# Dice probability help [closed]

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?

## closed as off-topic by Sahiba Arora, TheGeekGreek, Glorfindel, Namaste, Claude LeiboviciJul 29 '17 at 13:56

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Sahiba Arora, TheGeekGreek, Glorfindel, Namaste, Claude Leibovici
If this question can be reworded to fit the rules in the help center, please edit the question.

• Is 2 D6 short for 2 six sided dice? – yoshi Jul 29 '17 at 11:27
• Yes, i mean 2 six sided dice. – jcandli Jul 29 '17 at 11:28
• were the dice fair ? – user451844 Jul 29 '17 at 11:30
• Dice were fair. – jcandli Jul 29 '17 at 11:32
• Order doesn't matter : (5-2) equals (2-5) here. – 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 $$\left(\frac {11}{216}\right)^2=\frac {121}{46656}\approx .0026$$

• 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) – user451844 Jul 29 '17 at 11:41
• @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? – lulu Jul 29 '17 at 11:42
• @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. – lulu Jul 29 '17 at 11:43
• almost all of them are even in the 21 cases the only ones that aren't are when a=b. – user451844 Jul 29 '17 at 11:46
• @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. – lulu Jul 29 '17 at 11:48