Simple dice problem

**FatherChaos** · October 11th 2009, 11:45 AM

Hi guys, I'm really bad at probabilities, so here goes my problem:

We simulate a poker game with 5 identical dice (with 6 different faces each). What are the probabilities of rolling the dice and getting :

1) 5 different numbers
2) A pair
3) Two pairs
4) Three of a kind

I got #1 and #2 right; I'm however having problems with the latter two, so I'm not too sure if I have the right way of thinking the first two.

For the first one, I did the following reasoning: There are 6 ways to get a first card, 5 ways to get a second card, 4 ways to get a third card, 3 ways to get a fourth card and 2 ways to get a fifth card. Thus the probability is $\frac{6*5*4*3*2}{6^5}$ .

For the second one : there are 6 ways to choose which card will be doubled, thus 1 way to get the second (doubled) card, and then 5 ways to get the third card, 4 to get the fourth and 3 to get the fifth. The probability is therefore $\frac{6*1*5*4*3}{6^5}$ .

However, I have two contradicting reasonings for #3, none of which are correct according to my answer book. I can think like this: there are 6 ways to choose which one will be doubled first, then there are 5 ways to choose which one will then be doubled, and 4 ways to choose the last card, bringing the probabilities to $\frac{6*5*4}{6^5}$ .
I also have this reasoning: there are $\binom{6}{2}$ ways to get the two cards who will be doubled, and 4 ways to choose the last card, giving probabilities of $\frac{\binom{6}{2}*4}{6^5}$ .

I'm thoroughly confused. Can anybody help?
Thanks!

**Plato** · October 11th 2009, 12:23 PM

This has been answered.

**FatherChaos** · October 11th 2009, 12:27 PM

Originally Posted by Plato

This has been answered.

This doesn't help me at all. I want to know what's right/wrong with my reasoning. Besides, the link in that post doesn't even work.

**Plato** · October 11th 2009, 04:25 PM

Originally Posted by FatherChaos

Besides, the link in that post doesn't even work.

The link works in my browser.
What you are asking of us takes a good deal of careful analysis.
If you cannot see what is wrong from that website, I don’t think we can help.
Sorry.

**FatherChaos** · October 11th 2009, 04:36 PM

Originally Posted by Plato

The link works in my browser.
What you are asking of us takes a good deal of careful analysis.
If you cannot see what is wrong from that website, I don’t think we can help.
Sorry.

All right, the website works now. However, I already have the answers

**Grandad** · October 11th 2009, 10:22 PM

Hello FatherChaos

Originally Posted by FatherChaos

Hi guys, I'm really bad at probabilities, so here goes my problem:

We simulate a poker game with 5 identical dice (with 6 different faces each). What are the probabilities of rolling the dice and getting :

1) 5 different numbers
2) A pair
3) Two pairs
4) Three of a kind

I got #1 and #2 right; I'm however having problems with the latter two, so I'm not too sure if I have the right way of thinking the first two.

For the first one, I did the following reasoning: There are 6 ways to get a first card, 5 ways to get a second card, 4 ways to get a third card, 3 ways to get a fourth card and 2 ways to get a fifth card. Thus the probability is $\frac{6*5*4*3*2}{6^5}$ .

For the second one : there are 6 ways to choose which card will be doubled, thus 1 way to get the second (doubled) card, and then 5 ways to get the third card, 4 to get the fourth and 3 to get the fifth. The probability is therefore $\frac{6*1*5*4*3}{6^5}$ .

However, I have two contradicting reasonings for #3, none of which are correct according to my answer book. I can think like this: there are 6 ways to choose which one will be doubled first, then there are 5 ways to choose which one will then be doubled, and 4 ways to choose the last card, bringing the probabilities to $\frac{6*5*4}{6^5}$ .
I also have this reasoning: there are $\binom{6}{2}$ ways to get the two cards who will be doubled, and 4 ways to choose the last card, giving probabilities of $\frac{\binom{6}{2}*4}{6^5}$ .

I'm thoroughly confused. Can anybody help?
Thanks!

Your reasoning is fine for 1) and 2), but for 3) you need to do a little more thinking. It's not just the number of choices you need, but the number of arrangements of those choices. So there are, as you say, $\binom62\times4=60$ choices of 2 pairs from 6 plus one other different 'card'. But these have then got to be arranged in order before dividing by $6^5$ to get the probability.

The number of arrangements of 5 items with 2 'like' pairs is $\frac{5!}{2!2!}=30$ . So the probability is:

$\frac{60\times30}{6^5}=\frac{10\times5}{6^3}=\frac {25}{108}$

4) For three of a kind, do it in the same way as 3):

How many choices are there of the three of a kind?

How many choices are there of the remaining 2 'cards'?

How many arrangements are there of 5 items with one group of 3 'like' items?

Can you complete it now?

Grandad

**FatherChaos** · October 12th 2009, 06:35 AM

Originally Posted by Grandad

Hello FatherChaosYour reasoning is fine for 1) and 2), but for 3) you need to do a little more thinking. It's not just the number of choices you need, but the number of arrangements of those choices. So there are, as you say, $\binom62\times4=60$ choices of 2 pairs from 6 plus one other different 'card'. But these have then got to be arranged in order before dividing by $6^5$ to get the probability.

The number of arrangements of 5 items with 2 'like' pairs is $\frac{5!}{2!2!}=30$ . So the probability is:

$\frac{60\times30}{6^5}=\frac{10\times5}{6^3}=\frac {25}{108}$

4) For three of a kind, do it in the same way as 3):

How many choices are there of the three of a kind?

How many choices are there of the remaining 2 'cards'?

How many arrangements are there of 5 items with one group of 3 'like' items?

Can you complete it now?

Grandad

Thanks for your thorough answer, Grandad. I still don't understand why the order is important in this problem. I mean, what does it matter if a certain pair is placed before the other or if their placement is reversed? As I understand it, the order does not matter in this kind of problem... I'm confused because, in the first two problems, I did not compensate for the lack of order by multiplying with the number of possible permutations...

**Grandad** · October 12th 2009, 07:02 AM

Hello FatherChaos

Originally Posted by FatherChaos

Thanks for your thorough answer, Grandad. I still don't understand why the order is important in this problem. I mean, what does it matter if a certain pair is placed before the other or if their placement is reversed? As I understand it, the order does not matter in this kind of problem... I'm confused because, in the first two problems, I did not compensate for the lack of order by multiplying with the number of possible permutations...

You are asking three questions here:

Why does the order matter?
How did I get numbers 1) and 2) correct?
Why did we do 3) and 4) in a different way?

The answers are:

Why does the order matter?

Imagine that the dice are different colours: red, blue, yellow, green, white. Then when you say that there are $6^5$ ways in which they could land altogether, you are (correctly) taking into account the fact that the same set of 'cards' could appear many times over, because the same set of cards could appear on different coloured dice. You could have, for example, red showing 1, blue showing 2, yellow 3, green 4 and white 5; and this is a different result from, say, swapping over red and blue so that you get red showing 2, blue 1, yellow 3, green 4 and white 5.

You don't normally have to think about this: you just say: there are 6 ways in which the first die can land, 6 ways in which the second can land, ... and so on. So there are $6^5$ possible ways in which all five can land in total. The fact that you are using words like first, second, ... takes the order into account automatically.

How did I get the first two right?

Perhaps you can see this now: you used words like first, second, ... as you worked out the number of possibilities. In other words, you (unknowingly perhaps) arranged the dice in order as you made the choices.

So, why do 3) and 4) differently?

Because there are repeated items, it's easier to choose first, and then arrange, that's all. You can't expect to solve every problem in the same way!

Grandad

**FatherChaos** · October 12th 2009, 08:03 AM

Thank you, Grandad. It all makes much more sense now.

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