You and your friend play a game in which you and your friend take turns rolling a fair six-sided die and keep a running tally of the sum of the results of all rolls made. Play continues until either player wins if, after the player rolls, the number on the running tally is a multiple of 7. Should you start first or it is better if you let your friend rolls the die first.


This game has two states.

Starting: The total is 0, and whatever the first player rolls, the turn passes to the next player and the state changes to standard.

Standard: The total is greater than 0, but not a multiple of 7. The player rolling has a 1/6 chance to win; otherwise, the turn passes to the other player.

Does that make the strategy clear?

  • $\begingroup$ I see.But how would we go about to result to the numerical probability that person 2 wins? $\endgroup$ – John Chang Nov 16 '12 at 1:26
  • $\begingroup$ I got it already. Thanks! :) $\endgroup$ – John Chang Nov 16 '12 at 1:31
  • 1
    $\begingroup$ John, if you have it, let me encourage you to write it up and post it, for the benefit of those who come after us. $\endgroup$ – Gerry Myerson Nov 16 '12 at 1:46

I thought I'd answer this question as another opportunity to learn some math that's way out of my league in hopes that some of my usual favorites will evaluate it and see if I'm on the right track.

The first thing I did was consider a few trials:

  • Player 1 has a 0/6 chance of winning
  • Player 2 has a 1/6 chance of winning
  • Player 1 has a 1/6 chance of winning
  • Player 2 has a 1/6 chance of winning

The probability of player 1 winning a 1 round game is 0. The probability of player 1 winning a 2 round game is $\mathbb{P} = \mathbb{P}(\text{Winning Round 1}) + \mathbb{P}(\text{Rounds 1 and 2 Losses}) \cdot \mathbb{P}(\text{Winning Round 3}))$.

I figured that continues infinitely with the outcomes becoming less and less likely as the number of trials approaches infinity.

So, I came up with this: $$\mathbb{P}(\text{Player 1}) = \dfrac{0}{6} + \dfrac{6}{6} \cdot \dfrac{5}{6} \cdot \dfrac{1}{6} + \dfrac{6}{6} \cdot \left(\dfrac{5}{6}\right)^3 \cdot \dfrac{1}{6} + \ldots $$

$$\mathbb{P}(\text{Player 1}) = 0 + \dfrac{5}{6} \cdot \dfrac{1}{6} + \left(\dfrac{5}{6}\right)^3 \cdot \dfrac{1}{6} + \ldots = \dfrac{1}{6} \cdot \left( \dfrac{5}{6} + \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right) $$

So, then I had to figure out how to sum an infinite series: $\left( \dfrac{5}{6} + \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right) $

I did some research on how exactly that's done and the best I could come up with is to multiply the entire series by $\left(\dfrac{5}{6}\right)^2$ and subtract that from the original series. So, I think this is valid:

$$\left(\dfrac{5}{6}\right)^2 \cdot \left( \dfrac{5}{6} + \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right) = \left( \left(\dfrac{5}{6}\right)^3 + \left(\dfrac{5}{6}\right)^5 + \ldots \right)$$

If I subtract the new series from the original, all of the elements but the first should cancel out and I should have this:

$$\mathbb{S} - \left(\dfrac{5}{6}\right)^2 \cdot \mathbb{S} = \dfrac{5}{6}$$ $$\mathbb{S} \cdot \left[1 - \left(\dfrac{5}{6}\right)^2\right] = \dfrac{5}{6}$$ $$\mathbb{S} = \dfrac{\dfrac{5}{6}}{\left[1 - \left(\dfrac{5}{6}\right)^2\right]}$$

If I plug that into my original formula, I get this: $$\mathbb{P}(\text{Player 1}) = \dfrac{1}{6} \cdot \dfrac{\dfrac{5}{6}}{\left[1 - \left(\dfrac{5}{6}\right)^2\right]} = \overline{.45}$$

Because there are only two players this should be true: $\mathbb{P}(\text{Player 2}) = 1 - \mathbb{P}(\text{Player 1}) = \overline{.54} $

Unsure of my original conclusion, I thought I'd calculate the probability of Player 2 winning to verify.

$$\mathbb{P}(\text{Player 2}) = \dfrac{6}{6} \cdot \dfrac{1}{6} + \dfrac{6}{6} \cdot \left(\dfrac{5}{6}\right)^2 \cdot \dfrac{1}{6} + \dfrac{6}{6} \cdot \left(\dfrac{5}{6}\right)^4 \cdot \dfrac{1}{6} + \ldots $$

$$\mathbb{P}(\text{Player 2}) = \dfrac{1}{6} + \dfrac{1}{6} \cdot \left(\left(\dfrac{5}{6}\right)^2 \cdot \left(\dfrac{5}{6}\right)^4 + \left(\dfrac{5}{6}\right)^6 + \ldots \right) $$

To sum that series I thought: $\mathbb{S} = \dfrac{25}{36} + \left(\dfrac{25}{36}\right)^2 + \left(\dfrac{25}{36}\right)^3 + \ldots $

So, $\mathbb{S} - \dfrac{25}{36}\mathbb{S} = \dfrac{25}{36}$ and $\mathbb{S} = \dfrac{\dfrac{25}{36}}{1 - \dfrac{25}{36}}$

I plug that into the original formula and find that $\mathbb{P}(\text{Player 2}) = \overline{.54}$ which is what I expected from before.

  • $\begingroup$ That's a lot of work! Alternate solution: since round 1 can never win and future rounds are all identical, the best position to be in is the first player to play round 2. Alternate solution giving probability: for any round but the first, consider the next two turns. The player starting the round (not the game) has 6 ways to win. The other player has 5 ways to win. Anything else leads back to the same state and thus can be ignored, giving one player 6/(6+5) and the other player 5/(6+5). $\endgroup$ – Charles Nov 16 '12 at 15:55
  • $\begingroup$ I thought you were ultimately looking for the actual probabilities. :) $\endgroup$ – D. Patrick Nov 16 '12 at 15:57
  • $\begingroup$ I don't know if John needed to know that or not. The series method you give is correct, I just thought I'd mention an easier (potentially) approach. $\endgroup$ – Charles Nov 16 '12 at 16:02
  • $\begingroup$ Oops. Sorry. I mixed up who was whom. Well, thanks for the feedback. Like I mentioned, I answer these questions primarily as a way to learn more about math (turns out I like it more as an adult than I ever did in school). $\endgroup$ – D. Patrick Nov 16 '12 at 16:05

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