I have just started a game with my wife. We rolled $2$ six sided dice twice in this order:

  1. She rolled $\{5,6\}$.
  2. Then i rolled $\{5,6\}$ as well.
  3. Then she rolled $\{2,5\}$.
  4. Then i rolled $\{2,5\}$ as well.

What were the odds of me getting the same numbers twice?


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  • $\begingroup$ Is 2 D6 short for 2 six sided dice? $\endgroup$ – yoshi Jul 29 '17 at 11:27
  • 1
    $\begingroup$ Yes, i mean 2 six sided dice. $\endgroup$ – jcandli Jul 29 '17 at 11:28
  • $\begingroup$ were the dice fair ? $\endgroup$ – user451844 Jul 29 '17 at 11:30
  • 1
    $\begingroup$ Dice were fair. $\endgroup$ – jcandli Jul 29 '17 at 11:32
  • 1
    $\begingroup$ Order doesn't matter : (5-2) equals (2-5) here. $\endgroup$ – jcandli Jul 29 '17 at 11:32

The key point is to determine the probability of a single match. Since "match" disregards order, that takes a computation. Rolls $(a,b)$ with $a\neq b$ have two matches, but rolls $(a,a)$ have only one. Thus there are two ways to get a match:

Either the first player throws a double and then the second throws the same double, $\frac 16\times \frac 1{36}$. Or the first player throws a non-double and then the second throws a match up to order, probability $\frac 56\times \frac 2{36}$

Thus the probability of a single match is $$\frac 16\times \frac 1{36}+\frac 56\times \frac 2{36}=\frac {11}{216}\approx .051$$

To do it twice we must square, so the final answer is $$\left(\frac {11}{216}\right)^2=\frac {121}{46656}\approx .0026$$

  • $\begingroup$ except with the order not mattering there's only 21 possibilities: (1,1),(1,2),(1,3),(1,4),(1,5),(1,6),(2,2),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6),(4,4),(4,5),(4,6),(5,5),(5,6),and (6,6) $\endgroup$ – user451844 Jul 29 '17 at 11:41
  • $\begingroup$ @RoddyMacPhee Not following. If you throw a $(2,5)$ then there are two things I can throw to match you. Thus the probability of my matching you is $\frac 2{36}$, yes? $\endgroup$ – lulu Jul 29 '17 at 11:42
  • $\begingroup$ @RoddyMacPhee I am working with all the rolls, so each roll is equi-probable. Therefore I need to say that there are two rolls which match the non-double. $\endgroup$ – lulu Jul 29 '17 at 11:43
  • $\begingroup$ almost all of them are even in the 21 cases the only ones that aren't are when a=b. $\endgroup$ – user451844 Jul 29 '17 at 11:46
  • $\begingroup$ @RoddyMacPhee So what? With problems like these you have two ways to proceed. Either you can work with equi-probable rolls (as I did) in which case you can have multiple matches or you can work with the un-ordered rolls in which case there is only one way to match but the rolls are not equi-probable. Personal choice, but I always prefer to work with equi-probable events when possible. $\endgroup$ – lulu Jul 29 '17 at 11:48

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