gokeeffe Joined: Jan 12, 2012
• Posts: 3
January 12th, 2012 at 10:24:40 AM permalink
I'm currently revising for a maths exam that contain contains questions on probability. I was wondering if anyone could help me with either of these questions.

Question: If two cards are drawn from a standard pack, what is the probability that the second is an ace given that the first is a diamond? Discuss the calculation of this probability in terms of conditional probabilities.

Question: A six-sided dice is rolled six times. What is the probability that the total score is 10?

Any help would be greatly appreciated. Thanks
DJTeddyBear Joined: Nov 2, 2009
• Posts: 10504
January 12th, 2012 at 11:07:23 AM permalink
For the card question, if you don't know the rank of the first card, then it is irrelevant. The chance of the second card being any particular rank is 4 in 52. The fact that the first card was a Diamond is meaningless since we don't know it's rank and don't care what suit the ace is.

I'm not sure about the dice question.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ������������������������������������� Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
gokeeffe Joined: Jan 12, 2012
• Posts: 3
January 12th, 2012 at 11:28:59 AM permalink
But isn't there the probability that the first card could have been the ace of diamonds, doesn't this affect the question?
DJTeddyBear Joined: Nov 2, 2009
• Posts: 10504
January 12th, 2012 at 11:39:30 AM permalink
It would affect it as much as if it had been any other ace, and have the opposite affect if it had been any other diamond.

I.E. Since all you're concerned with is the rank of the second card, knowing the suit of either is irrelevant, and not knowing the rank of the first card makes it have no affect on the outcome of the second card.

Look at it this way.

Say you dealt the entire deck, and you know that at least 10 diamonds are in the discard pile, but have no idea of thier rank. What's the chance that the last card is an ace? 4 in 52.

For what it's worth, if you know that all 13 diamonds are in the discards, then the chance becomes 3 in 39.

But that's the same thing: 1 in 13.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ������������������������������������� Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
dwheatley Joined: Nov 16, 2009
• Posts: 1246
January 12th, 2012 at 11:50:39 AM permalink
The first question with conditional probabilities:

P ( A | d ) = P (A | Ad ) * P (Ad | d) + P ( A | d not A) * P (d not A | d)
= 3 / 51 * 1/13 + 4 / 51 * 12 / 13
= 1/221 + 16/221
= 1/13

DJ is right, of course, but I think the question wants to see this math.

The 2nd question is annoying. You can do it in excel, or create a multinomial distribution, or the generating function, or just crunch the #s.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
gokeeffe Joined: Jan 12, 2012
• Posts: 3
January 12th, 2012 at 12:04:35 PM permalink
Cool guys thanks very much.

Yea the second question is really annoying, I'v answered similar questions before but they seemed to be much easier than this particular example. Thanks again for the help :)
ThatDonGuy Joined: Jun 22, 2011
• Posts: 4927
January 12th, 2012 at 12:10:32 PM permalink
First answer: break the initial condition into two sub-conditions:
(a) The first card is the ace of diamonds (probability 1/13);
(b) The first card is not the ace of diamonds (probability 12/13).
In both cases, there are 51 cards remaining; in (a), three of them are aces, and in (b) four of them are aces.
P = 1/13 x 3/51 + 12/13 x 4/51

Second answer: first, determine how many different ways there are to roll a die (note: "dice" is plural - the singular is "die") six times; this is 66 = 46656.
Now, look at all of the possible ways to get a total of 10 from six rolls.
I can think of two ways to do this; either by listing every permutation, or by listing every combination and then counting the number of permutations that exist for each combination. The second way is faster when doing it yourself.
1,1,1,1,1,5 has 6 positions for the 5 and, for each of those, 1 for the 1s, so there are 6 x 1 = 6 permutations
1,1,1,1,2,4 has 6 positions for the 4; for each one, there are 5 remaining for the 2; for each one, there is 1 for the 1s, so there are 6 x 5 x 1 = 30
1,1,1,1,3,3 has 15 pairs of positions ({1,2}, {1,3},..., {5,6}) for the two 3s; for each one, there is 1 for the 1s, so there are 15 x 1 = 15
(1,1,1,1,4,4 > 10, so go to 1,1,1,2)
1,1,1,2,2,3 has 6 for the 3; for each one, there are 10 for the 2s; for each one there is 1 for the 1s, so there are 6 x 10 x 1 = 60
(1,1,1,2,3,3 > 10, so go to 1,1,1,3)
(1,1,1,3,3,3 > 10, so go to 1,1,2)
1,1,2,2,2,2 has 10 for the two 1s; for each one, there is 1 for the 2s, so there are 10 x 1 = 10
(1,1,2,3,3,3 > 10, so go to 1,1,3)
(1,1,3,3,3,3 > 10, so go to 1,2)
(1,2,2,2,2,2 > 10, so go to 2)
(2,2,2,2,2,2 > 10; all combinations have been checked)
The total number of permutations = 6 + 30 + 15 + 60 + 10 = 121, and the probability = 121 / 46656.
(This is irreducible, as 121's only prime factor is 11 and 46656's only prime factors are 2 and 3.)
NowTheSerpent Joined: Sep 30, 2011
• Posts: 417
January 14th, 2012 at 7:43:43 AM permalink
Quote: ThatDonGuy

First answer: break the initial condition into two sub-conditions:
(a) The first card is the ace of diamonds (probability 1/13);
(b) The first card is not the ace of diamonds (probability 12/13).
In both cases, there are 51 cards remaining; in (a), three of them are aces, and in (b) four of them are aces.
P = 1/13 x 3/51 + 12/13 x 4/51 = 51/(13x51) = 1/13.

1,1,1,1,1,5 has 6 positions for the 5 and, for each of those, 1 for the 1s, so there are 6 x 1 = 6 permutations
1,1,1,1,2,4 has 6 positions for the 4; for each one, there are 5 remaining for the 2; for each one, there is 1 for the 1s, so there are 6 x 5 x 1 = 30
1,1,1,1,3,3 has 15 pairs of positions ({1,2}, {1,3},..., {5,6}) for the two 3s; for each one, there is 1 for the 1s, so there are 15 x 1 = 15
1,1,1,2,2,3 has 6 for the 3; for each one, there are 10 for the 2s; for each one there is 1 for the 1s, so there are 6 x 10 x 1 = 60
1,1,2,2,2,2 has 10 for the two 1s; for each one, there is 1 for the 2s, so there are 10 x 1 = 10

The total number of permutations = 6 + 30 + 15 + 60 + 10 = 121, and the probability = 121 / 46656.
(This is irreducible, as 121's only prime factor is 11 and 46656's only prime factors are 2 and 3.)

for the (1,1,2,2,2,2) case, wouldn't combin(6,2) = 15 be the correct formulation? In form this appears identical to the (1,1,1,1,3,3) case, which has 15 perms.

If so, then the total perms adding to 10 would be 126, with a prob of 7/2592.