Re: What are the odds?
There are ten different straight flushes for each suit,
or 40 alltogether.
The number of different ways you can be dealt 5 cards is
52 * 51 * 50 * 49 * 48/(1 * 2 * 3 * 4 * 5)
The probability of drawing a straight flush
is the ratio = 40/(52 * 51 * 5 * 49 * 4) =
1/(26 * 51 * 49) = 1.53 E -5 approx
This assumes no other players, no draw, etc.
.
Relevant Pages
- Re: Another poker probability problem
... >three spade cards can fill any of three different straight flushes: ... 52 cards in the deck. ... there are 6 which have been dealt but that you haven't seen. ...
(sci.math) - Re: what are the odds?
... >> a deck of cards and carefully shuffle it. ... >> dealt a deck, one of the 8x10^{67) arrangements, all equally ... >Let's suppose I lay out the cards exactly as you have explained. ... Nope, the probability that all the necessary mutations took place is exactly 1 *after* they have happened, no matter how low their probability was before the event. ...
(talk.origins) - Re: Horror story playing the best 24 hands ....
... pair, only 4 that will yield two specific suited cards, ... total number of possible card combinations that a player can be dealt, ... hands you listed as if they each had the same probability of occurring, ...
(rec.gambling.poker) - Re: prob. problem
... The key is that the probability of the deuce of clubs ... coming out of the deck before the first ace is zero. ... You are assuming that you will see all the cards that came out before the ... that none of the dealt cards are aces, ...
(rec.gambling.poker) - Re: prob. problem
... The key is that the probability of the deuce of clubs ... coming out of the deck before the first ace is zero. ... You are assuming that you will see all the cards that came out before the ... that none of the dealt cards are aces, ...
(rec.gambling.poker) |
|
