Expert: Josh - 7/19/2004

Question
Hi Josh,

Thanks for being an AllExperts expert and for contributing to the knowledge of others.  My question is a pretty simple one for math-competent people, which I am not. My question is this:

If a person has to guess the color of 6 differently colored flags (black, blue, yellow, white, red, green) that are randomly selected by another person without the guesser being able to see those flags, what would the 'random' chance be that they would get any of them correct if they had 10 chances at guessing? (e.g. 1 out of 10 chance or whatever) Also, what would the chance be that that they would get 5 out of 10, 7 out of 10, 9 out of 10 or 10 out of 10 correct?  

I know this is a wierd question, but it is very applicable and important to my business.  I really appreciate your help and consideration Josh.  Thanks for being an Expert for me. :)

Warmest Regards,

Nate Zeleznick

Answer
Hi Nathan,

Central to the following discussion is the assumption that each guess is statistically uncorrelated with the outcomes from the past. That is to say, guessing right the previous time does not make it more likely or less likely that you guess right again the next time. You also have to accept that a hat trick, getting it right 10 times in a row (however unlikely) is nonetheless theoretically possible.

Intuitively, we know that a person has a one in six chance of guessing right each time. Let "p"=1/6 denote the probability of successfully guessing the flag. Let "N"=10 be the total number of repetitions.

The probability of getting any of them correct is equivalent to finding the cumulative probability that a person guesses right on ONE or MORE occasions.

Solving this problem requires an understanding of  combinatorial and binomial expansion. I won't go into the details.

Case 1: Guessing right on precisely one occasion.
There are C(10,1)=10 ways of guessing right once.
See the pictorial illustration below, where "R"=correct, "W"=wrong.
1 2 3 4 5 6 7 8 9 10
R W W W W W W W W W
W R W W W W W W W W
W W R W W W W W W W
W W W R W W W W W W
W W W W R W W W W W
W W W W W R W W W W
W W W W W W R W W W
W W W W W W W R W W
W W W W W W W W R W
W W W W W W W W W R
The probability of (1R & 9W)=p*(1-p)^9.

Case 2: Guessing right on two occasions.
There are C(10,2) ways of guessing right once.
1 2 3 4 5 6 7 8 9 10
R R W W W W W W W W
R W R W W W W W W W
R W W R W W W W W W
R W W W R W W W W W
R W W W W R W W W W
R W W W W W R W W W
R W W W W W W R W W
R W W W W W W W R W
R W W W W W W W W R OR
W R R W W W W W W W
W R W R W W W W W W
W R W W R W W W W W
W R W W W R W W W W
W R W W W W R W W W
W R W W W W W R W W
W R W W W W W W R W
W R W W W W W W W R OR
W W R R W W W W W W
W W R W R W W W W W
W W R W W R W W W W
W W R W W W R W W W
W W R W W W W R W W
W W R W W W W W R W
W W R W W W W W W R
and so forth....(other patterns are possible, we can account for this exactly, but too many to draw here)

The probability of (2R & 8W)=(p^2)*(1-p)^8.

Continuing this argument,
the final probability is given by
P = C(10,1)*p*(1-p)^9 + C(10,2)*p^2*(1-p)^8 + C(10,3)*p^3*(1-p)^7 +C(10,4)*p^4*(1-p)^6 + C(10,5)*p^5*(1-p)^5 + C(10,6)*p^6*(1-p)^4 + C(10,7)*p^7*(1-p)^3 + C(10,8)*p^8*(1-p)^8 + C(10,9)*p^9*(1-p) + C(10,10)*p^10

where C(N,m)=N!/[m!(N-m)!],
definition of factorial, n!=n*(n-1)*(n-2)*(n-3)*...*3*2*1,
(1-p)=1-(1/6)=5/6 is the probability of guessing wrong during any selection.

======Below is a program that computes these things=====

N=10;
f[0]=1;
for i=1 to N
 f[i]=i*f[i-1];
end

for j=0 to N
 C[j]=f[N]/(f[j]*f[N-j]);
end

p=1/6;
Final_probability=0;

for n=1:N
 probability_of_n_rights=C[n]*(p^n)*(1-p)^(N-n);
end

Final_probability = sum(probability_of_n_rights)
===========================================================
Outputs:
Factorials
0!=1,
1!=1,
2!=2,
3!=6,
4!=24,
5!=120,
6!=720,
7!=5040,
8!=40320,
9!=362880,
10!=3628800.

Combinations, C(j) from j=0 to j=10
1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1
======================================================
Individual proabilities of guessing n out of N correct (from n=1 to n=10)
p(1) = 32.30%,
p(2) = 29.07%
p(3) = 15.50%
p(4) = 5.43%
p(5) = 1.30%
p(6) = 0.22%
p(7) = 0.02%
negligible beyond this.

Final_probability =  0.8385 (83.85%)

Interpretations:
This number means that regardless of how well or poorly someone is doing, we always give them 10 chances (no more and no less). So, a person can get 1 right, 2 rights, 3 rights.....or even 10 rights [odds are 16 in a billion (10^9)].

A person can end up guessing one right 32.3% of the time. If you stop this person from guessing once they get one right and you ask them to walk away (or award them a price if you are nice), 32.3% is the percentage to quote.

If you change your rule, a person can get a MAXIMUM of 10 guesses and may collect up to two prices for guessing two rights, then, the total probability is p(1)+p(2)=61.37%
SO the original rules of "no less and no more than 10 guesses and getting at least one right" has a 83.85% likelihood of happening.

The person has more than a fair chance of getting lucky.
OK, this is what you are looking for.

SUMMARY:
getting EXACTLY 1/10 right: 32.30%     (around 1 in 3)
getting AT LEAST 1 right: 83.85%   (17 out of 20 people)
getting exactly 2/10 right: 29.07%
getting at least 2 right: 51.54%       (around 1 in 2)
getting exactly 3/10 right: only 15.50%
getting at least 3 right: only 22.47%  (around 1 in 5)
getting exactly 5/10 right: only 1.30%
getting at least 5 right: around 1.54%
getting exactly 7/10 right: only 0.0248%
getting at least 7 right:              (267 in a million)

The gambling industry has a very interesting theory on what constitutes random events. I don't recommend using these calculations in Vegas.

I hope this answers your question.

Cheers,

Josh