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1 die. Up to 3 rolls. Your winnings are equal to the value of a single roll. You can stop after roll 1 or roll 2. If you proceed to the next roll (i.e the 2nd or 3rd), you forfeit the previous value (no memory). Therefore, if you roll 3 times, the value of the 3rd roll is what you win.

What is the fair price of this game?

Strategy Since the expected value of a single roll is 3.5, I will stop on any roll if I get a 4, 5, or 6.

If Fair Price = Expected Value
And Expected Value = P(i_th roll) x E(i_th roll) i=1 to 3
Then Fair Value = P(1 roll) E(1st roll) + P(2 rolls) E(2nd roll) + P(3 rolls) * E(3rd roll)

E(i_th roll) = 3.5 for any roll
P(only 1 roll) = 1/2 :: must roll 4|5|6 P(only 2 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 4|5|6 P(3 rolls) = 1/2 * 1/2 :: must roll 1|2|3 then 1|2|3

$\frac{1}{2} * 3.5 + \frac{1}{4} * 3.5 + \frac{1}{4} * 3.5 = $

$\frac{1}{2}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} + \frac{1}{4}*\frac{7}{2} = $

$\frac{7}{4} + \frac{7}{8} + \frac{7}{8} = $

$\frac{14}{8} + \frac{7}{8} + \frac{7}{8} = $

$\frac{28}{8} = 3.50$

Whn trying to solve this before, I think I made a mistake and end up with something like $4\frac{5}{8} = 4.625$

Should 3.5 be the answer, or should the 1st and 2nd rolls be "wieghted" by what values you must roll to move on (1,2 or 3) instead of the general 3.5 expected value for a single roll?

VISQL
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  • Shouldn't the answer be:

    $$\frac{1}{2} \times 5 + \frac{1}{4} \times 5 + \frac{1}{4} \times 3.5=4.625 ?$$ What is wrong with this calculation?

     – Li Kwok Keung Dec 01 '22 at 14:30
  • Please explain. You don't need to take all 3 rolls. It's "up to". I therefore assert that your best strategy is to stop if you get 4,5, or 6 on roll 1, or 4,5 or 6 on roll 2. Right?

    You only rolls 3 times is the first 2 rolls were 3 or under. You roll only 2 times if the 1st was less than 3, and the 2nd is more than 3. You roll only once if it is 4 5 or 6.

     – VISQL Dec 01 '22 at 14:53
  • Please see my answer. – Li Kwok Keung Dec 01 '22 at 15:15

3 Answers3

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If you only have one throw left, then the expected amount you will have after the throw, if you do use it, is $\ \frac{7}{2}\ $. Therefore, if your previous throw was $3$ or less, you should use your last throw, but if it was $4$ or more you should not use your last throw, but accept the value of the previous one.

Thus, when you have two throws left, your expected winnings, if you decide to take the next throw, will be $$ \frac{1}{2}\times\frac{7}{2}+\frac{4+5+6}{6}=\frac{17}{4}\ . $$ Thus, if your first throw was four or less you should throw again, but if it was $5$ or $6$ you should not throw again, but accept the value your first throw. Your expected winnings, and the fair price of the game, is therefore $$ \frac{2}{3}\times\frac{17}{4}+\frac{5+6}{6}=\frac{14}{3}\ . $$

lonza leggiera
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  • "Expected winnings" for 2 throws left is 17/4? Why are we expecting a 4,5 or 6? If I have 2 throws left, there's a 50% chance of taking the third throw, right? $Ah. Interesting.$ – VISQL Dec 01 '22 at 15:06
  • The probability of each of $4$, $5$ and $6$ is $\ \frac{1}{6}\ $, in which case you will accept that amount and not throw again. Otherwise (with probability $\ \frac{1}{2}\ $), you will throw again for an expected gain of $\ 3.5\ $. – lonza leggiera Dec 01 '22 at 15:14
  • In the first equation. If there's a 50% chance to go to 3 throws, where is the counterpart 50% chance of not going to 3 throws? We need that, right? – VISQL Dec 01 '22 at 15:15
  • The $50%$ chance of not going to $3$ throws is precisely that of throwing a $4$, $5$ or $6$ on your second (given that you took it). – lonza leggiera Dec 01 '22 at 15:19
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There is $\frac{1}{2}$ chance that we get $4, 5$ or $6$ in the first throw and $\frac{1}{2}$ chance that we get $1, 2$ or $3$.

Thus for exactly one throw , the expected value is $$\frac{1}{2} \times \frac{4+5+6}{3}=\frac{1}{2} \times 5$$

For exactly two throws , we must get $ \{1, 2 \text { }\mathrm{or} \text { }3 \}$ in the first throw and $ \{4, 5 \text { }\mathrm{or} \text { }6 \}$ in the $2$nd throw.

The expected value for exactly $2$ throws is: $$\frac{1}{2} \times 5$$

For exactly 3 throws , we must get $ \{1, 2 \text { }\mathrm{or} \text { }3 \}$ in the first $2$ throws and accept whatever we get in the $3$rd throws.

So the expected value for exactly $3$ throws is $$\frac{1}{4} \times 3.5$$

Thus the final expected value is

$$\frac{1}{2} \times 5 + \frac{1}{4} \times 5 + \frac{1}{4} \times 3.5 = 4.625$$

Li Kwok Keung
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  • The 4 ,5 or 6 might be what we want, but still have a 1/6 chance or being rolled. If you use $XP(X)$ it's 1/6 (4+5+6) this is 15/6 or 5/2 as you've written.

    However, for exactly 2 rolls you didn't use $1/6 * (1+2+3) + 1/6 * (4+5+6) = 1 + 5/2 = 7/2$

    I don't see how the expected value for exactly 2 rolls can become 5/2. We force 1 rolls to have an average of 2, and one to have an average of 5. This is 7, over 2 rolls.

     – VISQL Dec 01 '22 at 16:00
  • I think I have misread the question. Thank you! – Li Kwok Keung Dec 02 '22 at 00:15
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The value of a single roll is $3.5$
With two rolls, keep the first if it is more than $3.5$, so the value is $$\frac12\times5+\frac12\times3.5=4.25$$ With three rolls, keep the first if it is more than $4.25$

Empy2
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  • Where are you getting 5 from? You get up to 3 rolls, but you can stop at any point. You would only move on to s subsequent roll is you were 3 or below. – VISQL Dec 01 '22 at 14:51
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    If you have two rolls, then you stop if you get $4,5$ or $6$, whose average is $5$. So it is $\frac16\times4+\frac16\times5+\frac16\times6+\frac36\times3.5$. So two rolls is worth $4.25$. Then with three rolls, you only keep $5$s and $6$s. A $4$ in the first roll is not as good as $4.25$ from the other two rolls. – Empy2 Dec 01 '22 at 14:58