math question poker prob

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.

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.

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.

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

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.

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.

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

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.

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.