# Math Help - Probability question i dont understand

1. That's how I used to solve things like this and if you look closely, it's the same thing.

51! = 51 x 50! = 51 x 50 x 49! = ...

2. I'll explain using smaller numbers.

Let's say you have ABCDEF, 6 letters.

The total number of combinations is 6!.

Now, A and B should be adjacent.

(AB)CDEF

Make as if A and B were stuck together; they form one item, for a total combination of 5!

But now, A and B could be either in the order AB or BA, so, you multiply by 2, the number of combinations with A and B together; 2 x 5!

Probability becomes $\dfrac{2\times 5!}{6!}$

Sorry for my previous mistakes... it's 10 past midnight now, and I'm really tired. I'll answer tomorrow if there is still some unclear things.

3. @Sneaky,
I don't get the last line of your statement. If you tried to picture a single ace and put the other one to the left or to the right of it, I think you might have overlooked the 'edges' (top and bottom of the deck) where the ace has only one neighbouring site?

4. ok but i have one more question

Suppose we roll 3 fair die, what is the chances that 2 match and one is different.
i put for this
there are 216 outcomes
chance of getting a number on die1 = 6/6
chance of getting the same number as on die1 in die2 = 1/6
chance of getting a different number as die1 in die3 = 5/6
6/6 * 1/6 * 5/6 = 5/36
Is this right?

5. Perfectly right.

6. yay, still dont completely understand the concept on how unknown008 gets 1/26 (2/52) .

7. Originally Posted by Sneaky
ok but i have one more question
Suppose we roll 3 fair die, what is the chances that 2 match and one is different. i put for this there are 216 outcomes
chance of getting a number on die1 = 6/6
chance of getting the same number as on die1 in die2 = 1/6
chance of getting a different number as die1 in die3 = 5/6
6/6 * 1/6 * 5/6 = 5/36
Is this right?
Originally Posted by needsomemathhelp
Perfectly right.
Both of those replies are incorrect.
There are only $90$ triples that contain exactly one pair.
So the probability is $\dfrac{90}{216}=\dfrac{5}{12}$

8. i am completely lost in your answer.
how do u get 90?

9. Let's try it this way:
(i) There are 50! possibilities to get the aces of spades in the first and the ace of clubs in the second slot. Right?
(ii) There are 50! possibilities to get the aces of clubs in the first and the ace of spades in the second slot. Right?
(iii) There are 51 possible positions of the pair in the deck. Ok?
(iv) There are 52! possible arrangements of cards in the deck in total.
(v) So in the end we have 2 * 51 * 50! possibilities to have the pair from 52! possibilities in total. This gives a probability of 2 * 51 * 50! / 52! = 2 / 52.

Which step is unclear to you?

10. ok i see now, but i don't get plato's answer, how they got 90.

11. A triple with exactly one pair looks like $(X,X,Y)$
Now the $X's$ can have six values and the $Y's$ can have five.
That is $30$. But each triple can be arranged is $3$ ways. That is $90$ in all.

12. That's true, shame on me. I forgot to arrange the triple.

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