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In this question If I consider first total no of outcomes as $6^5$ , then I divided it be ${6\choose5} *5!$ since there are 5 similar dices so the outcome (1 2 3 4 5 ) will be similar to (5 4 3 2 1) .

Now what's wrong with this approach ?

Rowan
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radhika
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    One problem is that while $(1,2,3,4,5)$ is overcounted $119$ times, the throw $(1,1,1,1,2)$ is only overcounted four times, and $(1,1,1,1,1)$ isn't overcounted at all. The solution method to this problem is called "stars and bars", and can be freely googled (there are probably more than a few explanations even on this site). – Arthur Nov 26 '15 at 16:15
  • Are they rolled simultaneously. – Archis Welankar Nov 26 '15 at 16:17
  • In die rolling problems, it is usually easiest to think of the dice being distinguishable (or rolled in a specific order), even if that is not actually the case. The reason is that then the outcomes are all equally probable. As others have noted, if we don't consider order, then some outcomes are more likely than others (e.g. 1, 2, 3, 4, 5 (in any order) is more likely than 1,1,1,1,1). – paw88789 Nov 26 '15 at 16:43

4 Answers4

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The answer is $6 + 5 - 1 \choose 5$. The reason is, to exclude re-counting equivalent outcomes, we look at the number of dices that land on $1, 2, 3, 4, 5, 6$. Now, there are $6$ possibilities. There are $5$ dices. Initially, you might think "Aha! The answer is just $6 \choose 5$." But it's a little more complicated than that. Because two dices could land on the same number. So we use the well known, tried and tested, stars and bars method see stars and bars to count allowing for these conincidences, and the answer is as above.

Colm Bhandal
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  • Nice to mention stars and bars theory. But does this case qualify this theorem? IMO, it seems that you miss a $\sum$ – Rowan Nov 26 '15 at 16:46
  • @Rowan. As I interpret the question, we are counting the number of distinct possibilities, ignoring the identities of the individual dices. That is, we just care about the number of dices that land on $1$, the number of dices that land on $2$ etc. all the way up to $6$. In this case, theorem 2 of the stars and bars does the trick. – Colm Bhandal Nov 26 '15 at 16:54
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Answer expanded

Choosing various combos of numbers, and permuting them in the sequence, we get

$5$ of a kind: $\binom61*\frac{5!}{5!} = 6$

$4-1$ of a kind: $\binom61\binom51*\frac{5!}{4!}=150$

$3-2$ of a kind: $\binom61\binom51*\frac{5!}{3!2!} = 300$

$3-1-1$ of a kind: $\binom61\binom52*\frac{5!}{3!}=1200$

$2-2-1$ of a kind: $\binom62\binom61*\frac{5!}{2!2!} = 1800$

$2-1-1-1$ of a kind: $\binom61\binom53*\frac5{2!} = 3600$

$1-1-1-1-1$ of a kind: $\binom65$*5! = 720

It adds up to 7776 = 6^5, as it must

Conclusion
There are $7$ types of outcomes adding to $252$ (which could have been obtained more simply using "stars and bars",) but the types are not equiprobable, and their distribution over an equiprobable sample space has been worked out, as the question is not very clear as to what exactly is meant here by an outcome

true blue anil
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  • Sir for calculating 4 of 1 kind , first we shall chose one number and then we have 5 places so we can put it in 5C4 ways ,now the remaining 1 place can be filled in 5C1 ways , and then we shall permute them as well so dividing it by 4! , what's wrong in this ? – radhika Nov 27 '15 at 08:19
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    You will unnecessarily get confused, I gave a simple way. In your way, you need to consider all 5 slots, 4-1-0-0-0, choose and place the 4 of a kind (say four 6's) in $\binom61\binom51 = 30$ ways, choose and place the 1 of a kind (say, 3) in $\binom51\binom41 = 20$ ways, then divide by $\binom52 = 10$ (the ways the 2 numbers can occupy the 5 slots), but why even think of such a tortuous way ? – true blue anil Nov 27 '15 at 09:08
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You are assuming 5 distinct values. There are more than a few cases to consider here assuming 6 sided dice where things can get tricky:

  1. 5 distinct values. In this case there are 6 possible combinations as it is just a matter of which value isn't showing up though there are 120 ways to roll the 5 dice to get that outcome.

  2. A pair with 3 distinct other values, e.g. 1 1 2 3 4.

  3. 3 of a kind with 2 distinct other values, e.g. 1 1 1 2 3.

  4. A pair of pairs with one other value, e.g. 1 1 2 2 3.

  5. A pair with 3 of a kind, e.g. 1 1 1 2 2.

  6. 4 of a kind with one other value, e.g. 1 1 1 1 2.

  7. All 5 dice show the same value which would be known as a Yahtzee! in that game.

These are all different cases that you'd have to add together.

JB King
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  • Sir I did 1) 2 same values= 65C25P3 ways 2) 3 same values= 65C35P2 ways 3)4 same values= 65C45P1 ways 4) 5 same values= 65C55P0 ways – radhika Nov 27 '15 at 08:34
  • Now I summed these all and subtracted from 6^5 but didn't get the answer , where am I wrong ? – radhika Nov 27 '15 at 08:35
  • You only note 4 cases where I listed 7! 3 same values has more than one case as (11223) and (11123) could be re-arranged differently that is something to note that I'm not sure you get here. You have to identify the patterns and how often would each occur. – JB King Nov 27 '15 at 15:30
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My explanation is that the probability of 1 set of the same value is 1/(number of possible combinations) simply. We are talking about 5 dice with six possible values so the calculation would be 1/(6^5).

In more detail, there are 6 combinations for 1 dice and there are 5 dice so you times the 6 combos by itself 5 times to get to the total combinations possible in the set.

6^5=7776 so the answer is 1/7776.

To calculate the probability from a certain number of throws add the probabilities of separate throws. Let’s say there is one combination (1, 1, 1, 1, 1). We already know that the probability of getting this is 1/7776, but what about in two throws?

In two throws getting 5 of a kind (for example: 1, 1, 1, 1, 1) you can think of two sets of combinations which give the value of one single combination (for example: 1,1 and 1,1,1). The probability of each part added will give the total and you will find that for this particular example it is 1/250 because it is 1/((6^2)+(6^3)).

However, for the second possible case (1, and 1,1,1,1) the probability is lower because increasing powers don’t follow a sequence and four 1s in one throw has a much lower probability than 1,1,1. This case has a probability of 1/1302.

Otis Webb
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