes the payoff, so $$(\frac{1}{6} 3+\frac{1}{6} 4+\frac{1}{6} 5+\frac{1}{6} 6) + \frac{2}{6}(3.5-1) = 3.83$$ $\endgroup$
    – Michal
    Oct 28, 2016 at 2:41
  • $\begingroup$ Sometimes they will lose money, but on average they will net $4$. See my answer below. (There is no "bankrupt" here, since you have not specified how much money the player has to begin with.) $\endgroup$ Oct 29, 2016 at 20:39

1 Answer 1

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Your calculation is correct.

I'd do it this way.

Let $E_n$ be the expected net if we roll until we get $n$ or greater.

Suppose $n=6$. We roll once. If it's a $6$, we are done. If not, we are in exactly the situation we started in, except that our "first" roll isn't free: it cost $1$ dollar. Hence $$ E_6 = \frac{1}{6}(6) + \frac{5}{6}(E_6-1).$$ Solving, we find $E_6=1$. (Alternatively: on average it takes $6$ rolls to get a $6$, and all but the first roll is free, so $E_6=6-5=1$.)

Similarly, \begin{align} E_5 &= \frac{2}{6}\left(\frac{5+6}{2} \right) + \frac{4}{6}(E_5-1) & \text{ so } &E_5=\frac{7}{2}=3.5 \\ E_4 &= \frac{3}{6} \left( \frac{4+5+6}{3} \right) + \frac{3}{6}(E_4-1) & \text{ so } &E_4=4 \\ E_3 &= \frac{4}{6} \left( \frac{3+4+5+6}{4} \right) + \frac{2}{6}(E_3-1) & \text{ so} &E_3=4 \\ E_2 &= \frac{5}{6} \left( \frac{2+3+4+5+6}{5} \right) +\frac{1}{6}(E_2-1) & \text{ so } &E_2=\frac{19}{5} = 3.8 \\ \end{align} and, of course, $E_1=3.5$.

So the best strategy appears to be "roll until $4$ or more" with an expected net of $4$. Sometimes you will roll, say, 10 times with this strategy and lose money, but the average net is $4$.

Here is the result of $10^6$ simulated plays with the $4$ strategy:

net/number of occurrences

-21 1

-20 0

-19 0

-18 0

-17 0

-16 1

-15 0

-14 1

-13 5

-12 6

-11 8

-10 19

-9 32

-8 80

-7 146

-6 293

-5 579

-4 1122

-3 2272

-2 4517

-1 9147

0 18256

1 36474

2 72912

3 145747

4 291215

5 250332

6 166835

with an average net of $4.000515$.

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  • $\begingroup$ @ Matthew Conroy I think you have a mistake in $E_2$ (in case of 2 you keep rolling, so 2 shouldn't be in the first bracket). The expectation keeps going up, asymptotically tending to 4 as n increases. on the other hand is it not at odds with common sense? any strategy with more than 7 rolls is loosing money, the higher n the bigger loss $\endgroup$
    – Michal
    Oct 28, 2016 at 10:58
  • $\begingroup$ @ Matthew Conroy secondly, why in your calculation the options you decide to pay off are decreasing with every next roll? for the 2 roll case E(x) value of the second roll is 2.5 so you pay off straight away after first roll if 3,4,5,6 come. for the 3 roll case and the next ones the E(x) is below 4 so you keep rolling in case of 1,2,3 and pay off in case of 4,5,6, correct? $\endgroup$
    – Michal
    Oct 28, 2016 at 14:10
  • $\begingroup$ My $E_2$ is the expected value if we roll until we get $2$ or greater. In this case, we stop when we get a $2$. $\endgroup$ Oct 28, 2016 at 22:13
  • $\begingroup$ Regarding your claim "the higher n the bigger loss", consider: if I've rolled a 1 (for example), it is always to my advantage, on average, to roll again, regardless of how many rolls I've taken already. Cheers! $\endgroup$ Oct 29, 2016 at 4:44
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    $\begingroup$ If you roll $1$ six times in a row, then you will lose money. However, on average, you will lose less money if you continue to roll for a higher number. The same goes for rolling $2$s. The $E_2$ strategy is not optimal, as the calculation above shows. It is still a strategy. If you still don't agree with these calculations, I recommend you write a simulation and try out these strategies yourself. Sometimes the $4$ strategy loses money, but on average the net is $4$ dollars. Cheers! $\endgroup$ Oct 29, 2016 at 20:05

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