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bernie
04-29-2003, 10:15 AM
holdem.

was wondering what the odds were for a str8 flush over str8 flush with both cards playing would be.

b

bernie
05-02-2003, 10:30 AM

Cyrus
05-02-2003, 08:12 PM
"[I] was wondering what the odds ... for a str8 flush over str8 flush with both cards playing would be."

One example : Player A has cards AB and the board is CDExx. (The board can come xDxEC or any other way; the order of the deal is unimportant.) Cards ABCDE make a SF. You want the probability of someone else holding cards FG which combined with CDE also make a SF. There are 45 cards remaining which can be combined into 2-card hands 990 ways. If you're playing heads up, your opponent has a 1/990=0.1% chance of having been dealt the unique hand FG. If you have 9 opponents, BruceZ's "exact approximation" would give a 0.9% probability of running against that FG hand.

RocketManJames
05-02-2003, 08:45 PM
Another example:

Player A has AD. Board has BCEGH. Player B must have FI. I think that this would need to be considered also, though the additive probability is extremely small. Maybe more examples too.

RMJ

Carl_William
05-03-2003, 04:27 PM
Bernie wrote: (regarding Jackpots in holdem)

“holdem.

was wondering what the odds were for a str8 flush over str8 flush with both cards playing would be.

Hi Bernie,

What you ask for is not so straight forward. ( Before I forget I will mentioned I just ordered Wilson’s Holdem software and should have it by next Friday. ) After rec’ing software, I will then attempt to get an answer for you and present in an understandable manner. How I ( at this time) will attempt is by just counting how many times a straight flush is beat Total_SF_Beat for NTrials (where NTrials = number of deals in the computer simulation). Then by probability enumeration (trying to take all possible combinations into account), I will try to accurately determine SFBEAT_Both_Hole, where SFBEAT_Both_Hole = fraction of the time that both Straight Flushes play both hole cards. This last part can be tricky – but maybe it can be determined to a fair degree of accuracy. The answer you desire is calculated by:

Odds_str8_flush_jackpts_% = NTrials/ ( (SFBEAT_Both_Hole)*(Total_SF_Beat)) - 1.0

Where (repeating my self):
NTrials = number of deals or iterations
SFBEAT_Both_Hole = fraction of times that both hole cards are pertinent for jackpot
Total_SF_Beat = total flushes beat for NTrials

There are many little loopholes to take in to account. For instance:
Suppose player A holds (7c, 8c); player B holds (Ac,2c) and the board is:
3c 4c 5c 6c Kd. As you know, this is not a Jackpot because player B cannot play the Ac. I’m sure you can guess some of the other tricky loopholes….

I will analyze various scenarios: when eyerybody plays every hand (i.e., showdown), and also loose, average and tight games. I will also vary the amount of players in a game – maybe 5, 7, 8, 9 and 10 players. As you know:
Jackpots odds increase by the following factors per number of players:
# of Players Jackpot Increase Factor

2 1
3 3
4 6
5 10
6 15
7 21
8 28
9 36
10 45

Most players are not aware of this factor

Carl_William
05-03-2003, 04:42 PM
Slight correct to my last post -- hopefully it is the last)

(# Players) (JackpotIncrease Factor)

2 ---------------- 1
3 ---------------- 3
4 ---------------- 6
5 ---------------- 10
6 ---------------- 15
7 ---------------- 21
8 ---------------- 28
9 ----------------- 36
10 ----------------- 45

Thus the percentage of hitting a jackpot in a ten handed compared to an eight handed game (other things assumed equal) is:

(45/28)*100 - 100 = 60.7 % (pecent greater)
As we realize, this is quit a significat factor.

Most players are not aware of this factor.

DPCondit
05-03-2003, 10:27 PM
Carl William Wrote:" Most players are not aware of this factor. "

In fact, some players seem to think it is easier to hit with less players (for who-knows-what ridiculous reason). I can't say I had ever calculated out a "jackpot increase factor", but it only makes logical sense that it would be much easier to hit with more players.

Don

Carl_William
05-04-2003, 03:59 PM
:" Most players are not aware of this factor. "

"(In fact, some players seem to think it is easier to hit with less players (for who-knows-what ridiculous reason). I can't say I had ever calculated out a "jackpot increase factor", but it only makes logical sense that it would be much easier to hit with more players.

Don )"

Hi Don -- it's a very simple calculation:
It takes two people to make a jackpot. Therefore what I defined as the "jackpot increase (multipication) factor" is simply n players taken two at a time. Example: for 9 players it is:

(9!/(7! * 2!)) = (9*8)/2 = 36

This factor is very convenient to save time (speed up things) when doing Monte Carl trials with Wilson poker PC computer software. A simulation looking into the frequency of straight flushes getting beat can be done for just two players. The frequency predicted for a straight flush getting beat in a two handed holdem games can then be multiplied by 36 to get a pretty accurate result for a nine handed game. Saves lot of computer time when running a large amout of iterations(trials). I have verified this by comparing results for two players and nine players.

DPCondit
05-04-2003, 05:33 PM