maths: probability

[amy]

(i) how many different poker hands can be dealt to a player?

(ii) how many hands include the Ace of spades?

(iii) how many hands do not contain include the ace of spades?

(iv)how many hands include all 4 aces?

(v) how many hands contain exactly 3 aces?

 

 

so i know that there are 52 cards in a pack and that a poker hand is 5 cards, so the answer to (i) is 52 choose 5 ⁵²C_{5 ,} using the calculator i get 2598960.

 

and i know that there is only one ace of spades in a pack of cards but i'm not quite sure how to work out part (ii) and if i knew the answer to this then obviously i could work out the answer to part (iii)

 

could anyone give me a bit of help please? thanks.

Edited March 22, 2013 by amy

[John]

Since the ace of spades must be in the hand, the number of cards left to deal is one smaller, as is the number of cards left in the deck.

Edited March 23, 2013 by John

[imatfaal]

wikipedia has a great page on poker probability - although it might be best to work thru the problems yourself before turning to the answers

[ydoaPs]

You have 52 cards for the first card. How many for the second? How many for the third, forth, and fifth?. Multiplying those will give you the correct (I). For (ii) you'll need to figure out P(A₁vA₂v...vA₅) then multiply by the total number of hands. From that number, (iii) should be easy. The last ones are similar reasoning.

[imatfaal]

You have 52 cards for the first card. How many for the second? How many for the third, forth, and fifth?. Multiplying those will give you the correct (I).

Surely that would give you the number of unsorted poker hands? It is normally assumed for questions like this that 10spades 4diamonds Jackhearts 5 clubs 5spades is the same hand as 4diamonds Jackhearts 5 clubs 5spades 10spades.

 

52 Choose 5 (which I believe is the correct answer) is 52! divided by 47! and 5! ie (52.51.50.49.48)/(5.4.3.2.1). The division by 5! removes the hands which are the same cards in different order.

 

For (ii) you'll need to figure out P(A₁vA₂v...vA₅) then multiply by the total number of hands. From that number, (iii) should be easy. The last ones are similar reasoning.

On similar reasoning I would say two is 51choose4 - ie, as John said above, how many ways can you accompany the ace of spades

 

edit or maybe 51choose4 divided by 5 for the ace of spades

second edit - actually I will stick to my first.

 

I think part 3 is easier than part two. If the number of hands of a 52 card pack is 52C5 - then what is the number of hands of a 51 card pack? Part two is the complementary hands to part 3.

[John]

Surely that would give you the number of unsorted poker hands? It is normally assumed for questions like this that 10spades 4diamonds Jackhearts 5 clubs 5spades is the same hand as 4diamonds Jackhearts 5 clubs 5spades 10spades.

 

52 Choose 5 (which I believe is the correct answer) is 52! divided by 47! and 5! ie (52.51.50.49.48)/(5.4.3.2.1). The division by 5! removes the hands which are the same cards in different order.

 

On similar reasoning I would say two is 51choose4 - ie, as John said above, how many ways can you accompany the ace of spades

 

edit or maybe 51choose4 divided by 5 for the ace of spades

second edit - actually I will stick to my first.

 

This is all correct (I'm assuming we're assuming amy's already figured out her answers, since we're freely discussing the specific solution). More completely, we know one card has to be the ace of spades. We can either conceptualize that as the equivalent of removing the ace of spades from the deck and calculating the number of four-card hands we can get from the remaining cards, which is C(51,4), or be fancy and multiply the number of ways to choose the ace of spades by the number of ways to choose the other cards, i.e. C(1,1)C(51,4). The latter method is especially useful for number 5

 

I think part 3 is easier than part two. If the number of hands of a 52 card pack is 52C5 - then what is the number of hands of a 51 card pack? Part two is the complementary hands to part 3.

 

Yeah, I was going to mention in my earlier post that starting with number 3 might be easier, but decided to relegate that to a later post if the OP didn't mention it herself. It's sad how many times I've gotten caught up trying to directly calculate some hairy combinatorics or probability result in situations where simply subtracting a previous result from the total (or 1) would give the correct answer much more easily. But that comes with being a derp, I suppose. :[

Edited March 26, 2013 by John

[imatfaal]

As a fellow derp - I will admit that I only realised after the fact.