Mathematicians - Engineers - Wizards Help with Odds (1 Viewer)

[Thomacetti]

depend...is the hero yelling 'one time' or not...and if so, when was it called ?

 

[Thomacetti]

But this is too precise for poker. In a game I'd just round this up to 25%.

this...and everybody knows that LAG player round up everything to a flip
I like these odds....sidebet it hits ?

 

[asian bino]

The odds for a single card hitting the board is:
1 - 47/52 = 0.0961

If you do it 10 times, then it’s 0.961. Or a 96.1% chance that the card hits the board in 10 hands.

In 100 hands, it should happen 9.61 times.

No troll my friend, just maffs

So if I do it *11* times, my chances are OVER 100%, and thus, it HAS to happen? Is that something we can wager on?

 

[Geremie]

So if I do it *11* times, my chances are OVER 100%, and thus, it HAS to happen? Is that something we can wager on?

On average, it will happen once per 10-11 hands. Not 100% guaranteed

 

[BGinGA]

Guys, guys......It always happens.....until you bet something of value on it. Some weird variation of Murphy's Law.

 

[v1pe]

So if I do it *11* times, my chances are OVER 100%, and thus, it HAS to happen? Is that something we can wager on?

This is why slot machines are effective money makers

 

[bigdonkey]

This is why slot machines are effective money makers

Because they pay out over 100%? I'm not seeing the logic.

 

[asian bino]

The odds for a single card hitting the board is:
1 - 47/52 = 0.0961

If you do it 10 times, then it’s 0.961. Or a 96.1% chance that the card hits the board in 10 hands.
On average, it will happen once per 10-11 hands. Not 100% guaranteed

My point was that you said the odds were 9.6%, and so if you do it 10 times, the odds are 96%, and that is NOT how it works....

If the odds of something occurring is 9.6%, and you do it 10 times, it still only has about a 65% chance of occurring.

You take the chance of it *NOT* occurring, (90.4), raise it to the power of trials (10) and subtract this from 100 percent.

96% is a LOT different than 66%.

 

[fish72s]

The odds over 10 hands are the 1 minus the odds of it not hitting in any of the 10 boards :

1 - (1-0.0961)^10 = 0.63591409695

*assuming I am not wrong.
** the odds of me being wrong are 35.31%
*** unless the odds of me being wrong are wrong.

 

[fish72s]

Just like always, I was the last one done taking the test because I just had to check my maff.

The odds over 10 hands are the 1 minus the odds of it not hitting in any of the 10 boards :

1 - (1-0.0961)^10 = 0.63591409695

*assuming I am not wrong.
** the odds of me being wrong are 35.31%
*** unless the odds of me being wrong are wrong.

 

[Geremie]

My point was that you said the odds were 9.6%, and so if you do it 10 times, the odds are 96%, and that is NOT how it works....

If the odds of something occurring is 9.6%, and you do it 10 times, it still only has about a 65% chance of occurring.

You take the chance of it *NOT* occurring, (90.4), raise it to the power of trials (10) and subtract this from 100 percent.

96% is a LOT different than 66%.

My bad...I meant on average it will occur 0.961 times with 10 boards

 

[Legend5555]

Others have done this already, but hopefully you understand the explanation here. Revisiting some high school math, we use the factorial. 5! = 5*4*3*2*1 = 120. 10! = 10*9*8*7*6*5*4*3*2*1 = 3628800. We also use the combination XcY = X!/(Y![X-Y]!). We choose Y things from X things and the order of those Y things doesn't matter. So a group of letters like "abcd" is counted the same as "dbca." Just like how the order of the flop cards doesn't matter.

Assuming you know NOBODY's hole cards including your own:

The probability that the Kh hits the board is equal to (1 - the probability it doesn't hit).

There are 52c5 combinations of the board or 52!/(5!47!) = 2598960.
There are 51c5 combinations that don't include the Kh, or 51!(5!46!) = 2349060
Making the probability of the Kh NOT hitting the board is: 2349060/2598960 = 0.904

So probability that the Kh will hit the board is: 1 - 0.904 = 0.096 or 9.6%.

Another way to think of it is that it does hit within the 5 cards somewhere. So you only need to figure out the number of combinations of the other 4 cards.
Which is 51c4 or 51!/(4!47!) = 249900. And 249900/2598960 = 0.096.

So on average about 1 in every 10.41 full board run outs will contain the Kh.

The probability that at least 1 board out of 10 will have the Kh is (1 - the probability none of the 10 boards have the Kh).

Probability that 10 boards DO NOT have the Kh is: .904^10 = 0.3645

So probability that at least 1 of the 10 boards has the Kh is: 1 - .3645 = 0.6355 or 63.55%.

 

[links_slayer]

50/50....always 50/50

 

[MatB]

If you're a true degenerate, you don't need math. is you're big toe tingling? then it will hit. Simple.

Math is great for poker and can dial in your game, but theres a reason when they land a rocket on the moon they dont call poker players in to figure it out.

 

[Darthhoodie]

 

[asian bino]

My bad...I meant on average it will occur 0.961 times with 10 boards

Again, *NO* it won't.

 

[Highli99]

Others have done this already, but hopefully you understand the explanation here. Revisiting some high school math, we use the factorial. 5! = 5*4*3*2*1 = 120. 10! = 10*9*8*7*6*5*4*3*2*1 = 3628800. We also use the combination XcY = X!/(Y![X-Y]!). We choose Y things from X things and the order of those Y things doesn't matter. So a group of letters like "abcd" is counted the same as "dbca." Just like how the order of the flop cards doesn't matter.

Assuming you know NOBODY's hole cards including your own:

The probability that the Kh hits the board is equal to (1 - the probability it doesn't hit).

There are 52c5 combinations of the board or 52!/(5!47!) = 2598960.
There are 51c5 combinations that don't include the Kh, or 51!(5!46!) = 2349060
Making the probability of the Kh NOT hitting the board is: 2349060/2598960 = 0.904

So probability that the Kh will hit the board is: 1 - 0.904 = 0.096 or 9.6%.

Another way to think of it is that it does hit within the 5 cards somewhere. So you only need to figure out the number of combinations of the other 4 cards.
Which is 51c4 or 51!/(4!47!) = 249900. And 249900/2598960 = 0.096.

So on average about 1 in every 10.41 full board run outs will contain the Kh.

The odds that at least 1 board out of 10 will have the Kh is (1 - the probability none of the 10 boards have the Kh).

Probability that 10 boards DO NOT have the Kh is: .904^10 = 0.3645

So probability that at least 1 of the 10 boards has the Kh is: 1 - .3645 = 0.6355 or 63.55%.

Without doing any maff I was going to guess that 60% of the time it will hit every time. I would have been close! My poker maff is like a broken clock, but I still claim victories.

 

[Frogzilla]

My bad...I meant on average it will occur 0.961 times with 10 boards

Again, *NO* it won't.

Oh yes it will. And this is what I thought the OP is actually asking. His answer was phrased incorrectly first time, but this one is spot on.

 

[MatB]

I think the issue is saying ‘it will occur”

 

[Geremie]

Math is easy, words are tough

 

[allforcharity]

You're all assuming that every hand is going to be dealt to the river. If the hand is over before seeing a turn more than 50% of the time, then you could go way more than 1-2 orbits on average before seeing a particular card on the board.

 

[Legend5555]

You're all assuming that every hand is going to be dealt to the river. If the hand is over before seeing a turn more than 50% of the time, then you could go way more than 1-2 orbits on average before seeing a particular card on the board.

Yes, but that's not what was originally asked.

 

[WedgeRock]

I’d say:

1 - (51/52 * 50/51 * 49/50 * 48/49 * 47/48)

Agreed, but cancel out common numerators and denominators... Make it easy on your self...

1 -
(51/52*
50/51*
49/50*
48/49*
47/48)

=

1-(47/52)

=0.0972, or about 10% of the time, if all five streets hit the board.

 

[toynoob]

Seems to be confusion, i am not looking for poker strategy/odds etc. Just need to know how often a king of hearts hits he board if dealt 10 hands to the river. That's it. If it's 10% great, that's all I needed. Now if you have stats that show how often a hand never makes it to river and can relate it to my question that's fantastic. Obviously every crew is different I know that, I don't need to be to the 1000 of a percent dead nuts. Thanks for those of you who got to the point and gave me a hard number so we can move on to the next obstacle.

 

[WedgeRock]

Seems to be confusion, i am not looking for poker strategy/odds etc. Just need to know how often a king of hearts hits he board if dealt 10 hands to the river. That's it. If it's 10% great, that's all I needed. Now if you have stats that show how often a hand never makes it to river and can relate it to my question that's fantastic. Obviously every crew is different I know that, I don't need to be to the 1000 of a percent dead nuts. Thanks for those of you who got to the point and gave me a hard number so we can move on to the next obstacle.

That's gonna depend on the stage of the tournament, in my experience.

But, rather than a zombie cure at the final table (or the final 3), it seems that if fewer hands that go to the flop, it actually works to minimize zombie penalty late...

 

[fish72s]

Seems to be confusion, i am not looking for poker strategy/odds etc. Just need to know how often a king of hearts hits he board if dealt 10 hands to the river. That's it. If it's 10% great, that's all I needed. Now if you have stats that show how often a hand never makes it to river and can relate it to my question that's fantastic. Obviously every crew is different I know that, I don't need to be to the 1000 of a percent dead nuts. Thanks for those of you who got to the point and gave me a hard number so we can move on to the next obstacle.

I think you got your answer:
A Kh will hit in at least 1 of 10 boards 63.591409695 % of the time.
Unless you want to know probability of a Kh hitting on EXACTLY one of 10 boards. That's an entirely different question.
What's the next obstacle?

The odds over 10 hands are the 1 minus the odds of it not hitting in any of the 10 boards :

1 - (1-0.0961)^10 = 0.63591409695

5/52 = 9.61% chance Kh on board if dealt to river. how many players are in the hand is irrelevant, its like counting outs; you don't care that someone else might be holding them.

p(Kh hitting the board in one orbit) = 1 - p(Kh never hits the board in an orbit) = 1 - (1 - (5/52))^10 = 63.6%

My point was that you said the odds were 9.6%, and so if you do it 10 times, the odds are 96%, and that is NOT how it works....

If the odds of something occurring is 9.6%, and you do it 10 times, it still only has about a 65% chance of occurring.

You take the chance of it *NOT* occurring, (90.4), raise it to the power of trials (10) and subtract this from 100 percent.

96% is a LOT different than 66%.

Others have done this already, but hopefully you understand the explanation here. Revisiting some high school math, we use the factorial. 5! = 5*4*3*2*1 = 120. 10! = 10*9*8*7*6*5*4*3*2*1 = 3628800. We also use the combination XcY = X!/(Y![X-Y]!). We choose Y things from X things and the order of those Y things doesn't matter. So a group of letters like "abcd" is counted the same as "dbca." Just like how the order of the flop cards doesn't matter.

Assuming you know NOBODY's hole cards including your own:

The probability that the Kh hits the board is equal to (1 - the probability it doesn't hit).

There are 52c5 combinations of the board or 52!/(5!47!) = 2598960.
There are 51c5 combinations that don't include the Kh, or 51!(5!46!) = 2349060
Making the probability of the Kh NOT hitting the board is: 2349060/2598960 = 0.904

So probability that the Kh will hit the board is: 1 - 0.904 = 0.096 or 9.6%.

Another way to think of it is that it does hit within the 5 cards somewhere. So you only need to figure out the number of combinations of the other 4 cards.
Which is 51c4 or 51!/(4!47!) = 249900. And 249900/2598960 = 0.096.

So on average about 1 in every 10.41 full board run outs will contain the Kh.

The probability that at least 1 board out of 10 will have the Kh is (1 - the probability none of the 10 boards have the Kh).

Probability that 10 boards DO NOT have the Kh is: .904^10 = 0.3645

So probability that at least 1 of the 10 boards has the Kh is: 1 - .3645 = 0.6355 or 63.55%.

 

[toynoob]

I think you got your answer:
A Kh will hit in at least 1 of 10 boards 63.591409695 % of the time.
Unless you want to know probability of a Kh hitting on EXACTLY one of 10 boards. That's an entirely different question.
What's the next obstacle?

Do we want the zombie to bleed a small blind or a big blind when that Kh hits the board

 

[MatB]

I too, am very impressed with all the number wizardry

But I didn’t go to that high school so can you explain it like i’m 5?


52 cards ( assume no one can see anyone’s cards and you can’t see your own)
5 cards dealt on the board.
Before the 5 cards are dealt every card has a 2% chance so chance of KH coming out in the 5 cards is 10%?

 

[fish72s]

Do we want the zombie to bleed a small blind or a big blind when that Kh hits the board

You might need to constrain the problem a bit more.

Where do these bleeders lie on the zombie scale?
Are we talking "Shawn of the Dead" or "World War Z" ?

 

[Legend5555]

I too, am very impressed with all the number wizardry

But I didn’t go to that high school so can you explain it like i’m 5?


52 cards ( assume no one can see anyone’s cards and you can’t see your own)
5 cards dealt on the board.
Before the 5 cards are dealt every card has a 2% chance so chance of KH coming out in the 5 cards is 10%?

The way you want to think about it are how many boards are possible, and how many boards contain the .

Say you have 6 items labeled ABCDEF. If you choose 2 of them, then how many different ways can that happen? Think of it like there are 2 blank spaces to fill __ __.
How many choices do you have for the 1st letter? 6 (ABCDE or F)

Let's assume you choose A.

After you put that letter into the first spot, you now have 5 left to choose from. So you can get AB, AC, AD, AE, AF. That's 5 possibilities. But you could have done that with any of the 6 letters as the first letter. Since there are 6 choices for the first letter, and each of those choices has 5 choices for the 2nd letter, that's 6*5 = 30 permutations of 2 of the 6 letters.

Now let's assume that the order of the letters doesn't matter, much like the order of the flop in poker. So AD is the same as DA for all we care. How do we discount all those possibilities? Easy, how many ways can you arrange two letters? Just 2. So 30/2 = 15. So there are 15 combinations of 2 of the 6 letters.

What if we took 4 letters from the 6? 6 choices for the 1st letter, 5 for the 2nd, 4 for the 3rd, 3 for the 4th. So 6*5*4*3 = 360 permutations.

What if we don't care about the order? Well, how many ways are there to arrange 4 things? That's just another permutation with 4 things and us picking all 4 of them. So 4*3*2*1 = 24. So the number of combinations of choosing 4 things out of 6 things is 360/24 = 15. There is a reason why this is the same as the number of combinations of 2 things out of 6 things, but it's not important for this discussion.

So we do the same thing for a deck of 52 cards. Since we want the we just assume we already have it. So now there are 51 cards left, and we need 4 more.

So: 51*50*49*48 (number of permutations of 4 cards from 51 cards)
and 4*3*2*1 (number of ways to arrange those 4 cards)

5997600/24 = 249900 different boards that have the .

But how many different boards are possible?

(52*51*50*49*48) / (5*4*3*2*1) = 2598960

So 249900 boards out of 2598960 will have the , or 249900/2598960 = 0.096153846

Or approximately 9.6%.