pocket ¢hange average.

[rcbought]

So I was bored tonight and after today's episode was wondering what the average amount of change a contestant gets in Pocket Change.  So I went to random.org and ran 100 trials drawing 4 numbers from 20 and this was the data I got.

Zero(1,2,3)           52
Nickel(4,5,6)           62
Dime(7,8,9,10,11,12)   113
Quarter(13,14,15,16)   87
Fifty(17,18)           38
Seventy Five(19)   28
2 Dollars(20)           20


And with the extra 25 cents that the contestant starts with...  This gives an average of about $1.41  

(I think the exact mathematical average is $1.35)

So I guess my point is.  What does everyone think of that number.  With $1.35 as the average that gives an average of 4 mistakes to be able to win the car.  Which I think is about right.

Wow, that seems a lot higher than what I would have expected, but I'm guessing that my experience is skewed by the shows I have seen, which I believe have generally been a final total of $1.00 or less.

However, the math and reasoning seem right to me.  That's really neat.

[hmtriplecrown]

Two years ago, when Pocket Change debuted, I ran 10,000 trials of the game on a statistical program.  The results of my experimenting is that the median value earned on Pocket Change was $1.10.

I posted the results of my trials somewhere back in the day but I don't remember where, and I don't have the files on my desktop any longer.

EDIT:  Median value was $1.10 (half the results at $1.10 or higher, half at $1.10 or lower).  The average value was higher, thanks to the weights of 75 cent and $2 cards.

[Joe_Mello]

Has anyone tried to get an Expected Value yet?  I would think that's a better indicator than random trials with abritrary positions.

[LiteBulb88]

Just ran it 1,000,000 times in MATLAB;  the average total, including the initial quarter, was $1.3504, verifying Mike's result above.  For MATLAB programmers, here's the code:

val = [zeros(1,3) .05*ones(1,3) .1*ones(1,6) .25*ones(1,4) .5*ones(1,2) .75 2]; % The envelope values
total = 0;
for i=1:1e6 % Loop 1 million times
    tempperm = randperm(20); % Randomly permute the numbers 1-20 
    total = total + (0.25 + sum(val(tempperm(1:4))));  % Take the first four values of the permutation
end

[Joe_Mello]

What the hell?  I'l take my own advice.

.00*3/20=.00
.05*3/20=.0075
.10*6/20=.03
.25*4/20=.05
.50*2/20=.05
.75*1/20=.0375
2.0*1/20=.1
Expected Value=.275 per pick, $1.10 after 4, and $1.35 total.  (At least that's what I think it is)

So why does this game have little room for error?

[LiteBulb88]

Because while the mean of the values is $0.275, the standard deviation is $0.45, which means that picks are likely to stray quite far from the mean.  In other words, the $0.75 and $2 envelopes really make a difference;  without those two, the average envelope value becomes about $0.15.  There's about a 63% chance the contestant does not pick either of those two, which means 63% of the time, the contestant has an average of $0.85--2 mistakes--to work with.  The other 37% of the time, the mean jumps to $2.28, which means that picking the $0.75 or $2 envelope pretty much amounts to an auto-win.

[Joe_Mello]

Yeah, but I can live with two mistakes.  However, most people seem like they can't even get that much.  It's often the case that 2 mistakes means game over.