Math problem with playing cards

[SuperPickle]

My boss was arguing about odds with his buddies over a late night card soiree last weekend. The question was whether it is more ?difficult? (worse odds) to get three of a kind in a hand of only three cards or if it is more ?difficult? to get four of a kind in a hand of five cards. These guys were just trying to argue logically and none had the math skills to back up their arguments. I too, have skills far too rusty to be confident in a proof.

The scene: One deck of 52 playing cards; five people, each getting one card in succession. Player 1 gets a card; Player 2 gets a card, etc. until all players have three cards or five cards. No exchanging/drawing cards after the deal. The denomination of cards for the hand of three/four-of-a-kind is unimportant (aces, kings, sevens).

This is not a homework question, I promise. Rather, it?s a pseudo-intellectual argument among people without skills enough to fortify their respective claims.

Anyone willing to help us out determining which is more unlikely?

 

[DainBramaged]

We used to play a lot of cards in Vietnam, but I don't know the answer to your question.

 

[Juice Box]

3 of a kind has better odds IIRC

 

[Siva]

4 of a kind is harder

/math








btw i actually have no clue

 

Originally posted by: Juice Box
3 of a kind has better odds IIRC

Did you read the entire question? Its 3 of a kind in a hand of 3 versus a 4 of a kind in a hand of 5. He is looking for people to back up their "facts" with proof.

 

[hypn0tik]

4 of a kind trumps a 3 of a kind simply because the probability of being dealt a 3 of a kind is higher.

 

[Juice Box]

Originally posted by: FallenHero

Originally posted by: Juice Box
3 of a kind has better odds IIRC

Did you read the entire question? Its 3 of a kind in a hand of 3 versus a 4 of a kind in a hand of 5. He is looking for people to back up their "facts" with proof.

yeah I read the quetsion....and I am simply stating my opinion from what I remember about odds......

 

[dullard]

3 of a kind with three cards
1) First card doesn't matter. Whatever number the card is, will determine what the other two must be.
2) 51 cards remain. Three of them match your first card. Thus your odds are 3/51 to get the first two to match.
3) 50 cards remain. Two of them match your first two cards. Thus your odds are 2/50 to get the third card to match.

Overall odds: 3/51 * 2/50 = 1 out of 425.

4 of a kind with five cards
1) There are 672 possible four of a kind hands
2) There are a total of 2,598,960 possible hands.
3) 672/2,598,960 = 1 out of 4165 odds of getting a four of a kind.

 

[Juice Box]

Originally posted by: dullard
3 of a kind with three cards
1) First card doesn't matter. Whatever number the card is, will determine what the other two must be.
2) 51 cards remain. Three of them match your first card. Thus your odds are 3/51 to get the first two to match.
3) 50 cards remain. Two of them match your first two cards. Thus your odds are 2/50 to get the third card to match.

Overall odds: 3/51 * 2/50 = 1 out of 425.

thats what I thought....but dont you have to take into consideration that the other ppl are also getting cards given to them?

 

[PawNtheSandman]

Odds of getting 3 of a kind = 1 in 34.8
Odds of getting 4 of a kind = 1 in 4100

 

[zendari]

I'll do my best though I don't have a calculator and have given up math anyway.

You have to use combinations.

For the 3 of a kind.

Player 1 gets a card, an A. Players 2-5 have to pick up, out of 51 cards, the 48 cards that aren't A's.

So thus far we have (48/51 * 47/50 * 46/49 * 45/48) as the odds of that happening.

Player 1 has to pick another A. Multiply above by 3/47.

Players 2-5 have to pick up non A's again. (43/46 * 42/45 * 41/44 * 40/43)

Now mutiply by 2/42 and you have your number. I know this is way off because there are 4 aces in the deck not 3, ill stop talking out of my ass and defer to the mathmeticians. But in short I think the 3 of a kind is easier because you can "lose an Ace" and still succeed.

 

[Siva]

nm, i didn't see you were up against other players, i still think 3 of a kid is easier

 

[royaldank]

Originally posted by: Juice Box
thats what I thought....but dont you have to take into consideration that the other ppl are also getting cards given to them?

I don't believe so. It's the same as the cards sitting on the bottom of the deck.

 

[dullard]

Originally posted by: Juice Box
thats what I thought....but dont you have to take into consideration that the other ppl are also getting cards given to them?

Not if you don't know what the cards are (typical for card games). If you do know what the cards are (where you can see all hands), then you can adjust the odds for this additional information. However, the odds then vary with each deal and is useless for this type of conversation.

 

[ActuaryTm]

This thread is nothing if not entertaining (at least, the responses are).

Originally posted by: dullard
3 of a kind with three cards
1) First card doesn't matter. Whatever number the card is, will determine what the other two must be.
2) 51 cards remain. Three of them match your first card. Thus your odds are 3/51 to get the first two to match.
3) 50 cards remain. Two of them match your first two cards. Thus your odds are 2/50 to get the third card to match.

Overall odds: 3/51 * 2/50 = 1 out of 425.

That would be true (for the most part), if only one player. As the original post mentions, there are five.

 

[dullard]

Originally posted by: ActuaryTm
Overall odds: 3/51 * 2/50 = 1 out of 425.

That would be true (for the most part), if only one player. As the original post mentions, there are five.

Like I just said, no it doesn't matter how many players there are (if you can't see the other player's hands).

The remaining 49 cards can be in the deck. Or the remaining 49 cards can be split among however many players you want. What you do with the other 49 cards does not affect your 3 cards.

 

[ActuaryTm]

Originally posted by: dullard
Like I just said, no it doesn't matter how many players there are (if you can't see the other player's hands).

From a pure probability standpoint, it doesn't. From a conditional probability standpoint, it does.

 

[dullard]

Originally posted by: ActuaryTm
From a conditional probability standpoint, it does.

If you don't know the conditions (if you can't see the other cards), then no you cannot adjust for the unknown conditions.

By the way, I edited my original post with the 4 of a kind math.

1/425 for 3 of a kind in 3 cards.
1/4165 for 4 of a kind in 5 cards.

Thus one of the two considered situations is ~10 times more rare.

 

[Tommunist]

Originally posted by: SuperPickle
My boss was arguing about odds with his buddies over a late night card soiree last weekend. The question was whether it is more ?difficult? (worse odds) to get three of a kind in a hand of only three cards or if it is more ?difficult? to get four of a kind in a hand of five cards. These guys were just trying to argue logically and none had the math skills to back up their arguments. I too, have skills far too rusty to be confident in a proof.

The scene: One deck of 52 playing cards; five people, each getting one card in succession. Player 1 gets a card; Player 2 gets a card, etc. until all players have three cards or five cards. No exchanging/drawing cards after the deal. The denomination of cards for the hand of three/four-of-a-kind is unimportant (aces, kings, sevens).

This is not a homework question, I promise. Rather, it?s a pseudo-intellectual argument among people without skills enough to fortify their respective claims.

Anyone willing to help us out determining which is more unlikely?

there are 52 choose 3 ways to be dealt 3 cards which = 22100 ways
out of these possibilities there are 4 ways (4 choose 3) for each value (Ace, 2, 3, etc) to have 3 of a kind. this is a total 13*4 = 52 ways to get 3 of a kind.

52/22100 = 0.235 %

there are 52 choose 5 ways to be dealt 5 cards which = 2598960 ways
out of these possibilities there are 48 ways for each value to have 4 of a kind (for instance 4 aces and then there are 48 possible cards left for the 5th card). this is a total of 13*48 = 624 ways to get 4 of a kind with 5 cards.

624/2598960 = 0.02401 %

it appears that 3 of a kind out of 3 cards is more likely for an individual. i just did this without thinking too hard and my probability is a bit rusty so i may be wrong...

 

[dullard]

Originally posted by: Tommunist
52/22100 = 0.235 %

624/2598960 = 0.02401 %

Thank you for backing me up.

0.235% = 1/425.
0.0401% = 1/4165.

 

[royaldank]

Originally posted by: Tommunist
there are 52 choose 3 ways to be dealt 3 cards which = 22100 ways
out of these possibilities there are 4 ways (4 choose 3) for each value (Ace, 2, 3, etc) to have 3 of a kind. this is a total 13*4 = 52 ways to get 3 of a kind.

52/22100 = 0.235 %

there are 52 choose 5 ways to be dealt 5 cards which = 2598960 ways
out of these possibilities there are 48 ways for each value to have 4 of a kind (for instance 4 aces and then there are 48 possible cards left for the 5th card). this is a total of 13*48 = 624 ways to get 4 of a kind with 5 cards.

624/2598960 = 0.02401 %

it appears that 3 of a kind out of 3 cards is more likely for an individual. i just did this without thinking too hard and my probability is a bit rusty so i may be wrong...

This is how I went about solving it and it looks good to me.

 

[ActuaryTm]

Originally posted by: dullard
If you don't know the conditions (if you can't see the other cards), then no you cannot adjust for the unknown conditions.

The "(if you can't see the other player's hands)" must have been added during the edit.

I adore how FuseTalk makes adds addenda in that manner.

 

[Tommunist]

Originally posted by: dullard

Originally posted by: ActuaryTm
From a conditional probability standpoint, it does.

If you don't know the conditions (if you can't see the other cards), then no you cannot adjust for the unknown conditions.

By the way, I edited my original post with the 4 of a kind math.

1/425 for 3 of a kind in 3 cards.
1/4165 for 4 of a kind in 5 cards.

Thus one of the two considered situations is ~10 times more rare.

you are correct dullard - if this was indeed a hw question the fact that there are 5 players would be thrown in as extra info to confuse the student. since we don't know what the other players have it is irrellevant. if the question is what are the odds that at least ONE person out of the 5 will end with the desired result will have different probabilities but the 3 of a kind should still be more likely I would guess.

 

[skdsp]

the probability will be so low with that many players. ive played 5 card draw for a long time and not once have i gotten a 4 of a kind

 

[dullard]

Forget it.