# Dice throwing calculations

I made some calculations on dice throwing probabilites (they are fair dices) and I want to know if they are correct:

Let $(X_1, X_2)$ be the throwing of $2$ dices. Then the probability of having 2 of a kind $\{(1,1),(2,2),(3,3),(4,4),(5,5),(6,6))\}$ is

$$P(X_1=X_2)=P(X_2=x\mid X_1=x)=\frac{P(X_2 = x\cap X_1=x)}{P(X_1=x)}=\frac{6}{1}\cdot \frac{1}{36}=\frac{1}{6}$$

Let $(X_1,X_2)$ be the same as before. Then the probability of having successively two 2 of a kind in 5 trials (trial = two dice throwing) meaning a successful event looks like this: $$\{(x_1,x_2),(x_3,x_4),(x_5,x_5),(x_6,x_6),(x_7,x_8)\}$$ where $x_i\neq x_j \forall i \neq j$:

The probability of throwing two 2 of a kind in a row: $$P(X_{1}=X_{2})^2=\frac{1}{6^2}=\frac{1}{36}$$ Then we have $5-1=4$ chances to have two 2 of a kind: $$\sum\limits_{i=1}^4 P(X_{1}=X_{2})^2=\frac{4}{36}$$

• I don't think your answer to the second question is correct. Multiplication is needed, not addition
– joeb
Commented Feb 19, 2017 at 19:43
• @joeb Well I have multiplication for the probability of throwing 2 times a 2 of a kind $P(X_1 = X_2)^2 = \frac{1}{36}$. Then my thought is that I have 4 times the chance to throw 2 times a 2 of a kind. Commented Feb 19, 2017 at 19:46
• @Ekesh Extend the logic. Then the probability of throwing two 2-of-a-kind in a row in 10 trials would be $9/36$. And in 20 trials would be $19/36$. And in 40 trials would be $39/36$. This can't be! Commented Feb 19, 2017 at 21:50
• @Ekesh Try this: In three trials you would have 2 chances, so your prediction of the probability would be $2/36\approx0.0556$. But there are three ways to achieve two consecutive 2-of-a-kinds: $YYY$, $YYN$ and $NYY$. The probabilities are $P(YYY)=1/216$, $P(YYN)=5/216$ and $P(NYY)=5/216$, for a total probability of $11/216\approx0.0509$. You are overestimating the probability. Commented Feb 19, 2017 at 22:16

Let's consider all the possible ways we can get 2 consecutive 2-of-a-kind out of 5 trials.

I denote a successful 2-of-a-kind as $Y$ and a lack of 2-of-a-kind as $N$.

We could get $YY$, which is two 2-of-a-kinds right away, with probability $(1/6)(1/6)=1/36$.

We could get $NYY$ with probability $(5/6)(1/6)(1/6)=5/216$.

We could get $YNYY$ with probability $(1/6)(5/6)(1/6)(1/6)=5/1296$ or $NNYY$ with probability $(5/6)(5/6)(1/6)(1/6)=25/1296$.

Finally, we can get $YNNYY$ with probability $(1/6)(5/6)(5/6)(1/6)(1/6)=25/7776$ or $NYNYY$ with probability $(5/6)(1/6)(5/6)(1/6)(1/6)=25/7776$ or $NNNYY$ with probability $(5/6)(5/6)(5/6)(1/6)(1/6)=125/7776$.

Summing all of these probabilities gives a total probability of $751/7776\approx0.0966$.

• But we also can get YYNNN, YYNYN, YYNYY, YYYYY...? Commented Feb 19, 2017 at 20:49
• @WaldoRozir But YYNNN means we achieved two 2-in-a-kind throws right away. Calculating YY covers all YYxxx possibilities. Commented Feb 19, 2017 at 21:47
• Someone downvoted your answer. Now I don't know whether it is correct or not. Commented Feb 20, 2017 at 8:33
• @WaldoRozir Well, I feel confident. Perhaps I'm misreading your original question, but I don't think so. In any event, good luck to you! Commented Feb 20, 2017 at 14:12

The first one is correct, assuming the dices do not have to take on a preset value.

The second one is also correct. The probability of rolling a two 2-of-a kind in a row is equal to $1/36$.

Now, note that there are 4 ways that this can happen

Therefore, the probability is equal to 4 * $\frac{1}{36}$ = $\frac{4}{36}$.

• You note in the three-roll situation that "we do NOT have the second roll equal to the first roll". In the four-roll situation, doesn't your calculation allow for something like rolling 2, 2, 4, 4 or 3, 3, 1, 1? Commented Feb 19, 2017 at 20:01
• Thanks for the answer. But I'm not sure if this is what I was looking for? This is the probability of getting two numbers in a row, right? I was looking for the probability of getting two 2 of a kind in a row. Meaning the probability of the event: $$\{(x_1,x_2),(x_3,x_4),(x_5,x_5),(x_6,x_6),(x_7,x_8)\}$$ where $x_i\neq x_j \forall i \neq j$ Commented Feb 19, 2017 at 20:02