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I am looking to find the probability of winning a basic dice roll-off using a 6 sided die if one of the players can re-roll their die. The main thing that is throwing me off is that player 2 can re-roll the die but doesn't have to, and if the first roll or the re-roll equals player 1's roll then the process restarts.

Example 1: Player 1 - Rolls a 2 Player 2 - Rolls a 4 (win)

Example 2: Player 1 - Rolls a 2 Player 2 - Rolls a 1, re-rolls and gets a 5 (win)

Example 3: Player 1 - Rolls a 5 Player 2 - Rolls a 5 At this point player 2 can call it a tie and start fresh, or use his re-roll to attempt and roll a 6, although this doesn't seem to be his best option to win.

Example 4: Player 1 - Rolls a 4 Player 2 - Rolls a 4 At this point player 2 decides to call it a tie, and then they both re-roll. Player 2 still has the ability to then re-roll his result in this round. Player 1 - Rolls a 3 Player 2 - Rolls a 2, re-rolls and gets a 1 (loss)

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  • $\begingroup$ You write: "player 2 can re-roll the die but doesn't have to." Why would player 2 not re-roll in the case he is losing on his first roll? Isn't your question really: "What is the probability at least one of two rolls of die B will beat a single roll of die A?" $\endgroup$ – David G. Stork Feb 19 '16 at 21:35
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I gather that only the last roll of player 2 "counts."

If player 2 is losing after her first roll, she will certainly roll again. If player 2 is winning after her first roll, she will certainly not roll again. If player 2 ties after her first roll, she will roll again if she has a 1 or 2 or 3 (since there is a greater chance she'll tie or win on her second roll) and she will not roll again if she has a 4, 5 or 6 (since there is a greater chance she'll lose on her second roll).

One can enumerate all these results in a decision tree and simply count the wins and losses.

Player 1: ${\bf 1}$

Player 2:

  • ${\bf 1}$ followed by ${\bf 1}$ or $\{ {\bf 2}-{\bf 6} \}$: P[tie] = 1/6; P[win] = 5/6
  • $\{ {\bf 2}-{\bf 6} \}$ P[win] = 1

So if Player 1 rolls a ${\bf 1}$, the Player 2 has the following probabilities:

Tie: $1/6$

Win: $5/6$

Continue through other cases of player 1 rolling ${\bf 2}$ through ${\bf 6}$, and be sure to weight each result by the probability of the rolls.

Then add up the results!

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P1 = Player 1. P2 = Player 2

If P2 wins in the first roll, P2 will not roll again.
If P2 loses the first roll, P2 will roll again.
If P2 ties the first roll, then P2 will roll again if P1 rolled $1, 2$ or $3$

Let P1 roll a die. By taking cases

Each case has the following format

The die that P1 rolled: Probability that P2 will lose / tie / win

$$1: 0 / \frac{1}{36} / \frac{35}{36} $$ $$2: \frac{1}{36} / \frac{3}{36} / \frac{32}{36} $$ $$3: \frac{4}{36} / \frac{5}{36} / \frac{27}{36} $$ $$4: \frac{9}{36} / \frac{9}{36} / \frac{18}{36} $$ $$5: \frac{16}{36} / \frac{10}{36} / \frac{10}{36} $$ $$6: \frac{25}{36} / \frac{11}{36} / 0 $$

Adding probabilities of P2 losing, each case having a weighting of $\frac{1}{6}$ we get $\frac{55}{216}$
Similarly the probability of P2 winning will be $\frac{122}{216}$

Now coming to the point, what if the match was a tie? Well they'll keep playing until one of them is the winner. In each round the probability of P2 winning and losing will be the same. So one can change the problem to: "what is the probability of P2 winning if given P1 or P2 wins the round."

Using the conditional probability:

$A$ is the event that P2 wins
$B$ is the event that P1 or P2 wins

$$P(A/B) = \frac{P(A\cap B)}{P(B)} = \frac{122/216}{122/216+55/216} = \frac{122}{177}$$

Hence the probability of P1 winning is $\frac{55}{177} = 0.3107344...$
and the probability of P2 winning is $\frac{122}{177} = 0.6892655...$

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  • $\begingroup$ Your calculations are a bit off, not accounting for a loss after rolling again on a drawn $2$ or $3$. The probabilities should be $2: \frac{2}{36} / \frac{2}{36} / \frac{32}{36}$ and $3: \frac{6}{36} / \frac{3}{36} / \frac{27}{36}$ giving a final result of $\frac{122}{180}=0.6777...$ $\endgroup$ – Logophobic Feb 20 '16 at 15:06
  • $\begingroup$ A better result can be obtained if P2 does not roll again on a drawn $3$. The probability then is $3: \frac{4}{36} / \frac{8}{36} / \frac{24}{36}$ giving a final result of $\frac{119}{175}=0.68$ $\endgroup$ – Logophobic Feb 20 '16 at 15:09

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