question about partys all red/black

[geo8o2]

collusion type question but not really.
it was discussed that party has about a 5% edge.
let's say some friends sat at a party table. and 2 of them had all red cards. or what about 3 friends have all red cards.

how many cards do i need to know the color of preflop to have an edge?

[OrangeKing]

[ QUOTE ]
collusion type question but not really.
it was discussed that party has about a 5% edge.
let's say some friends sat at a party table. and 2 of them had all red cards. or what about 3 friends have all red cards.

how many cards do i need to know the color of preflop to have an edge?

[/ QUOTE ]

I think you have an edge knowing just your own cards, but it's small (don't feel like doing the math right this second). I calculated earlier that with two people - or one person at an Omaha table - the player edge is 5% when you know all 4 cards are of the same suit.

Before you get your hopes up - you have to bet before you see your cards. [img]/images/graemlins/smile.gif[/img]

[threeonefour]

i thought you could only bet before you got dealt cards. making those kind of decisions impossible

[stoli]

5% edge to party? Try again.

[tpir90036]

I think you only need to know 3 or 4... but you bet before the hole cards are dealt so it doesn't really matter. It should be mostly obvious that they thought of this before putting it online for millions of people to bet on.

P.S. Please do not respond by giving me examples where people put things online that could be exploited for money. I won't be returning to read this thread.

[Jeebus]

what exactly are we talking about here. I m confused. [img]/images/graemlins/confused.gif[/img]

[Quicksilvre]

[ QUOTE ]
what exactly are we talking about here. I m confused. [img]/images/graemlins/confused.gif[/img]

[/ QUOTE ]

[_Kevin_]

Here is my take, someone please confirm or correct.

Note: I'm using the notation for combinations of nCm = m!/(m-n)!*n!. So the number of combinations of 3 things taken 2 at a time is 3C2 = 3!/((1!)*2!) = 3.

Let's assume you have 2 red cards. There are 50 unknown cards in the deck. From the 50 unknown cards, there are 50C3 = 19,600 possible flops (ignoring order)

The number of those flops that are all black are 26C3 = 2600.

Therefore, the flop will come all black 2600/19600 = 13.265% of the time. If you prefer, you compute this as a probability as (26/50)*(25/49)*(24/48) = 13.265% as well.

From this, the EV for a bet on all black when you hold 2 red cards is EV = ($8-$1)*(.13265) - $1*(1-.13265) = $0.0612

If you don't know any of the cards, the probablity of getting an all 1 color flop is (26/52)*(25/51)*(24/50) = 11.765%. This has an EV = -0.0588.

Have I made a mistake anywhere?

If not, it seems you should always place the bet on the other color whenever you have 2 cards of the same suit. Is the catch you must place your bet before you are dealt cards?

[Xhad]

[ QUOTE ]
what exactly are we talking about here. I m confused. [img]/images/graemlins/confused.gif[/img]

[/ QUOTE ]

Party now has a sidebet feature that lets you bet that the entire flop will be all red cards or all black cards. If you win you are paid 8-to-1. The chance of winning the bet is a little over 11%, which is about 9-to-1, so it's just another casino sucker bet.

EDIT: I just realized I made a slight error in my math this post, the chances of winning are more like 11.7%, but it needs to be 12.5% to be profitable so there's still an edge for the "house".

[DrSavage]

When you do red/black bet it works for the next hand that would be dealt after you've made a bet. So you can't get any additional info.