A new (?) colored marbles in a bag/jar problem

[Silent A]

The idea is as follows:

You have a number of beads of different colors. For the sake of an example, let's say you have 3 reds, 2 blues, and 1 green.

You now withdraw one marble at a time until none are left. The results are categorized as follows:

Colors are ranked in reverse order of when their last marble was removed. In other words, the last marble color remaining finishes 1st and the first color to have all its marbles removed finishes last.

For example, if the above marbles are removed in the order: "RBGRBR" then the final rankings are 1st: Red, 2nd: Blue, 3rd: Green.

So the question is: what is the probability that each color will finish 1st, 2nd, or 3rd?

Also, can you come up with a relatively simple way to do this calculation for any arbitrary set of marbles (any number of colors and any number of marbles per color)?

Finally, can you show that the probabilities for 3 reds, 2 blues, and 1 green are different from the probabilities for 30 reds, 20 blues, and 10 greens?

Note: I don't know the answer to any of these questions (short of solving the first one by brute force via all 720 permutations).

[Siegmund]

Congratulations. You have just reinvented the Independent Chip Model for poker tourney finishes.

First place is easy: each marble is equally likely to be the last one left, therefore each colour's chance of being first is simply the proportion of that colour in the bag at the start.

For second place, you imagine the sack with all of the marbles of the first-place colour removed, and calculate your chance of hypothetically coming in first in a game played with the reduced bag. You have to do this for every possible first-place colour and combine the answers.

Number of cases to evaluate is the factorial of the number of colours, not the factorial of the number of marbles.

In your simple example with 3 red 2 blue and 1 green:

1st place: red 3/6, blue 2/6, green 1/6

2nd place: red = 2/6*3/4+1/6*3/5 = 1/4+1/10 = 7/20 (2/6 chance that blue is first, and 3 out of 4 non-blue marbles are red, plus a 1/6 chance that green is first, and 3 of 5 non-green marbles are red)
blue = 3/6*2/3+1/6*2/5 = 1/3+1/15 = 2/5
green = 3/6*1/3+2/6*1/4 = 1/6+1/12 = 1/4

3rd place: red=2/6*1/4 + 1/6*2/5 = 1/12+1/15= 3/20 (blue first green second + green first blue second) or, alternatively, 1 - 1/2 - 7/20 (1- chance that red is first or second)
blue = 4/15
green = 7/12

There are a few computational shortcuts in special cases - but in general, you can't get around doing approximately that many calculations. (There's a reason why people use ICM for SnGs extensively, but rarely use it for more than the top few places in a big tournament. ICM on a 100-player game would take an eternity.)

[jay_shark]

Ex 1 )

P(R1) = 1/2 ; P(B1)= 2/6 ; P(G1) = 1/6

For R to come in second place , either B comes in first or G comes in first .

P(R2|B1) + P(R2|G1) = P(R2)
2/6*1/4 + 1/6*3/6 = 1/6 =P(R2)

Similarly we have P(R3|B1) + P(R3|G1) = P(R3)
2/6*3/4 + 1/6*3/6 = 1/3 = P(R3)

You can do the others in the same way .

Ok let me explain what I did .
To solve for P(R2) it's worthwhile if you do the following :

_R_R_R_ and notice that there are 4 lines dividing the R's .

So for P(R2|B1) we have B finishing in first which is 2/6 .However , there are 4 possible locations for the green ball . For the green ball to come in last , it must occupy the first division line . So we have 2/6*1/4 = 1/12

To solve for P(R2|G1) we divide the lines the same way

_R_R_R_ . There are 2 blue balls and 4c2 different locations for them . Only three of them give R the second place finish . So 1/6*3/6 = 1/12 . Now just add them .

You can solve all these problems this way and the solution will come quickly .

[jay_shark]

Hey Siegmund , this isn't exactly an ICM problem . Actually that is what I was thinking the first time I read it but the answers are not the same .

[Silent A]

I'm getting different numbers from you guys.

I agree with 1st place: Red = 1/2, Blue = 1/3, and Green = 1/6.

I then calculated the odds of Red coming in last this way:

In order for red to come in last, the first 4 marbles must come in one of the following ways: RRRx, BRRR, RBRR, or RRBR.

RRR: (3*2*1)/(6*5*4) = 1/20
(BRRR, RBRR, RRBR) = 3*(2*3*2*1)/(6*5*4*3) = 1/10

P(R3) = prob red is 3rd = 0.15

so, P(R2) = 1 - 0.5 - 0.15 = 0.35

In order for Green to come in last, you need one of the following combinations: G, RG, RBG, BRG, RRG, RRBG, RBRG, BRRG

G = 1/6
RG = (3*1)/(6*5) = 1/10
RRG = (3*2*1)/(6*5*4) = 1/20
(RBG, BRG) = 2*(3*2*1)/(6*5*4) = 1/10
(RRBG, RBRG, BRRG) = 3*(3*2*2*1)/(6*5*4*3) = 1/10

P(G3) = 31/60 ~ 0.52

so, P(G2) = 1 - 1/6 - 31/60 = 19/60 ~ 0.32

and, P(B2) = 1 - P(R2) - P(G2) = 1 - 0.35 - 19/60 = 1/3
P(B3) = 1 - P(R3) - P(G3) = 1 - 0.15 - 31/60 = 1/3

And these are definitely different from ICM.

Congrats on immediately realizing what I was going for though.

[Silent A]

I'll add this:

It's pretty easy to show that any color with an average number of marbles will always have a 1/n chance of finishing at each position (where n is the number of different colors). This is definitely not what ICM predicts.

Also, I strongly suspect that the relative odds for 30, 20, 10 marbles are not the same as for 3, 2, 1 marbles. I'm basically hoping that this is a way to account for the blind size in chop calculations.

EDIT: Damn. I just did a quick comparison of 2, 1, 1 vs. 4, 2, 2 and they're exactly the same. Oh well, so much for that.

EDIT #2: I also just realized that the way I thought it could be easily shown that an average color is always 1/n doesn't actually work.

[Silent A]

Damn, I just noticed that I forgot the BG combination.

BG = (2*1)/(6*5) = 2/30 = 4/60

so P(G3) = 31/60 + 4/60 = 35/60 (exactly the same as ICM)

P(G2) = 1 - 1/6 - 35/60 = 1/4 (same as ICM)
P(B3) = 1 - 0.15 - 35/60 = 16/60 (same as ICM)
P(B2) = 1 - 1/3 - 16/60 = 2/5 (same as ICM)

So I guess Siegmund is correct, this is just another way of describing ICM.

[pzhon]

[ QUOTE ]
Congratulations. You have just reinvented the Independent Chip Model for poker tourney finishes.

[/ QUOTE ]
Yes. One way to describe the ICM is that you shuffle the chips, and rank the players by the order of their highest chips. You can describe this by looking at the higher chips, or by removing the lower chips one by one.

[ QUOTE ]

ICM on a 100-player game would take an eternity.

[/ QUOTE ]
Actually, I did some computations with 100 player tournaments. See this past thread. Inductively maintain distributions for the highest chip of the player in each place, and update these as you add additional players.

[jay_shark]

Hey Silent ,I made a small blunder on one of my calculations but the idea is still the same .

p(R2) = 2/6*3/4 +1/6*6/10 =0.35
P(R3) = 2/6*1/4 + 1/6*4/10 =0.15

P(B2) = 3/6*1/3 + 2/6*1/4 = 1/4
P(B3) = 3/6*2/3 + 2/6*3/4 = 7/12

P(G2)= 3/6*2/3 + 1/6*4/10 4/10
P(G3) = 3/6*1/3 + 1/6*6/10 = 4/15

I will explain one of the examples but you probably know it anyway . The calculations are the same as ICM .

Here is how you solve for P(R2) .

There is a 2/6 chance that the blue ball finishes first . Given that this happens , the probability that the red ball finishes second is 3/4 . _R_R_R_ . There are 3 spots that satisfy the condition that blue finishes last and 4 equally likely spaces .

Moreover , there is a 1 /6 chance that the green ball finishes in first . Given that this happens , there is a 6/10 chance that the red ball finishes in second .

BBRRR, RBBRR, RRBBR, RRRBB, BRBRR, BRRBR, BRRRB, RBRBR,RBRRB,RRBRB ; of which 6 of them are good choices .

[DrVanNostrin]

[ QUOTE ]
Hey Siegmund , this isn't exactly an ICM problem . Actually that is what I was thinking the first time I read it but the answers are not the same .

[/ QUOTE ]
Go on...