3CP drawing probablity

Hi everyone,

What is the probability that one card from a 52 card deck ends up to one of 4 players or the dealer in 3 card poker?

Example: Probability of the Ace of diamonds to show up in any hand of the 4 players (4*3=12 cards) or the dealer's (3 cards), while the rest of the deck (37 cards) remains undealt?

Thank you

Quote: gunbj

Hi everyone,

What is the probability that one card from a 52 card deck ends up to one of 4 players or the dealer in 3 card poker?

Example: Probability of the Ace of diamonds to show up in any hand of the 4 players (4*3=12 cards) or the dealer's (3 cards), while the rest of the deck (37 cards) remains undealt?

Thank you

Pretty sure it's as straightforward as it looks.

1/52 x 15 cards dealt. 15/52, = .28846, which is 28.846%, or 1 in 3.4666 hands.

Yes it is, thanks.

What about the probability of that card ending up in the dealer's hand only and not to the other players?

Quote: gunbj

Yes it is, thanks.

What about the probability of that card ending up in the dealer's hand only?

Same process, right? 3/52, or 1 in 17.3333 hands.

Thank you, really just asking as confirmation. Last question: probability of 2 and 3 cards ending to the dealer and not the players?

Quote: gunbj

Thank you, really just asking as confirmation. Last question: probability of 2 and 3 cards ending to the dealer and not the players?

That question, I'm not following. Can you be more specific?

If they're 2 specific cards, and any third card, it's 1/52 x 1/51 x 6 ways to receive them (since order doesn't matter) x 50/50 other cards. So 1 in 442 hands.

If instead it's, say, a pair of any suit, and any third card, it happens much more often. 2/52 x 6 ways to pair x 13 ranks x 6 ways to receive them x 50/50 other cards. So 1 in 18 hands. Note that this includes the third card making trips. If it is ONLY a pair, the trips would have to be subtracted out.

Anybody else on here feel free to correct my calculations throughout this, please.

Yes, sorry, I didn't explain correctly.

I meant the total probability of the dealer receiving EITHER of 2 specific cards or BOTH. Same question with 3 specific cards (receiving any one of the 3, plus 2 of the 3 at the same time, plus all of those 3 cards at the same time).

PS: the value of the card is irrelevant. I'm not trying to calculate whether it would be a pair, or trips etc...

For 2 specific cards:
-dealer will have either or both 11.312% of the time.
= 1- c(50,3)/c(52,3)

For 3 specific cards:
- dealer will have one or more of those 3 cards 16.6335% of the time.

= 1- c(49,3)/c(52,3)

gunbj, how about treating us like something more than a mathematical version of Monica Lewinski? Do you want to say something about what you are calculating or offer any kind of chat?

Thank you, that makes sense.

I'm trying to learn to calculate probabilities and combinations but don't have the means to confirm my calculations apart from posting here. 3cp seemed like the easiest game for me to start and once I understand the basics I'll hopefully be able to analyse other games too.