anniea111
anniea111
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November 26th, 2009 at 11:12:02 PM permalink
There is a promotion being advertised by a Las Vegas card room.
Make a flush in all four suits and you get $400.
There is a time period of 5 hours...(I think)
You have to have 2 of the suit in your hand.

Assume if 5 clubs come on the board and you have 2c 3c in your hand, that it still counts as a club flush for the promotion.

Assume you play all suited hands.
(Assume 35 hands per hour are being dealt)

Can someone give a good estimate as to the number of hands it might take to accomplish a flush in all four suits.
Very interesting debate we have going on, but NONE of us has a lick of math ability.
Thanks
Annie


I posted this under the help section, before I realized that this was the correct place.
Sorry for the double post.
dk
dk
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November 26th, 2009 at 11:44:33 PM permalink
I'm assuming the game you're playing is Texas Hold'em.

I believe it will take about 552.5 hands on average (a little under 16 hours at 35 hands per hour) for any given individual to make a flush in all four suits assuming that person plays every suited hand dealt and stays as long as they have any chance of making a flush. Here's the math:

Odds of making a flush given 2 suited cards are 135,597/2,118,760 ~ 0.064

Total # of possible starting hands - 1,326

Odds of being dealt any suited hand are 312/1,326 ~ 0.235
So it will take about 1/(0.064*0.235) ~ 66.5 hands to make the first flush.

Odds of being dealt a suited hand in one of the 3 remaining suits are 234/1,326 ~ 0.176
So it will take about 88.8 additional hands to make the second flush.

Odds of being dealt a suited hand in one of the 2 remaining suits are 156/1,326 ~ 0.118
So it will take about 132.4 additional hands to make the third flush.

Odds of being dealt suited cards in the last remaining suit are 78/1,326 ~ 0.059
So it will take about 264.8 additional hands to make the fourth flush.

66.5 + 88.8 + 132.4 + 264.8 = 552.5 total hands to make a flush in all 4 suits on average.
The ratio of people to cake is too big.
dk
dk
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November 26th, 2009 at 11:53:02 PM permalink
Note: If the 5 hour clock starts after your first flush, then the number of hands to make the final 3 flushes would be 486 hands.

As for strategy in this game, I would not change my play until I'd made at least flushes in 2 different suits. At that point, I might be more inclined to play suited hands in the other 2 suits, but I wouldn't go out of my way too much. If I had 3 flushes made, I'd try to see a flop if I was dealt suited cards in the final suit, but even then I wouldn't get too crazy.
The ratio of people to cake is too big.
anniea111
anniea111
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November 27th, 2009 at 12:16:50 AM permalink
DK,
Thank you so much for very quick reply!
Very much appreciated.

So it looks like a promotion that is not likely to be paid in the short qualifying sessions.
The idea is fun, but needs some tweaking to be inviting to players.

Thanks again,
Annie
dk
dk
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November 27th, 2009 at 12:53:09 AM permalink
Quote: anniea111

The idea is fun, but needs some tweaking to be inviting to players.


Actually, I disagree. I would love to play in this game. Most poker players would not know the math, and would be going out of their way to win the bonus by playing many more suited hands than usual. This would make for a more lively game, which is always a good thing for the more solid players.
The ratio of people to cake is too big.
Wizard
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November 27th, 2009 at 1:05:08 AM permalink
dk, I agree completely with your math.

If the player were to foolishly play every hand until the end, or there were no hope for a flush, then from my probabilities in poker page, we see the probability of a flush with 7 cards is 3.03%. So it will take 1/0.03025=33.05 hands to get the first one. For the second it will be 75% of 3.03% = 2.27% per hand, or 1/0.02269=44.07 hands to get the second suit. By the same logic, 66.10 hands to get the third, and 132.21 hands to get the fourth, for a total of 275.44 hands.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
dk
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November 27th, 2009 at 1:11:30 AM permalink
Quote: Wizard

dk, I agree completely with your math.

If the player were to foolishly play every hand until the end, or there were no hope for a flush, then from my probabilities in poker page, we see the probability of a flush with 7 cards is 3.03%. So it will take 1/0.03025=33.05 hands to get the first one. For the second it will be 75% of 3.03% = 2.27% per hand, or 1/0.02269=44.07 hands to get the second suit. By the same logic, 66.10 hands to get the third, and 132.21 hands to get the fourth, for a total of 275.44 hands.


Wizard,

I think you are counting boards with four to a flush where the player has only one of the suit, and boards with a five flush where the player has 0 or 1 of the suit, which is why your number of hands is lower than mine. If the game were 7-stud, I would agree with your numbers.
The ratio of people to cake is too big.
anniea111
anniea111
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November 27th, 2009 at 1:17:56 AM permalink
The idea is fun, but needs some tweaking to be inviting to players.


Actually, I disagree. I would love to play in this game. Most poker players would not know the math, and would be going out of their way to win the bonus by playing many more suited hands than usual. This would make for a more lively game, which is always a good thing for the more solid players.

Also note that while it's unlikely for any one specific player to win, if there are enough players all playing to make flushes, it's not unlikely that someone will win.



OK, I agree about loving to play in a game like this. I should have been more clear.
By inviting to play, I meant a better chance for a player to hit the promo $$.
I would like to introduce a variation of this as a possible promotion in our card room, but I would like to make it so that there is a reasonable chance of it getting hit during the qualifying period,
OR making it so that if nobody makes all four suits....all the players with 3 suits chop the money.
(Think cap on)
Wizard
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November 27th, 2009 at 2:55:49 AM permalink
Sorry, I didn't even notice the requirement that both hole cards hand to be part of the flush. My bad.

Let me pose the question, once you make the first flush, what is the probability that you make the other three within five hours? The assumption of 35 hands per hour was already given, and a five-hour timeframe. So 35*5=175 hands. The mean time to completion, as already stated is 486 hands from the first flush.

I don't think we can say that the time until completion of the three other flushes is a Poisson process. What distribution it does take, I'm not sure. I could easily simulate this, but that is always so unsatisfying. Let me get back to this, it is 3 AM, and I'm too tired to think about it. If there are any mathematicians better than me out there, please chime in.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
dwheatley
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November 27th, 2009 at 7:03:41 AM permalink
Mentioning the Poisson process got me thinking about a continuous approximation. Define X1, X2 and X3 to be random variables representing how many hands it takes to get each flush, and define Y, the number of hands it takes to get the 3 specific flushes, as:

Y = max{X1, X2, X3}

Then the probability of getting all 3 flushes in less than t hands is:

P{Y < t} = P{max{X1, X2, X3} < t}
= P{X1 < t}*P{X2 < t}*P{X3 < t}

If you assume that the occurrence of a specific promo flush is exponentially distributed with a rate of 0.059 * 0.064 = 0.003776 per unit of time t (or hand), then

P{X1 < t} = P{X2 < t} = P{X3 < t} = (1 - e^(-0.003776*t))

so

P{Y < t} = (1 - e^(-0.003776*t))^3

Note this distribution is not exponential, or Poisson...

So, for t = 176, I get P{Y
My estimate is there is a 11.44% chance you will hit the last 3 flushes in 175 hands or less.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
DJTeddyBear
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November 27th, 2009 at 9:37:27 AM permalink
Quote: anniea111

Assume if 5 clubs come on the board and you have 2c 3c in your hand, that it still counts as a club flush for the promotion.

Wow. Considering how tight the rules are for the Bad Beat jackpots, this is quite liberal.

Then again, since you'd have to call bets all the way to the showdown even with a losing hand just to qualify for this bonus, it may make for a more lively game.

But that leads me to wonder.

Suppose you DID play a 2 3 suited to get your qualifier for that suit. Or played equally bad suited cards. Suppose you played SEVERAL potential qualifiers longer than you normally would have and they ended up only being 4-flushes.

How much money have you thrown away, chasing that $400?
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ 覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧覧 Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
Wizard
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November 27th, 2009 at 10:06:25 AM permalink
Quote: dwheatley

My estimate is there is a 11.44% chance you will hit the last 3 flushes in 175 hands or less.



Bravo. Very nice. I should have thought of that. I just did a simulation and got a probability of 11.18%. The difference is probably due to rounding somewhere. This will make a good "ask the wizard" question.

Here is my code. The maximum value for a random number is 4,294,967,296.


void FlushPromotion(void)
{
int clubflush,heartflush,spadeflush,nummin,handcount;
unsigned int rn;
__int64 count,TotWin;
time_t curtime,endtime;
cerr << "enter number of minutes: ";
cin >> nummin;
curtime=time(NULL);
endtime=time(NULL)+60*nummin;
count=0;
TotWin=0;
do
{
count++;
clubflush=0;
heartflush=0;
spadeflush=0;
handcount=0;
do
{
handcount++;
rn=genrand_int32();
if (rn<16168855)
clubflush=1;
else if (rn<32337709)
heartflush=1;
else if (rn<48506564)
spadeflush=1;
}
while ((clubflush==0)||(heartflush==0)||(spadeflush==0));
if (handcount<=175)
TotWin++;
if (count%20000==0)
{
curtime=time(NULL);
cerr << "minutes remaining = " << (float)(endtime-curtime)/60.0 << "\n";
}
}
while (curtime printf("total count=\t%I64i\n",count);
printf("total wins=\t%I64i\n",TotWin);
printf("prob win=\t%f\n",(double)TotWin/(double)count);
}
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
Wizard
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November 27th, 2009 at 1:38:17 PM permalink
Here is how I plan to answer this in my next "Ask the Wizard" column. Thanks for the question, and to dwheatley for his solution, which helped me think of this one.

Let's say your first flush is in spades. At 35 hands per hour, in five hours 175 hands could be played. You then have 175 hands to make a flush in hearts, diamonds, and clubs. I'm going to assume the player never folds a hand that has a possibility of attaining a flush in one of the suits he needs.

The probability of a flush of a specific suit, let's say hearts, using both hole cards is combin(13,2)ラ[combin(11,3)ラcombin(39,2) + combin(11,4)ラ39 + combin(11,5)]/(combin(52,2)ラcombin(50,5)) = 10576566/2809475760=0.003764605.

In the next 175 hands the probability of missing a heart flush would be (1-0.003764605)175=0.51682599.

It would be incorrect to say the probability of failing to make the other three suits would be pr(no heart flush)+pr(no dimaond suit) + pr(no club flush), because you would double counting the probability of faling to make two of them. So you should add back in pr(no heart or diamond flush) + pr(no heart or club flush) +pr(no club or diamond flush). However, that would incorrectly over-subtract the probability of not making all three flushes. So you should add back in pr(no club, diamond, or heart flush).

The probability of going 175 hands and never get either of two specific suits is (1-2*0.003764605)175=0.266442448.

The probability of going 175 hands and never getting any of the three suits left is (1-3*0.003764605)175=0.137015266.

So the answer is 1-3ラ0.51682599 + 3ラ0.266442448 - 0.137015266 = 0.111834108.
Last edited by: Wizard on Feb 11, 2010
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
anniea111
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November 27th, 2009 at 6:56:12 PM permalink
Sorry...removed
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