Probability

Yeah, I got the same as miplet, too.

Quote: statman

The number of ways 52 cards could be dealt any which way is 52!, so the final formula is

SUM(x = 1 to 13, x^4)/52^4

I get 1.1067817962E-63

Is that what you guys got?

1.106e-63 is very small number, so I started looking at your formulas. I think I must be missing something. Sum(x=1 to 13, x^4)=1+16+81+256+625+1296+2401+4096+6561+10000+14641+20736+28561=89271, and 52^4=7311616. 89271/7311616=.012209, which is a very long way from 0.000000000000000000000000000000000000000000000000000000000000001106781796.

Quote: statman

Quote:

13 players are each given 4 cards from a 52 card deck. what is the probability that each player has one card of each suit?

In the first hand, the diamond could be any one of 13 cards; the same for the club, heart, spade, so the number of ways the first hand could be formed is 13^4.

. ..

Is that what you guys got?

There are also 4! or 24 ways to arrange those 4 cards which I believe is where your error is.

Quote: statman

Quote:

13 players are each given 4 cards from a 52 card deck. what is the probability that each player has one card of each suit?

In the first hand, the diamond could be any one of 13 cards; the same for the club, heart, spade, so the number of ways the first hand could be formed is 13^4.

The second hand could be formed in 12^4 ways.

For the third hand get 11^4 ...

For the next to last hand get 2^4

For the last hand it is 1

so the number of ways of forming the hands is

SUM(x = 1 to 13, x^4)

The number of ways 52 cards could be dealt any which way is 52!, so the final formula is

SUM(x = 1 to 13, x^4)/52^4

I get 1.1067817962E-63

Is that what you guys got?

There are three errors in your formula:
1) It's a product, not a sum
2) Miplet already pointed this out -- there are 4! ways to arrange each 4-card hand.
3) You quoted 52! as the denominator but then wrote 52^4 in your formula.

Fixing all these yields
PRODUCT(x = 1 to 13, 4!*x^4)/52!
which gives the correct answer, 1.63387571E-11 or 1 in 61204166001.13

Quote: MathExtremist

Fixing all these yields
PRODUCT(x = 1 to 13, 4!*x^4)/52!
which gives the correct answer, 1.63387571E-11 or 1 in 61204166001.13

Going by suits alone, 4!^13 / (52!/13!^4) .

Then i began to wonder, that for every ordering of the deck, there is at least one means of dealing which shall make each of those a solution. And intuitively, that some of these solutions might have more possibly-corresponding means of dealing than others. Then rendering this problem precisely-insoluble, without first observing the order of the cards before the means of dealing has been revealed, and the above number to merely an average over all means of dealing.

Even were the problem well-defined here... in practice, there is still the comparatively-likely chance of dealing (clockwise, the usual way,) from a new, "mistakenly" unshuffled "boxed rack" of cards (with the jokers removed).

Thank you for the correction, but I don't quite understand why the order of the cards in each hand is important.

Quote: statman

... but I don't quite understand why the order of the cards in each hand is important.

Eg, if the suits run in an order like spade, heart, diamond, club, on down through the deck, then you can (and might) deal out the cards in both most-typical manners to arrive at the solution: by a set of four to each the players; or singly, clockwise, one card at a time to each. Perhaps someone would check on that, as i just now ran it through my head quickly only once.

Besides that, understanding the differences from methods of shuffling to those of dealing can have implications in computer-program efficiencies, if not in also which is programed.

This event (getting 4 hands of 13 cards all of the same suit, known as 4 perfect hands in bridge) actually happened in the 1950's. The 4 ladies who got dealt the perfect bridge hand appeared as guests on both "What's My Line?" and "I've Got a Secret".