15 replies to this topic

[unreality]

CASINO GAME
There are five cups.

Under each cup is a paint chip, those little colored cards in hardware stores.

There are only RED and GREEN paint chips.

There is AT LEAST ONE red paint chip and NO MORE THAN THREE green paint chips.

Spoiler for ...

In other words:
Red: from 1-4 paint chips present
Green: from 0-3 paint chips present

Combinations that total five:
2 and 3
3 and 2
4 and 1

Thus the REAL limits are:
Red- 2,3 or 4 present
Green- 3, 2 or 1 present

Your goal is to pick two of one color card in a row. (Your first pick is revealed to you before you pick the second cup)

There is also an option in which you can pick one cup and the Host tells you what color paint chip is under it... but then you CANNOT pick that cup when you do your two picks.

Figure out the probability of winning (two same-color cards in a row) with and without that extra help option. Does it change?

Spoiler for Solution

The possibilties of the five cards (see the above hint):
4 RED, 1 GREEN
3 RED, 2 GREEN
2 RED, 3 GREEN

Each are equally probable. Let's look at them individually now to get the probabilities:

4 RED, 1 GREEN:
When drawing one card, it is a 4/5 probability it is red, then drawing a second card it is 3/4. Thus this possibility has 12/20=6/10= 3/5 chance of getting two cards in a row. Divide by three to get 1/5

3 RED, 2 GREEN:
Both green is 2/5 and then 1/4, thus that probability is 2/20 or 1/10
To get two red, it is 3/5 and then 2/4 (1/2). That is 3/10. Thus we have 3/10 plus 1/10 chance, or 4/10. Divide by three to get 4/30, which is 2/15.

The last scenario:
2 RED. 3 GREEN
Obviously this is the same as the above: 2/15


Add them up:
1/5 = 3/15

3/15 + 2/15 + 2/15 = 7/15

There is a 7/15 chance you will get two same-color cards without help.

WITH THE HELP: Since you can NOT use the drawn paint chip in your two-in-a-row-count, all this does is limit the total from 5 to 4

Before, the possibilities were:

4 RED, 1 GREEN
3 RED, 2 GREEN
2 RED, 3 GREEN

If you draw a RED, they are now:
3 RED, 1 GREEN
2 RED, 2 GREEN
1 RED, 3 GREEN

If you draw a GREEN, they are now:
4 RED
3 RED, 1 GREEN
2 RED, 2 GREEN

Remember, you know whether you picked GREEN or RED... what we want to figure out is how this changes your chances.

We'll do it by scenario, like before:

If you drew a RED:

3 RED, 1 GREEN:
3/4 and then 2/3 chance to get 2 reds: 6/12 or: 1/2 chance

2 RED, 2 GREEN:
1/2 and then 1/3 chance to get 2 of either color: 1/6 chance

1 RED, 3 GREEN:
Same as first (1/2)

To put them together, each of those scenarios has 1/3 chance, so multiply the three chances by 1/3 and then add them. 1/6 + 1/18 + 1/6 = 7/18

So, if you drew a red with the help, your chances are 7/18.... 7/18 is WORSE CHANCES than the 7/15 of not using the help!

But there is still the DIFFERENT chances for if you drew GREEN for the help:

If you drew GREEN:

4 RED: Your chances = 100%, 1/1, 3/3, 5zillion/5zillion, lol

3 RED, 1 GREEN AND 2 RED, 2 RED: these are the same as these same options for if you drew red. Thus 1/2 and 1/6

Combine these three chances after dividing by three: 1/3 + 1/6 + 1/18 = 10/18

Okay lets compare all of em. We have /18 and /15- we shall meet where they collide- 270

NO HELP: 7/15 = 126/270 chance of success

HELP, DREW RED: 7/18 = 105/270 chance of success

HELP, DREW GREEN: 10/18 = 150/270 chance of success

It's obvious that it matters which one you drew... so now we have to figure that probability. Let's go back to the opening chances:
4 RED, 1 GREEN
3 RED, 2 GREEN
2 RED, 3 GREEN
(each out of 5)

Combine all of them:

9 RED, 6 GREEN
(out of 15)

3/5 Draw Red for help
2/5 Draw Green for help

Now we re-do the overall chances:

NO HELP: 7/15 = 126/270 chance of success

(3/5 of help) HELP, DREW RED: 7/18 = 105/270 chance of success

(2/5 of help) HELP, DREW GREEN: 10/18 = 150/270 chance of success

multiply everything by 5, to combine the help options:

126/270 = 630/1350

105*3 = 315/1350
150*2 = 300/1350
Add together = 615/1350

DO NOT USE THE HELP: 630/1350
USE THE HELP: 615/1350

If you decline the HELP option, you have a 15/1350 higher chance of success than if you had. 15/1350 is 1/90.

*wow*... that's a really close margin, lol

I like probability, and I was doing this as I went- yes, I had no idea what the answer would be. So if you spot any errors in my reasoning, PLEASE tell me!

Thanks for reading this huge answer,
unreality

[bonanova]

CASINO GAME
There are five cups.
Under each cup is a paint chip, those little colored cards in hardware stores.
There are only RED and GREEN paint chips.
There is AT LEAST ONE red paint chip and NO MORE THAN THREE green paint chips.

Hmmm...? Interesting. No more than 3 green means at least TWO red.

Anyway, that aside, this gives 26 distributions of chips:

1 - 0 green, 5 red
5 - 1 green, 4 red
10 - 2 green, 3 red
10 - 3 green, 2 red - but stop here: no more than 3 green are present.

Specifically, the distributions are

Spoiler for ...

To make it easier to distinguish the colors, use 0 for red, and use 1 for green.

0 0 0 0 0 - 1 case of 0 green, 5 red

1 0 0 0 0 - 5 cases of 1 green, 4 red
0 1 0 0 0
0 0 1 0 0
0 0 0 1 0
0 0 0 0 1

1 1 0 0 0 - 10 cases of 2 green, 3 red
1 0 1 0 0
1 0 0 1 0
1 0 0 0 1
0 1 1 0 0
0 1 0 1 0
0 1 0 0 1
0 0 1 1 0
0 0 1 0 1
0 0 0 1 1

1 1 1 0 0 - 10 cases of 3 green, 2 red
1 1 0 1 0
1 1 0 0 1
1 0 1 1 0
1 0 1 0 1
1 0 0 1 1
0 1 1 1 0
0 1 1 0 1
0 1 0 1 1
0 0 1 1 1

Case 1:

Your goal is to pick two of one color card in a row. (Your first pick is revealed to you before you pick the second cup)

"In a row" suggests that perhaps all five are picked in sequence and you hope that two consecutive picks are the same color.
[With only two picks, they are always in a row.]
But the rest of the wording suggests you have only two picks.
I will assume you get only two picks.

Spoiler for ...

Since the 26 distributions are equally likely, and your picks are random, absent any clues,
all we need to do is see in how many of the outcomes the first two cups have the same color.
The result is the same if we look at cups 2 and 5, or any other pair of cups.

It turns out that in 12 cases the first two have the same color.

Probability is 12/26 = 0.461

Case 2:

There is also an option in which you can pick one cup and the Host tells you what color paint chip is under it... but then you CANNOT pick that cup when you do your two picks.

Here you pick one cup and are told what color it is.
Then you must pick two other cups.

Again since all outcomes are equally likely, we can assume we pick and are told the color of the first cup,
and then see in how many of the 26 outcomes cups 2 and 3 [or any other pair of cups] are the same color.
We already know that two specific cups have a 12/26 chance of being the same color.

Probability again is 12/26 = 0.461

Figure out the probability of winning (two same-color cards in a row) with and without that extra help option. Does it change?

Spoiler for ...

It makes sense that it doesn't matter about knowing what's under cup 1, since there is no way to make use of that knowledge in deciding which other two cups to pick.

Since there is an unequal likelihood of any specific cup being red or green, your chances of picking two of the same color after eliminating a green cup [5/11 = .454] is different from your chances after eliminating a red cup [7/15 = .467]. But it is more likely [by a 15 to 11 margin -- look again at the distributions] that a red cup is eliminated. Weighting the two probabilities in this way leads to the expected average result of 12/26 = .461

[unreality]

Oops i forgot to add to the very beginning there is at least 1 green as well... drat

well my solution stays the same, if you read it- i was thinking along the lines there was AT LEAST ONE GREEN as well... sorry bovanova.. lol.

However still then we would differ in our answers. You weighted the probabilities, I did not- each chance is the same equality... since we know nothing. Remember the paint chips are not there RANDOMLY... there isnt RANDOM CHANCES which would thus accordinly weight them... since their positions are chosen by the game show, we cannot base our initial setup on the random distribution probabilities but just on our sample space.

[bonanova]

In Case 2, you pick a cup -- and without knowledge this is a random pick -- and then eliminate that cup.
Since there are more red than green cups [in the sample space I used, where 5 reds were possible],
the odds of picking a red cup first are different from the odds of picking a green cup first, and
the odds of getting a pair after picking a red cup differ from the odds of getting a pair after picking a green

Not that that really matters, I just made a note of it.

In the space of all possible outcomes for Case 2, [that includes 15 times a red cup is picked first and 11 times that a green is picked first] the probability of getting a pair after one cup is eliminated is the same as for Case 1.

In either case, it's simply the odds that any two cups are the same color: 12/26 = .461

You could take away three cups at first then ask the odds that the remaining cups are a match.
Same result.

[unreality]

I dont think you got what I meant.... sure you pick a cup randomly, but the actual positions of the cards are 100% chosen by the casino/gameshow... so there isnt random weighting involved- each could be as equally likely as the next- when you take that into account, I believe my solution is correct (tho you might need to look over it )

[bonanova]

the actual positions of the cards are 100% chosen by the casino/gameshow... so there isnt random weighting involved- each could be as equally likely as the next

I think we are agreeing loudly on this point.

The positions of the colors are random [unknown to us] and the combinations of so many reds and so many greens --
it doesn't matter that we used different numbers -- permit an enumeration of all possible,
equally likely distributions.

What's not 50-50 is the chance that any one particular cup is red or green.
Count the number of times in my 26 equally likely color distributions that
the first cup [for example] is red. It's 15. Not 13. -- or is green. It's 11. Not 13.

Next, you see that in 7 of the 15 red-removed cases any 2 of the remaining 4 cups have the same color.
In 5 of the 11 green-removed cases any 2 of the remaining 4 cups have the same color.

If all you are saying is that 7/15 is different from 5/11, then the discussion is over and we can get some sleep.

[Perhaps erroneously] I took your Case 2 question to mean in all cases when a cup is picked and removed from the group
[not a particular case, like a red cup is picked first] does your chance of picking two matching colors
differ from your chance of picking two matching colors when all 5 cups are there.

My answer to that question is that the chances do not change.

This follows from the observation you can make by inspecting the 26 equally likely distributions of 2 colors among 5 cups.
Namely, that whatever two cups you inspect, 12/26 of the time they will be the same color.
That statement is true whether any or all of the other three cups were removed initially or not.

That is, there are no circumstances that force you to pick 2 particular cups that do not have a 12/26 chance of having the same color,
simply because every pair of cups in the 26 equally likely cases has a 12/26 chance of a color match.

Finally, it appears that if you eliminate the 5-red case [requiring at least 1 green]
you will decrease the total distributions from 26 to 25,
and the colors-will-match distributions from 12 to 11.

That would change the answer for both Case 1 and Case 2 to 11/25 = 0.44.

[unreality]

equally likely distributions.

Exactly. Equally likely UNWEIGHTED distributions. You weighted your distributions among 26 options instead of the 3 options that existed (just forget about the 5-0 as that was a typo error by me and 5-0 doesnt exist). As they are not distributed randomly, but chosen deliberately, each of the three options are equally likely as you agreed on with me earlier.

This means your decision to use 26 (well 25, forgetting the 5-0 thing) is wrong and should be 3, if you were indeed agreeing with me.

Sure, if there was 25 equally possible outcomes, 1/5 ? 1/8 or whatever you were saying, of course, but there is not 25 equally possible outcomes, as there is NO weighting!


To prove my point, i think...
roll 2 dice and add them together... you can list the numbers 2-12 and give probabilities, and the probabilities are DIFFERENT. But to get EQUAL things for each case, you have to extend the sample space to be multiple options of the same thing... for simplicity lets roll two 3 sided die...

total:
2 = 1/9
3 = 2/9
4 = 3/9
5 = 2/9
6 = 1/9

however to have all them have equal chances of being picked:
2,3,3,4,4,4,5,5,6

there is 9 equal options there, even though there were only 5 possibilities... the answers are WEIGHTED... some are worth more than others


[/demonstration]

i know thats what you were trying to do here, but there is NO WEIGHTING... there is EQUAL chances as the cups are NOT randomly distributed but PICKED

[roberto0]

This is an interesting discussion, unreality and bononova....too bad I'm coming in 3 months late to add to the discussion.

I think unreality is doing the math right on this one. While bonanova's method of calculating the overall odds of picking 2 of the same from the color from all possible combinations is correct, the critical point is whether you know what you've picked once you've picked it. Put simply, the more information you have, the more the odds change.

For example, when you pick the first chip and see that it is red or green, you now have more information about the remaining chips. You now must choose a chip from a smaller pool. unreality is doing the math right to figure it out.

I'm basing this on the Bayesean logic behind the popular Monty Hall Problem [url:cdca1]http://en.wikipedia....ty_Hall_problem[/url]

[bonanova]

when you pick the first chip and see that it is red or green, you now have more information about the remaining chips. You now must choose a chip from a smaller pool.

Absolutely correct.

[1] If no cup is removed, a particular pair of remaining cups match colors 12 of 26 times.
[2] If a red cup is removed, a particular pair of remaining cups match colors 7 of 15 times.
[3] If a green cup is removed, a particular pair of remaining cups match colors 5 of 11 times.

[4] If we remove a cup 26 times [say cup 1] a particular pair of remaining cups match colors 12 of 26 times.

Choose your interpretation of UR's 2nd question and you have your answer.

I'm basing this on the Bayesean logic behind the popular Monty Hall Problem [url:341ca]http://en.wikipedia....ty_Hall_problem[/url]

In the Monty Hall Problem you are able to take guidance from the added knowledge.
You can deduce that you'll win 2/3 of the time if you swap and 1/3 if you don't.

If you think UR's 2nd question is like the MH problem, then let's give it a try:
You pick a cup, say it's Cup 1, and I tell you it's green.

Based on your knowledge that Cup 1 was green, which of your 6 options are you guided to choose?

Cups 2 and 3
Cups 2 and 4
Cups 2 and 5
Cups 3 and 4
Cups 3 and 5
Cups 4 and 5

[unreality]

[1] If no cup is removed, a particular pair of remaining cups match colors 12 of 26 times.

Right there ^ is where you are wrong. It is not randomly distributed. There would be 26 different options if and only if the chips under the cups were distributed randomly.

They are not distributed randomly, but chosen, hand-picked, determined, pre-set, arranged, placed, put, set, ordered specifically, however you want to put it by the owner of the game. There is no random chances for the cups- no distribution and weighting theory, none of that, because the cups placement's are NOT random. More than that, you have no information about it.

I dont wanna re-explain myself, just read my last post before roberto0's last post (thanks for the backup man, lol, I didnt get around to reading on that Monty Hall thing or the wiki link, but i will later )