KasinoKing's SlotBeater Strategy

[Pulver]

edit.

 

[DiamondGeezer]

was that your back up guess Pulver?

 

[Pulver]

Excuse my French - but what total bollocks!

It's nothing to do with what colour the sides are, or which side you chose to show me.
You said there are 2 chips in the bag and you pick one at random.
The chance of picking either chip IS 1 in 2.

And people say my maths is way out!

KK

Yeah, but the important thing to remember is that you are shown one side, leaving 3 sides unknown, you know that 2 of the sides are red, and 1 of the sides are green, hence the odds are 1\3. (One out of 3 unknown sides are green)

If he did not show you any of the sides before you had to choose, the odds would be 1\2.

The tricky part is that you are shown 1 side.

 

[Pulver]

There is another riddle that deals with the same concept, all I remember is that it has something to do with that gameshow with the briefcases were one of them contains a million or something. (What was the name of that show again?).

Does anyone remember that riddle?

 

[3Dice]

Excuse my French - but what total bollocks!

It's nothing to do with what colour the sides are, or which side you chose to show me.
You said there are 2 chips in the bag and you pick one at random.
The chance of picking either chip IS 1 in 2.

And people say my maths is way out!

KK

KK, at least you are proving that you honestly don't understand what is wrong with your reasoning. Once you are shown that one side you KNOW it is red. Thats an absolute given. The odds for that side to be red are ONE at that point. You have to take that into account. If I showed you a side that was green, you would say the odds are 1 that I'm holding the chip with the green right - and not 1/2 ..

This is literally a schoolbook example by the way.

 

[Jasminebed]

Enzo, the question asked was "What are the odds that I picked the chip with the green ?"

I maintain two chips in the bag, 1/2 (or 50/50) odds that you picked the chip with the green.

Had the question been "What are the odds of you guessing correctly which chip I picked" that would have been a different answer.

 

[Pulver]

** to see why, put it like this .. I have 4 sides in my bag, 3 are red and 1 is green. I show you one side. its red, what is the chance that for any of the remaining sides to be green ?. one in three. Three sides left, one is green so one in three.

The crucial factor.

 

[elscrabinda]

There is another riddle that deals with the same concept, all I remember is that it has something to do that gameshow with the briefcases were one of them contains a million or something. (What was the name of that show again?).

Does anyone remember that riddle?

I'm reluctant to bring this up, as whenever I've tried it on people it ends up with them swearing blind (and often swearing) that they are right, so goodness knows what will happen on here! (Its worth it though for the penny dropping moment that does usually occur when they realise they're wrong though)


Three boxes on a game show. One has a prize in, the other two are empty. You, the contestant, chooses one, but before it is opened to see if you win the gameshow host opens one of the other boxes that he knows to be empty (it is important to understand that the host knows where the prize is). He then gives you the choice to stick to your original guess, or to switch to the other remaining closed box. What should you do, or does it make no difference?

 

[3Dice]

Enzo, the question asked was "What are the odds that I picked the chip with the green ?"

I maintain two chips in the bag, 1/2 (or 50/50) odds that you picked the chip with the green.

Had the question been "What are the odds of you guessing correctly which chip I picked" that would have been a different answer. This makes the assumption for the one in three answer to be correct that you have the ability to choose which side to show.

How is

"what are the odds that I picked the chip with the green"

different from

"what are the odds that the backside of this chip I picked, which I have not yet shown you the backside of is the one green side from the remaining three sides you didn't see ?"

Jasmine ?

The answer is 1/3, I was very careful in how I posed the question. There is no room for discussion. The answer is 1/3.

It is however contra-intuitive, like the type of statistics that we deal with all the time in gambling.

Cheers,

Enzo

 

[3Dice]

I'm reluctant to bring this up, as whenever I've tried it on people it ends up with them swearing blind (and often swearing) that they are right, so goodness knows what will happen on here! (Its worth it though for the penny dropping moment that does usually occur when they realise they're wrong though)


Three boxes on a game show. One has a prize in, the other two are empty. You, the contestant, chooses one, but before it is opened to see if you win the gameshow host opens one of the other boxes that he knows to be empty (it is important to understand that the host knows where the prize is). He then gives you the choice to stick to your original guess, or to switch to the other remaining closed box. What should you do, or does it make no difference?

I guess I'm not allowed to answer

 

[DiamondGeezer]

The thing is Jaz, if a feature has a 1 in 100 chance of hitting, and you have just done 300 spins without it, you gotta admit you've been pretty unlucky, no?
Do another 100 and still nothing, you've been very unlucky.
Another 100, you've been very very unlucky.
And so on...

I suppose it depends on just how unlucky you think you can be!

KK

This is where conceptually you are going wrong KK!

This has no bearing on your profit expectation. Your profit expectation is the formula Enzo posted in his long post above.

It is not hitting a feature on a raised bet that determines the profitibility of your system but the bonus and WR you do.

Enzo there are a few casino's I hope you never join . I hope you are happy at 3DICE and will stay forever

 

[DiamondGeezer]

There is another riddle that deals with the same concept, all I remember is that it has something to do with that gameshow with the briefcases were one of them contains a million or something. (What was the name of that show again?).

Does anyone remember that riddle?

The Monty Hall problem.

 

[RobWin]

Enzo, the question asked was "What are the odds that I picked the chip with the green ?"

I maintain two chips in the bag, 1/2 (or 50/50) odds that you picked the chip with the green.

Had the question been "What are the odds of you guessing correctly which chip I picked" that would have been a different answer.

Jazzy, if one chip were totally green and one chip were totally red then the answer would be 1/2 or 50/50 chance...
____
____

 

[Pulver]

The Monty Hall problem.

Bang! Thank you for that DiamondGeezer :notworthy

(There is a really long wikipedia post about this problem for those interested)

 

[Jasminebed]

Okay, let's talk semantics and not math.

The odds of you picking the chip with the green side are 1/2. Your choice of the word "I show" implies that you decided which side to reveal.

I'd have to say if you "showed" the green side, the odds were 100% you picked the one with the green side. But what would be the fun in that

 

[LaHutti]

I have to disagree on the Chip quiz thing as well.
If the question was "What are the odds, that the SIDE you can't see, on the chip you're holding is green" ? odds would be 1/3 since you have 3 unknown SIDES, and it can only be one of them.
The question was "What are the odds, that you picked the CHIP with one green side" and the odds of that is obviously 1/2, since you only have 2 CHIPS to choose from.

You know my username Enzo ... LOL

 

[3Dice]

Okay, let's talk semantics and not math.

The odds of you picking the chip with the green side are 1/2. Your choice of the word "I show" implies that you decided which side to reveal.

I'd have to say if you "showed" the green side, the odds were 100% you picked the one with the green side. But what would be the fun in that

It has little to do with semantics. I don't need to have decided which side to show.

What matters is that it was red. Whether random or picked makes no difference (in this sample). I'm pulling it random, we look at it at the same time and we simply see its red. That however is information you cannot simply ignore. Before we looked chances that it was red were 3/4 but after we looked, and saw it was red, the chance for it is 1.

This is the very mistake that inspires KK's thinking and strategy. Statistics are a very peculiar field in mathematics - one can be a genius at calculus, algebra, geometry .. and be a noob at statistics. Compared to other fields of maths it is very very very contra-intuitive. You cannot trust your common sense there you have to trust the numbers and the formula's.

KK, did someone mention the bomb example here yet ? The chance to have a bomb on an airplane is one in a million, the chance to have two bombs is 1 in (million * million). So I always take a bomb on the plane. I'd have to get super-unlucky then instead of normal unlucky . Of course if you take the bomb in the plane, the chance of that is 1, and the chance of a second bomb is 1 in a million again. If you search a random plane, and you find a bomb, the chance of a second bomb is also just one in a million. Before you start searching the chance of finding a second bomb is one in a (million*million).

If the chance of having an unlucky session is 9 in 10, then before you play the chance for two unlucky sessions is 81 in 100 or 8.1 in 10. So if you lose one we call it unlucky, if you lose both we call it very unlucky.

you play one session.

9 times this is unlucky. (9/10)
1 time its lucky (1/10)

we play a new session

9 times its unlucky (9/10)
1 time its lucky (1/10)

the combined outcomes will be p1 * p2

so 9/10 * 9/10 that both are unlucky so we are very unlucky.
and 1/10 * 1/10 that both are lucky
and 1/10 * 9/10 first lucky, second unlucky
and 9/10 * 1/10 first is unlucky, second is lucky.

between the two sessions, if you lost the first one, the chance of becoming very unlucky is 9/10 AND NOT 81/100. You already have unlucky, that is certain. you need another unlucky to become very unlucky .. coming at exactly the same odds as your first unlucky.

so before you play you have 81% chance of becoming very unlucky, but after one session this chance goes either to 0 (if you won the first we cant become very unlucky anymore), or 9/10 .. if we lost the first we now have 90% chance of becoming very unlucky ( and not 81% even tho only 81% of any two consecutive sessions would be very unlucky).

Cheers,

Enzo

 

[Scooter7]

I'm enjoying this thread now. It was a little tedious before, but has turned into a good discussion. Here is what Wiki has to say about the Monty Hall Problem:

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I liked the section that takes the number of doors/boxes to 1,000,000. The scale of that example as a type of limit equation gives more clarity.


Increasing the number of doors

It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goatsimagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors.

Stibel et al. (2008) propose working memory demand is taxed during the Monty Hall problem and that this forces people to "collapse" their choices into two equally probable options. They report that when increasing the number of options to over 7 choices (7 doors) people tend to switch more often; however most still incorrectly judge the probability of success at 50/50.

 

[KasinoKing]

Say I have a closed bag with in it two colored chips. One is red on two sides, and the other is green on one side and red on the other. I take one chip out of the bag and show you one side. You see red. What are the odds that I picked the chip with the green ?

This is the critical point, as Pulver said.

"I take one chip and show you one side" implies to me that you chose which side to show me.

I don't know what you are trying to achieve with these semantics, but I am not amused.

KK

 

[3Dice]

KK,

I'm trying to explain the flaw in your thinking. I'm not trying to offend you but to help you.

It doesn't matter if I pick or not. What matters is that you see it. You KNOW its red. that's what matters. What matters is that once you know it, its an event in the past with a probability of one.

If I chose the red side and show you the odds are 1/3
if I pick a random side and show you, and its red, the odds are still 1/3

makes no difference. the difference is showing you.

If I dont show you the odds are 1 in 2
if I show you its either 1 (if green showed, we're sure) or 1 in 3 (if red showed).

Cheers,

Enzo

 

[LaHutti]

I have to disagree on the Chip quiz thing as well.
If the question was "What are the odds, that the SIDE you can't see, on the chip you're holding is green" ? odds would be 1/3 since you have 3 unknown SIDES, and it can only be one of them.
The question was "What are the odds, that you picked the CHIP with one green side" and the odds of that is obviously 1/2, since you only have 2 CHIPS to choose from.

You know my username Enzo ... LOL

You can ignore it, but it won't go away ...lol

 

[3Dice]

Ok guys,

I'm not giving up yet ..

for those who are good in mathematics .. this may work better ..

Ai = chosing two color chip.
B = showing red.
Aj = chosing one color chip.

P(Ai after B) = P(Ai) * P(B after Ai) / ( P(Ai)*P(B after Ai) + P(Aj)*P(B after Aj) )

P(two color chip after red shown) =

P(choosing two color chip) * P(showing red if two color chip is chosen) / ( P (choosing two color chip) * P(showing red if two color chip is chosen) + P (choosing one color chip) * P(showing red if one color chip is chosen))

= 1/2 * 1/2 / ( 1/2 * 1/2 + 1/2 * 1)

= 1/2 * 1/2 / ( 3/4 )

= 1/2 * 1/2 * 4/3

= 4/12 = 1/3

 

[LaHutti]

ok Enzo....that's Russian to me, but if by math, I can make 2 coins 3, I'm done gambling, and will learn that, and do it all day

 

[3Dice]

La hutti.

two coins yes. But I didn't show you a coin. I showed a side. there are 4 sides.
3 you havent seen. 1 of those 3 sides you didn't see is green.

asking if I'm holding the two color chip is the same as asking if the side you didn't see yet is green.

the odds are NOT 1 in 2.

Cheers,

Enzo

 

[KasinoKing]

Post in haste - repent in leisure. Me that is!

I went for a shave, thought about it, and finally got it.
It's sort of a trick question due to the clever wording - and I fell for it!
(The key being it's past tense).

Before you show the side, the odds of the chip having 1 Green side are 2/4.
After showing it's red, you have removed a 1/1 chance of it being red.
Therefore 2/4 minus 1/1 = 1/3.

There is a 1 in 3 chance of the other side of the picked chip being Green.

Enzo, you are too clever for me!
Sorry for any offense.
KK