Your Odds?
Suppose there's fifty people in the room, and each gets to draw from a bag with 100 marbles. 99 are black, one is red. The one that draws the red marble will be the winner. Everyone gets to draw three marbles, examine them and put back two marbles. You are assigned to be the 10th to draw.
What are your odds of being the winner?
What are your odds of being the winner?
posted by Ragknot at
5:32 PM
20 Comments:
First person has 3/50 chance of winning.
Second person has (1 - 3/50) = 47/50 chance of getting to have a dip, then 3/49 chance of winning = 141/2450.
Third person has (1 - 141/2450) chance of having a go, then 3/48 chance of winning.
It gets too tedious to work out what the 10th person's chances are.
You can say, though, that the last two in line have no chance since the 48th person (if he gets there) draws the last three marbles.
To Anon,
Opps, You made a big mistake. There are 100 marbles and 50 people. There's odds that the red marble may not even be found!
The tenth persons odds are:
0.003333333, or .3333%.
Oi, it's not cool to steal someone's brilliant problem!!
That bag and marble thing is MINE!
3 in 82 surely? 27.33(recurring)
3 in 82 for the 10th person is a little high.
Here's a hint... just over 12% of the time, there's no winner, the red marble is never picked.
To James,
Sorry but your bag was to simple. Every one has a 2% chance of winning. I decided to put some computational effort into it with my bag.
I'll change mine to a box with tokens. 99 black, and 1 red, with 50 people, each person get to draw 3 tokens.
If he draws a red, the games over, if not he keeps one token and puts back two.
Your the 10 person to draw three tokens, what are your chances of drawing the red token?
I think this may to hard 97.5% of the people.
For the 10th person to still have a chance the first 9 have to all miss out.
The chance of this happening is 97/100 x 96/99 x 95/98 x .... x 89/92 x 88/91
= (91 x 90 x 89) / (100 x 99 x 98)
= 72.65% approx.
There are 90 marbles left. He has a 1 in 30 chance of picking the red one.
So his overall chance is 2.42%.
Oops, I miscalculated the chances of the first nine missing out (I actually calculated for the first ten).
It should have been 75.13%.
So the answer is 75.13/30 % = 2.5% (near enough).
That is very close. Actually I got this. When #10 draws, there are 91 left so he has 3/91... and
0.7513 * 3 / 91 = 0.0248 = 2.48%
Person 1 has 3.00%
Person 10 has 2.48%
Person 50 has 0.76%
That is very close. Actually I got this. When #10 draws, there are 91 left so he has 3/91... and
0.7513 * 3 / 91 = 0.0248 = 2.48%
Person 1 has 3.00%
Person 10 has 2.48%
Person 50 has 0.76%
100%vif you kill everyone>=)
What is this obsession with killing everyone? We have some troubled people in the room. It's a simple math problem, why are we killing people?
But yes I too got that he has a 3/91 chance if the red ball is still in the bag. Thus times the .7513 calculated above and got 2.48% chance.
I ran the scenario 6 million times
and got these percentages, which came close to the commputed prediction.
Person Percent won
1 3.007%
2 2.933%
3 2.873%
4 2.829%
5 2.767%
6 2.716%
7 2.648%
8 2.601%
9 2.522%
10 2.461%
11 2.403%
12 2.371%
13 2.333%
14 2.276%
15 2.217%
16 2.153%
17 2.114%
18 2.046%
19 1.970%
20 1.946%
21 1.861%
22 1.853%
23 1.816%
24 1.791%
25 1.697%
26 1.674%
27 1.599%
28 1.605%
29 1.535%
30 1.511%
31 1.459%
32 1.414%
33 1.360%
34 1.308%
35 1.302%
36 1.244%
37 1.196%
38 1.161%
39 1.141%
40 1.091%
41 1.065%
42 1.021%
43 1.009%
44 0.948%
45 0.915%
46 0.878%
47 0.836%
48 0.823%
49 0.780%
50 0.765%
no winner 12.154%
Not quite. You all forgot one factor. Someone could draw three marbles, including the red one, but keep the two black ones and put the red one back - since the problem states:
Everyone gets to draw three marbles, examine them and put back two marbles.So, you need to include the probability that this occurs one one or more of the first 9 draws. Now, we can come up with any number of reasons that someone might choose to return the red marble to the bag but let's just go with the observation that 5% to 8% of males are color blind. The problems says "people" so we include the less than 1% of females that are color blind. Genetics tells us these numbers really should be 5% and 0.25% respectively. That gives 5.25/2 or 2.625% of the total population.
So, if Pat is color blind and draws three marbles and puts back two randomly because they all look the same (and Pat's embarrassed to admit being color blind or whatever.) This reduces Pats chances of ending up with the red marble by 1/3rd. In other words, if Pat is first, then the odds of NOT ending up with the red marble are the odds of Pat either not choosing the red marble in the first place OR choosing the red marble but putting it back in the bag: 97/100 + 3/100*1/3.
But, the second part only comes into play IF Pat is color blind so we get:
97/100 + 3/100*1/3*0.02625
This gives us:
(98-i)/(101-i) +
3/(101-i)*1/3*0.02625
or:
(98-i)/(101-i) + 0.02625/(101-i)
as the probability that the i-th person "fails" to end up with the red marble IF the previous i-1 people also failed.
This brings the probability of the 10th player up to:
0.0250484001090026111
Assuming we do not know if the 10th player is color blind or not.
-h
To Hägar,
Sorry you are wrong.
The post says "The one that draws the red marble will be the winner."
The game is over, when we see the red marble drawn. The winner does not place any back, because the second it is drawn the game is over.
You have a point, however:
* It does NOT say the game ends when the red marble is drawn.
* it does say the person "drawing" will be the winner - which could be interpreted as will be the winner at the end of the drawing. My argument would be better if it said the one that ends up with the red marble...
* But, MORE importantly, it does say that everyone gets to draw... which would not make sense if everyone got to see the marbles as they are drawn.
I suspect either interpretation of the problem is arguable.
-h
Hagar you are wrong with your estimates anyway. What you are referring to as "colorblind" is red-green colorblind. I have that and rest assured I can distinguish reds and greens just fine. I would have no problem determining what ball is red and what is black. Now if you are wanting actually color blind (black and white) the genetic odds are very VERY low that you have someone who is truly colorblind.
Also true. But it's a very dimly lit room and somewhat difficult for even those of us fairly good eyesight to tell black from red...
Well, could be... Of course I'd try pretty hard since the winner (if any) gets ALL the marbles!
Actually, the color blindness was just an example of what one might come up with as a quantifiable reason for someone returning the red marble to the bag. But, probably not intended to be part of this problem.
-h
If 50 people draw three each, thats more than 100 marbles. but the first 33 people have a 33.333333 % chance
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