# Same number multiple times on a dice roll

I was playing a game last night that involved rolling ONE die/dice. Within about 10 rolls, the person I was playing with and I rolled a 5 or a 6 at least 50% of the time. It got me thinking.

I'm interested in knowing:

1. What is the probability of rolling any 2 numbers 50% of the time within 10 dice rolls (each time rolling one die/dice)?

2. What is the probability of rolling a specific number (e.g. 5) 50% of the time within 10 dice rolls? (each time rolling one die/dice)

• If I understood your first question correctly, then in order to not have any two numbers at least 50% of the time no number can appear more than twice (easy to validate). So $4$ numbers will appear exactly twice and two will appear once... can you pick it up from here? – Daugmented Aug 21 '16 at 13:05
• @Daugmented I don't do probability a lot, would you mind quickly taking a look over my answer to see if the method is correct? – Ovi Aug 21 '16 at 13:27
• It seems we understood the question differently. I thought the OP wanted any two numbers at least 50% of the time. I didn't check your final results (it would be much more useful for the OP if you described how you reached it) but I think you calculated the probability of a specific pair. – Daugmented Aug 21 '16 at 13:47
• @Daugmented Well I provided the formula, values for the variables, and links to Wolfram Alpha inputs showing the calculation. I'm not sure what else I could put, other than hand calculations. I think your interpretation is better. But I would've thought to calculate the probability of getting any pair could be calculated as ${6 \choose 2} \cdot P($getting one pair at least $50 \%$ of the time$)$, but this gives a result $>1$ – Ovi Aug 21 '16 at 14:12
• Thank you both. I'm not sure I fully understand the exchange but @Ovi has answered the question I intended to ask below. Forgive me if I was unclear. – Lara Aug 21 '16 at 15:30

I am going to answer Part $2$ first, then part $1$. We can use the Binomial Formula here:

$$P(x) = {n \choose x} p^x q^{n-x}$$

Where assuming a fair die,

$n = 10$, the number of trials

$x=5$, the number of successes, i.e. $50 \%$ of $10$

$p = \dfrac 16$, the probability of success i.e. rolling a $5$ (or any other specific number)

$q = \dfrac 56$, the probability of failure, i.e. NOT rolling a $5$ (or any other specific number)

Plugging it all in, we get $\approx .01302$, or about $1.302 \%$.

It's probably more accurate to find the probability of getting a $5$ at least $50 \%$ of the time, because you would be equally or even more impressed with any success rate $\ge 50 \%$. To do this, you want to find the probability of $(x=5)$ OR $(x=6)$ OR $\cdots$ OR $(x=10)$, which is equal to $P(5) + P(6) + \cdots P(10)$. Plugging this in, we get $\approx .01546$, or about $1.546 \%$.

To find the probability of getting any $2$ numbers, Daugmented gives a nice idea in the comments. But we can still brute force our way through with the Binomial formula:

For getting $2$ numbers exactly $50$ of the time, we have

$n = 10$, the number of trials

$x=5$, the number of successes, i.e. $50 \%$ of $10$

$p = \dfrac 26 = \dfrac 13$, the probability of success, i.e. rolling a $5$ or a $6$ (or any other pair of numbers)

$q = \dfrac 46 = \dfrac 23$, the probability of failure, i.e. NOT rolling a $5$ or a $6$ (or any other pair of numbers)

The probability of getting a pair of numbers exactly $50 \%$ of the time is $\approx .1366$, or about $13.66 \%$.

Calculating as in part $2$ above, the probability of getting a pair at least $50 \%$ of the time is $\approx .2131$, or about $21.31 \%$.

• Thank you very much Ovi. That's brilliant. Very very helpful exactly what I was looking for :) – Lara Aug 21 '16 at 15:29
• @Lara You're welcome! – Ovi Aug 21 '16 at 21:02