math question poker prob

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November 27th, 2009 at 9:37:27 AM permalink

DJTeddyBear

Member since: Nov 2, 2009
Threads: 105
Posts: 5737

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?

Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood?

November 27th, 2009 at 10:06:25 AM permalink

Wizard

Administrator
Member since: Oct 14, 2009
Threads: 313
Posts: 6795

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);
}

It's not whether you win or lose; it's whether or not you had a good bet.

November 27th, 2009 at 1:38:17 PM permalink

Wizard

Administrator
Member since: Oct 14, 2009
Threads: 313
Posts: 6795

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)×[combin(11,3)×combin(39,2) + combin(11,4)×39 + combin(11,5)]/(combin(52,2)×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 1-3×0.51682599 + 3×0.266442448 - 0.137015266 = 0.111834108.

It's not whether you win or lose; it's whether or not you had a good bet.

November 27th, 2009 at 6:56:12 PM permalink
anniea111 Member since: Nov 26, 2009 Threads: 2 Posts: 5	Sorry...removed

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