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**34 members have voted**

A casino dealer is working on a new Three Card Poker variant. She takes all the face cards from a standard deck and thoroughly shuffles them. She then deals 3 cards to Player #1, three cards to Player #2, three cards to Player #3, and the final three cards to Player #4. What is the probability that all four hands will contain a straight (J-Q-K of any suit)?

First player, first card - must be correct.

First player, second card - must be one of the 8 from 11 cards that don't match the first card.

First player, third card - must be one of the 4 from 10 cards that don't match the first or second cards.

So P(player 1) = 1 * 8/11 * 4/10.

Similarly P(player 2) = 1 * 6/8 * 3/7

Similarly P(Player 3) = 1 * 4/5 * 2/4

And then Player 4 must get a straight.

These are the same numbers as ChesterDog starts with but I then get 72/1925.

(I think this is because he needs 3*2*1 as the numerators for the 4th player.)

Quote:charliepatrickI'm just looking at the chances at each stage of dealing he nexr card correctly.

First player, first card - must be correct.

First player, second card - must be one of the 8 from 11 cards that don't match the first card.

First player, third card - must be one of the 4 from 10 cards that don't match the first or second cards.

So P(player 1) = 1 * 8/11 * 4/10.

Similarly P(player 2) = 1 * 6/8 * 3/7

Similarly P(Player 3) = 1 * 4/5 * 2/4

And then Player 4 must get a straight.

These are the same numbers as ChesterDog starts with but I then get 72/1925.

(I think this is because he needs 3*2*1 as the numerators for the 4th player.)

I guess your answer include the STRAIGHT FLUSH. I also get (4^3)/12C3 * (3^3)/9C3 * (2^3)/6C3 * (1^3)/3C3 = 72/1925 (include STRAIGHT FLUSH)

What if only consider STRAIGHT(WITHOUT any STRAIGHT FLUSH) ?

Quote:charliepatrick...These are the same numbers as ChesterDog starts with but I then get 72/1925.

(I think this is because he needs 3*2*1 as the numerators for the 4th player.)

Thanks! Now, I see that I missed the 3*3*1.

So, I agree it should be 12*8*4 * 9*6*3 * 6*4*2 * 3*2*1 / 12! = 72/1925.

Quote:ssho88...What if only STRAIGHT but WITHOUT any STRAIGHT FLUSH?

I guess that's the real trick of the puzzle. I get:

72/1925 - 12*2*1*9*2*1*6*2*1*3*2*1 = 72/1925 - 1/15400 = 23/616

Quote:ChesterDogQuote:charliepatrick...These are the same numbers as ChesterDog starts with but I then get 72/1925.

(I think this is because he needs 3*2*1 as the numerators for the 4th player.)

Thanks! Now, I see that I missed the 3*3*1.

So, I agree it should be 12*8*4 * 9*6*3 * 6*4*2 * 3*2*1 / 12! = 72/1925.

I guess that's the real trick of the puzzle. I get:

72/1925 - 12*2*1*9*2*1*6*2*1*3*2*1 = 72/1925 - 1/15400 = 23/616

If EXCLUDED any straight flush, I get 453/15400 = 0.0294

Quote:ssho88...If EXCLUDED any straight flush, I get 453/15400 = 0.0294

Thanks; I see that I made a mistake on excluding the straight flushes.

How do you calculate the correct answer?

Quote:ChesterDogThanks; I see that I made a mistake on excluding the straight flushes.

How do you calculate the correct answer?

Prob(all straight including straight flush) - Prob(1 SF + 3 ST) - Prob(2 SF + 2 ST) - Prob(4 SF) = 72/1925 - 2496/369600 - 432/369600 - 24/369600 = 453/15400 = 0.0294

EDITED : Here are the detail calculations

A) Case - All straight or straight flush for all player

Prob(ALL) = (4^3)/12C3 * (3^3)/9C3 * (2^3)/6C3 * (1^3)/3C3 = 72/1925

B) Case - 1 SF + 3 ST, there are four permutations, [SF][ST][ST][ST], [ST][SF][ST][ST], [ST][ST][SF][ST] and [ST][ST][ST][SF].

Let just consider [SF][ST][ST][ST],

i) Probability for player 1 get SF = 4/12C3

ii) When player 1 get SF(removed JsQsKs), cards left in the deck :-

JhQhKh

JcQcKc

JdQdKd

There are total 24 straight combinations(18 straight combinations with two type of suit and 6 straight combinations with three type of suit) for player 2.

a) when player 2 get straight with two type of suit(removed JhQhKc), then left 6 straight combinations for player 3

Prob case a) = 18/9C3 * 6/6C3 * 1/3C3

b) when player 2 get straight with three type of suit(removed JhQcKd), then left 8 straight combinations for player 3

Prob case b) = 6/9C3 * 8/6C3 * 1/3C3

So Prob for case [SF][ST][ST][ST] = 4/12C3 * (18/9C3 * 6/6C3 * 1/3C3 + 6/9C3 * 8/6C3 * 1/3C3)

There are four permutations, so Total Prob = 4 * 4/12C3 * (18/9C3 * 6/6C3 * 1/3C3 + 6/9C3 * 8/6C3 * 1/3C3) = 2496/369600

C) Case - 2 SF + 2 ST, there are six permutations, [SF][SF][ST][ST], [SF][ST][SF][ST], [SF][ST][ST][SF], [ST][SF][SF][ST], [ST][SF][ST][SF] and [ST][ST][SF][SF].

Let just consider [SF][SF][ST][ST],

Probability for player 1 get SF and player 2 get SF = 4/12C3 *3/9C3

Cards left in the deck :-

JcQcKc

JdQdKd

So there are 6 straight combinations for player 3,

Prob for case [SF][SF][ST][ST] = 4/12C3 *3/9C3 *6/6C3* (1^3)/3C3

There are six permutations, so Total Prob = 6 * 4/12C3 *3/9C3 *6/6C3* (1^3)/3C3 = 432/369600

D) Case - 3 SF + 1 ST, DOES NOT EXIST

Prob = 0

E) Case - 4 SF, only one permutations

Prob = 4/12C3 * 3/9C3 * 2/6C3 * 1/3C3 = 24/369600

So Prob(only straight) = Prob A) - Prob B) - Prob C) - Prob D) - Prob E) = = 72/1925 - 2496/369600 - 432/369600 - 0 - 24/369600 = 453/15400 = 0.0294

Assume that player #1 gets Ks, #2 Kh, #3 Kd, #4 Kc. Then look at how the Queens and Jacks might be distributed.

There are 24 ways to organise the Queens and also 24 ways to organise the Jacks.

4 players have SF : 1 way.

3 players have SF : no way (as the 4th would also have one!)

2 players have SF : 6 ways to pick the players, the other two have four ways to place their Q and J but one of these gives them a SF, so 3 ways left. Hence total ways = 6*3 = 18.

1 player has straight flush.

(i) This can be any one of the four players so first factor is 4.

(ii) Let's assume it's the first player so the remaining cards can be HDC HCD DHC DCH CHD CDH.

With HDC, all three Queens are correct so now all three Jacks have to be wrong, two ways DCH CHD. 2 ways

With HCD, DHC or CDH, one of the Queens is correct, so now only the specific Jack has to be wrong. 4 ways x 3 combinations.

With DCH or CHD all the Queens are wrong, so any combination of Jacks is fine. 6 ways x 2 combinations.

4 players = 1 way

2 players = 18 ways (6 x 3)

1 player = 104 (4 x (2+4x3+6x2), 4x26)

Total 123 ways

Leaving 576-123 ways where no players have a SF.

So Pr = 453/15400.