- Let's say we are paying a game. I have two standard dice and you have one. We rolls the dices, if the sum of the pips on two die roll is greater than twice the pips on your single roll dice, I get the difference and vice a versa. For ex, I roll a (3,6) and you roll 4 then I get (3+6-2*4 = 1) and vice a versa. What is the expected value of the payout of this game.
I calculated the expected value of the sum of two dice rolls to be 7 using the following guidelines:
Rolling a 2: one chance (1&1)
Rolling a 3: two chances (2&1)(1&2)
Rolling a 4: three chances (3&1)(1&3)(2&2)
Rolling a 5: four chances (4&1)(1&4)(3&2)(2&3)
Rolling a 6: five chances (5&1)(1&5)(4&2)(2&4)(2&2)
Rolling a 7: six chances (6&1)(1&6)(5&2)(2&5)(4&3)(3&4)
Rolling an 8: five chances (6&2)(2&5)(5&3)(3&5)(4&4)
Rolling a 9: four chances (6&3)(3&6)(5&4)(4&5)
Rolling a 10: three chances (6&4)(4&6)(5&5)
Rolling an 11: two chances (6&5)(5&6)
Rolling a 12: one chance (6,6)
And the expected value of twice the pips on a single roll as 7 (2*3.5). Then took the difference (zero in this case) to be the expected value of the earnings in this game. My first question is this correct?
- Now let's say we raise the stakes a little bit, With same structure as above, we decide the winner would get double the difference(so in the above (3,6) and 4 rolls, I would get 2$). Here's the catch though, I roll the first of my two dice and based on that decide whether to raise the stakes to double the difference or to continue the game as mentioned in the first part. i. What would my strategy be in this case? I thought if first dice shows 1,2,3 then I would continue the game as is because in that case my sum to be higher than your average(7) would have lower probability than when I roll a 4, 5, 6. Is this the right way to approach it? ii. What would be the expected winnings in this case? I couldn't solve this but here's my approach:
$\frac{1}{6}*0+\frac{1}{6}*0+\frac{1}{6}*0+\frac{3}{6}*{expected earnings in the high stakes game}$
Wouldn't the expected earnings in the high stakes game also be zero though? Because I double my earnings but the opponent does it too.
Your analysis in problem 1 is correct.
In problem 2, your analysis of the optimal strategy (don't double unless roll is greater than 3) is also correct.
However, in problem 2, assuming that you use the optimal strategy, you have a positive expectation. First, from an intuitive perspective, this must be the case because in problem 1, when you roll a 3 or less on your first die, you have a negative expectation that is perfectly balanced by your positive expectation when you roll a 4 or more on your first die.
As to the actual math, all I have to do is prove that you have a positive expectation assuming that you roll a 4 or more and assuming that the payout is as in problem 1.
When you roll a $4$, your chance is $(1/6)$ of having the two dice total 5,6,..., 10.
When you roll a $5$, your chance is $(1/6)$ of having the two dice total 6,7,..., 11.
When you roll a $6$, your chance is $(1/6)$ of having the two dice total 7,8,...,12.
Therefore, under the assumption that you roll $4$ or higher, the probabilities are
$5 : (1/18)~~$ Win (3 + 1), Lose (1 + 3 + 5 + 7) : Net = -12.
$6 : (2/18)~~$ Win (4 + 2 + 0), Lose (2 + 4 + 6) : Net = -6.
$7$ : $(1/6)~~$ Win (5 + 3 + 1), Lose (1 + 3 + 5) : Net = 0.
$8$ : $(1/6)~~$ Win (6 + 4 + 2 + 0), Lose (2 + 4) : Net = 6.
$9$ : $(1/6)~~$ Win (7 + 5 + 3 + 1), Lose (1 + 3) : Net = 12.
$10$ : $(1/6)~~$ Win (8 + 6 + 4 + 2 + 0), Lose (2) : Net = 18.
$11 : (2/18)~~$ Win (9 + 7 + 5 + 3 + 1), Lose (1) : Net = 24.
$12 : (1/18)~~$ Win (10 + 8 + 6 + 4 + 2 + 0) : Net = 30.
The above chart makes the math obvious.