A fun probability question to try. =]

[_henwee]

Hi People!

My cousin and I were playing 13 and was wondering the probability of getting a chop (4 of the same number, or 3 consecutive pairs).

So... What is the probability of drawing 13 cards from a standard deck and receiving 4 of a kind (except for the numbers 2 and 3)?

or

What is the probability of getting 3 consecutive pairs (without any two's)?

HAVE FUN! =]

PS. Please tell me that this is hard - we kinda took 3-4 hours trying to figure it out.
The answer to the first question was .107867

[iNerd]

Oh wow...I haven't played 13 since Year 9 with MHS azns

[ruchika5]

The probability of probability being fun is .0000000000000001

[|ll|lll|]

...someone teach me how '13' works please


It might be what I'm thinking, but I have another name for it I guess

[iNerd]

...someone teach me how '13' works please


It might be what I'm thinking, but I have another name for it I guess

Big 2?

3 of spades is the lowest card.
2 of hearts is the highest card.

Get rid of your cards by playing pairs or sequences.

[Andiio]

...someone teach me how '13' works please


It might be what I'm thinking, but I have another name for it I guess

Big 2?

3 of spades is the lowest card.
2 of hearts is the highest card.

Get rid of your cards by playing pairs or sequences.

In big 2, isn't the hierarchy like this? Spades > Hearts > Clubs > Diamonds

That's how my friends and I play it anyways

[iNerd]

...someone teach me how '13' works please


It might be what I'm thinking, but I have another name for it I guess

Big 2?

3 of spades is the lowest card.
2 of hearts is the highest card.

Get rid of your cards by playing pairs or sequences.

In big 2, isn't the hierarchy like this? Spades > Hearts > Clubs > Diamonds

That's how my friends and I play it anyways

No idea, I call it 13.

[Pixon]

...someone teach me how '13' works please


It might be what I'm thinking, but I have another name for it I guess

Big 2?

3 of spades is the lowest card.
2 of hearts is the highest card.

Get rid of your cards by playing pairs or sequences.

In big 2, isn't the hierarchy like this? Spades > Hearts > Clubs > Diamonds

That's how my friends and I play it anyways

Yeh, that's how I've always played.

[nacho]

I'm most likely wrong but, to get the first one:
Total combinations = 52.C.13
Combination wanted = 52.C.4 x 11
Probability = (52.C.4x11)/52.C.13

And question 2:
Total combinations = 52.C.13
If you count Queen, King, Ace a consecutive pair, then there are 52.C.10 ways of picking the three consecutive pairs
Probability= 22.C.10/52.C.13

Yea, i'm most likely way off, since this took like a 3 mins, where as you took longer... stupid combinatorics

[_avO]

I've never ever heard of big 2, only 13 been playing it since like 7

[pi]

...someone teach me how '13' works please


It might be what I'm thinking, but I have another name for it I guess

Big 2?

3 of spades is the lowest card.
2 of hearts is the highest card.

Get rid of your cards by playing pairs or sequences.

In big 2, isn't the hierarchy like this? Spades > Hearts > Clubs > Diamonds

That's how my friends and I play it anyways

Ive always played it as Hearts > Diamonds > Clubs > Spades

What else to do during boring geo classes?

[Andiio]

...someone teach me how '13' works please


It might be what I'm thinking, but I have another name for it I guess

Big 2?

3 of spades is the lowest card.
2 of hearts is the highest card.

Get rid of your cards by playing pairs or sequences.

In big 2, isn't the hierarchy like this? Spades > Hearts > Clubs > Diamonds

That's how my friends and I play it anyways

Ive always played it as Hearts > Diamonds > Clubs > Spades

What else to do during boring geo classes?

Geo = sleep :]

[|ll|lll|]

Ive always played it as Hearts > Diamonds > Clubs > Spades

Same here.

Anyway, this is going off-topic.

I'm glad there's not so much combinatorics in year 12 probability!

[pi]

I'm glad there's not so much combinatorics in year 12 probability!

Me too, my worst area of maths

[brightsky]

Hmm...I got:

1) Let x be the number of 4 of a kinds and Pr(X = x) be the probability of getting x number of 4 of a kinds.

Then the answer to your question would be Pr(X = 1) + Pr(X = 2) + Pr(X = 3).

Pr(X = 1) = (4C4 * 48C9)*11/52C13
Pr(X = 2) = (8C8 * 44C5)*11C2/52C13
Pr(X = 3) = (12C12 * 40C1)*11C3/52C13

Adding this apparently gives 0.2914569...but I'm doubting myself. :/