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Suppose player X has a 6 sided die and player Y has a 10 sided die. They each get two rolls and they can each choose to stop rolling on either one of the rolls, taking the number on that roll. Whoever has the higher number wins, when there is a tie X wins. What is the probability that Y wins this game?

I even couldn't start computing probabilities since X's decision is based on Y's decision. I assume that they are both rational. How can we proceed?

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  • $\begingroup$ Are they rolling simultaneously, or does it proceed in a specific order? $\endgroup$
    – Nick Peterson
    May 25, 2014 at 18:18
  • $\begingroup$ First X and Y roll simultaneously. Then they decide whether to keep it or roll it again. Then they roll $\endgroup$
    – neticin
    May 25, 2014 at 18:19
  • $\begingroup$ Okay... but when they decide whether or not to re-roll, does one decide first? Because if not, then neither of them knows the others' decision when they make theirs. $\endgroup$
    – Nick Peterson
    May 25, 2014 at 18:20
  • $\begingroup$ After first roll, they both look each other's dice and they decide whether to roll again or not. But they don't share their decisions with each other $\endgroup$
    – neticin
    May 25, 2014 at 18:24
  • $\begingroup$ You can't compute the probability without knowing their strategies. Presumably, you mean if they both choose optimal strategies? $\endgroup$
    – Thomas Andrews
    May 25, 2014 at 18:27

5 Answers 5

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The probability that X will win is 7/20 and the probability that Y will win is 13/20.

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  • $\begingroup$ The game would be fair if Y's die had 7 sides. $\endgroup$
    – Locke
    May 25, 2014 at 19:42
  • $\begingroup$ These are the probabilities if there is only one round. But that's not the game in the question. Each player can choose to reroll after seeing the initial results. $\endgroup$
    – Meni Rosenfeld
    May 26, 2014 at 9:50
  • $\begingroup$ It makes no difference if you roll again or not. The numbers are the same. $\endgroup$
    – Locke
    May 26, 2014 at 14:19
  • $\begingroup$ You seem to be thinking a player can choose to force both players to reroll. No, each player chooses whether to reroll his own die or not. The result will depend on the original rolls and each player's corresponding choice. In some cases the choice is easy, in others there's a mixed strategy equilibrium. $\endgroup$
    – Meni Rosenfeld
    May 26, 2014 at 15:20
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For concreteness, let's assume neither player shows the other their first roll, and neither player knows, before committing to keep or re-throw the first roll, whether the other player has decided to re-throw.

Then each player's strategy consists of choosing what is the minimum number he will keep if she rolls that on the first roll. The game matrix is in principle 6 by 10, but it can easily be shown that Y should always keep a 6 or better, and always discard a 3 or less, and that X should always re-roll a 3 or less. The remaining matrix is 3x3, and is 1/3600 times $$ \left( \begin{array}{c|c|c|c|c|} & & & Y & \\ & & 4 & 5 & 6 \\ \hline \\ & 6 & 925 & 834 & 899 \\ \hline \\ X & 5 & 950 & 888 & 988 \\ \hline \\ & 4 & 845 & 828 & 988 \\ \end{array} \right) $$ It is easy to see that keeping 5 dominates for Y, and then keeping 5 dominates for X. (This did not have to be the case; for some combinations of dice the proper strategies come out to be mixed.)

So the optimal strategies are that each player keeps a 5 or higher on the first roll and re-rolls a 4 or less. And the value of the game for X is $\frac{888}{3600} = \frac{37}{150}$ or very nearly 1/4. Not 7/20. Also, pertaining to the other comment, with a 7-sided die vs a 6-sided die, the correct strategies are again to keep 5 and up for both players, and the value is $\frac{45}{98}$ or about .458 -- not at all a fair game.

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  • $\begingroup$ But according to the OP's description, both players see both rolls before making the decision. $\endgroup$
    – Meni Rosenfeld
    May 26, 2014 at 9:51
  • $\begingroup$ If you re-roll on <3 then there is still a chance of getting <3, in fact the probability is still the same if i'm not mistaken. $\endgroup$
    – Locke
    May 29, 2014 at 2:14
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It all depends on their strategies.

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The best approach is to treat it as 60 unrelated games, one for each combination of results for the first throw. For each such game, each player has two strategies, keep or throw.

For each game we will write the payoff matrix, and find the mixed-strategy Nash equilibrium and the resulting value. Then the value for the entire game will be the mean of values for the 60 games.

Let us denote by a X's first throw and by b Y's throw. There are several cases:

Case 1. b > 6:

Y keeps and wins. (Value 0, Probability 40%)

Case 2. a < b <= 6:

The payoff (for X) is:

$$ \left( \begin{array}{c|c|c|c|} & & Y & \\ & & K & T \\ \hline \\ X & K & 0 & a/10 \\ \hline \\ & T & (7-b)/6 & 0.35\\ \end{array} \right) $$

Subcases:

b = 5,6:

Y keeps, X throws. Mean value (5 * 1/6 + 4 * 2/6) / 9 = 13/54, probability 9/60.

b <= 4 (so a < 4):

X throws. Y throws. Value is 0.35, probability 1/10.

Case 3. b <= a:

The payoff matrix: $$ \left( \begin{array}{c|c|c|c|} & & Y & \\ & & K & T \\ \hline \\ X & K & 1 & a/10 \\ \hline \\ & T & (7-b)/6 & 0.35\\ \end{array} \right) $$

Subcases:

a>3:

X keeps. Y throws. Mean value (4 * 4/10 + 5 * 5/10 + 6 * 6/10)/15 = 77/150. Probability 1/4.

a<=3:

This is the interesting case which requires finding a mixed-strategy Nash equilibrium. X keeps with probability (5 (-1 + b))/(25 - 3 a + 5 b), Y keeps with probability (3 (-10 + a))/(-25 + 3 a - 5 b). Value is (60 + a (-7 + b))/(-6 a + 10 (5 + b)). There are 6 possibilities in the range b<=a<=3, their mean value is 257191/280488 and probability 1/10.

So the mean value is: 0*4/10 + 13/54*9/60 + 21/60*1/10 + 77/150*1/4 + 257191/280488*1/10 = 12249131/42073200.

X has a probability of 12249131/42073200 to win, which is roughly 29.114%.

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The probability that Y wins in the first round by rolling 7-10 $= 0.40$

Probability that Y rolls winning number besides 7-10 $= 0.25$

Probability that Y rolls unwinning number besides 7-10 $= 0.35$

For the last two cases, Y may choose to go for the second round.

In the second roll,

Probability that Y rolls winning number from 7-10 = $0.40$

Probability that Y rolls winning number from 1-6 = $0.25$

Thus Probability that Y rolls winning number in the second round given he rolled 1-6 in the first round $= (0.4+0.25)*0.25$

In the first round Y does not roll a winning number from 1-6 $= 0.35$

Probability that Y rolls winning number (7-10) and winning number in (1-6) given Y does not roll winning number $= (0.4+0.25)*0.35$

Thus the probability that Y wins $$= 0.4+ (0.65)*0.25 + (0.65)*0.35 = 0.79$$

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  • $\begingroup$ Your calculation assumes that Y, and only Y, gets a mutual reroll if he loses on the first throw. That's not how the game works. Each player chooses whether to reroll his own throw. $\endgroup$
    – Meni Rosenfeld
    May 26, 2014 at 9:49
  • $\begingroup$ Ok, the calculation is more nuanced after the edit, but I still don't believe it's correct. According to the OP's description, you need to find the Nash equilibrium for each scenario. $\endgroup$
    – Meni Rosenfeld
    May 26, 2014 at 10:06

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