21 of 100: Two Ways to Roll Ten Dice

It is not necessary to calculate the exact probabilities of both situations. We can see that the ordering of the dice is much less strict in A as in B. More formally, if we list all the rolls in order, we see that any situation matching A would also match B (as an example, ABABABABAB would match both situations). However, many lists of rolls would match B but not A (for example, AAAAABBBBB). Therefore, the number of favorable outcomes is necessarily greater for B, and with the same number of possible outcomes (10 die rolls) the probability for situation B is greater .

Situation A: The probability of one roll of two dice coming up as 1 and 6 is: $\frac{1}{6}$ * $\frac{1}{6}$ * 2 = $\frac{1}{18}$ and since the rolls are independent, the probability of getting the same results 5 times is: $\frac{1}{18^{5}}$ .

Situation B: The probability is: $\frac{10!}{5!(10-5)!}$ * $\frac{1}{6^{5}}$ * $\frac{1}{6^{5}}$ = 252 * $\frac{1}{36^{5}}$ .

If we divide A by B, we get $\frac{2^{5}}{252}$ = $\frac{8}{63}$ < 1, which means Situation B is more likely than Situation A .

If you think of the question in terms of 5 different colors of dice. Scenario B with 2 of each color of dice would be equivalent to A if you required that you had a 1 of each color and a 6 of each color. Obviously this is more restrictive than just having any 5 be 1s. Therefore B is more likely.

Think of the given conditions as following-->

• Case 1

You roll two dice 5 times and, every time, one of the two comes up as 1 and the other as 6. *

You roll dice for the first time and you don't get the expected output i.e. 1 and 6, Game Over . You need not to make further move as it doesn't satisfy given condition.

• Case 2

You roll 10 dice all at once. 5 come up as 1s and the other 5 come up as 6s.

you roll 2 dice (2 out of 10, we will roll 2 dice at a time) and you get output 1 and 1. you can still continue , as here we need total five $1$ s and five $6$ s . So, even after rolling two dice, we still have chance for getting expected output after rolling all the dice.

Therefore we can clearly observe that Case 2 is more likely.

NO MATHS REQUIRED !!

Event B is more likely than event A, and this is more easily seen by modifying Experiment A for ease of comparison. In each of the following dice experiments, consider having 10 dice, numbered #1 to #10.

Event A' : You roll the 10 dice and place them in order. Each pair of dice (#1, #2), (#3, #4), (#5, #6), (#7, #8), and (#9, #10) come up as a 1 and a 6 (in either order).

We note the following:

Event B : You roll the 10 dice and place them in order. There are precisely five 1s and five 6s.

Now that we've restated Event A as Event A', we can more easily compare it to Event B by noting that

But there is at least one case of Event B (many, actually) that is not a case of Event A'. For example:

Thus the probability that Event B occurs is strictly greater than that of Event A' occurring.

Situation A : The probability of rolling two dice and getting a 1 and a 6 is $\frac{2}{36} = \frac{1}{18}$ since order doesn't matter. The probability of doing this 5 times in a row is $(\frac{1}{18})^5 = \frac{1}{18^5}$ .

Situation B : This problem is equivalent to dividing 10 similar objects (the dice) into 6 different boxes (the six faces of the die). You can use the "stars and bars" method to determine this. There are 15 spaces (10 objects and 5 dividers) in which to place 5 dividers. Therefore there are ${15 \choose 5}$ = 3003 ways to roll 10 dice. Only 1 of these ways is five ones and five sixes. The probability is $\frac{1}{3003}$

The probability of B > the probability of A.

Situation A can be one of the variations of situation B. In other words, situation B is a conclusion of situation A, and have less limits.

Outcome A with five pairs of 1,6 is a small subset of outcome B with five 1s and five 6s, so B is more likely.

Without calculating any probabilities, treat it as a permutation problem with five 1s and five 6s. Fully one hundred percent of the permutations will satisfy condition B, but only certain ones where the 1 and 6 are paired will satisfy A: 16,61,61,16,16 etc. Thus B is far more likely.

This doesn't take very long, though I started actually writing out an expression for A and realized that in A since you roll 2 at a time, you only have 2 ways to get 1 1 and 1 6 while in B you roll 10 at once, so you have 10 C 5 ways to get 5 1s and 5 6s and by what I think you know 10 C 5 is way larger than 2 so B has a lot more ways to occur

Because of in B the order doesn't matter B is more likely.