Probability and how it relates to RC Modeling
 

Question- what do geeks do to blow off steam? Answer- get into a debate about probability. Statistics and probability are the foundation of reliability engineering. That is what put a man on the moon.

Ok, here is the opening question. A certain world renound pilot states that no matter what competition he attends, he always draws fist in the pilot order. So here we go.

If there are 10 pilots in the finals, and its time to draw the flight order for the following day. Ten pieces of paper in a hat, each has a number (1 through 10) on each piece of paper. If this pilot does NOT want to be selected to be first up, which of the three choices should he pick?

1. It does not matter- every pilot who draws has 1/10 chance of drawing first.
2. It is better if he waits for a few people to draw before going up and picking a number.
3. He runs to the front of the room and draws first.

Hi Don - my answer is #1. Each pilot has a 1/10 chance of drawing the #1 regardless of if they draw first or last. The first pilot has a 1/10 prob of drawing #1. This leaves a 9/10 prob that the first to draw doesn't draw a #1. The second to draw then has a 9/10 x 1/9 prob (equals 1/10) of drawing a #1. Third to draw has a 8/10x1/8 prob (also equals 1/10) of drawing a #1 and so on.

Steve

I say #3. Of course, this is all under the assumption that the "1" remains in the hat as you go along. Once it's pulled, all bets are off!

The odds of pulling "1" increase as the number of pieces in the hat diminish. The odds start at 10% pulling "1". When there are 2 piecs left, the odds are at 50%.

Therefore, the earlier you draw, then then less likely to pull the "1". The longer you wait, the more your fate rests in the hands of others. I'm sure mathematics will have something to say about this, but then again, so does philosophy!

I studied and learned about this thing called a "Probability tree". Pretty simple, actually- a map of all possible solutions, with all the paths addiing up to 1.00. This scenario is WITHOUT replacement (otherwise wouldn't make much sense to do the draw in the first place).

My calculation is the first one to run to the front of the room and draw (note, I did that at the Don Lowe Masters back in 2004) had 1/10 chance of drawing the "1" out of the hat, and 9/10 chance it is not drawn. Second draw (next "Y" of the probablity tree), 1/9 chance of drawing the "1" multiplied by 9/10 (the probability that it was not drawn first) which comes out to 1/10 again, for the chance of drawing "1". While the probability of NOT drawing the second round is 8/9 times 9/10, which is 4 out of 5 it was not drawn. Now from a pilot persective, I already drew first, and did not get "1" so the next person that draws has twice the chances of drawing the "1" (1/10 plus 1/10) equals 2/10 or 1 out of five chances of drawing the one (note, it is this way because these are not independent events). The premise is "if-then", like Kevin said, its assuming the number 1 has not been drawn. Its a bit hard to imagine how the first pilot up has a one in 10 chance of drawing the 1, while the second pilot up has a 1 in 5 chance, the third pilot as a little better than 1 in 3 chance of drawing it! So it goes until the "1 is drawn".

Now, I noticed at many contests since, there are actually two drawings which equalize the probabilties quite a bit (so pilots like me don't game the draw by running up to be the first to draw his flight order number). A hat with all the 10 pilot's names are then mixed up, Dale pulls names out of the hat, then that pilot goes up to draw his number out of another hat for his place in the flight order.

Any way, I got fairly deep into several probability and statistics books thus the delay in response. If any of the above is not correct, please speak up. I'll continue studying the subject and will report back if I discover otherwise.

Best regards,
Don

Don, you're missing the picture. If you are drawing one pill then the chance of you drawing a "1" is always 1/10 no matter if you are first, last, or in the middle. However, if you were to draw two pills then the chance of you drawing a 1 is now 2/10. And if you were to draw 10 pills the chance is 10/10 (100%).

So if two men each draw one pill, each of them has a 1/10 chance of getting the "1" but together they have a 2/10 chance of getting it.

If 10 men each draw a pill they each have a 1/10 chance of getting the "1", but together they have a 10/10 chance of getting it....or in other words one of them is guaranteed to get it.

To elaborate further. The probably of drawing the "1" changes depending on what has already happened. Obviously if you draw last and nobody has drawn the "1" then you are 100% going to get it. However, if you draw last there is a 90% chance that someone else will already have drawn it.

At the beginning of the draw you don't know who is or isn't going to get the lucky number so the chance of you getting it is still 1/10 no matter which position you draw in.

Remember this is without replacement. (replacement is simply drawing one of the numbers, then putting it back in so the next person draws one of the 10 pieces of paper). Please cite your references (on line link, text book, etc.) which would be very helpful.

If you draw first, it is 1/10 you draw the number 1, and 9/10 its not drawn. The first person to draw is now out of the picture (so to speak)- we are assuming if that person did not draw the 1. There are now 9 pieces of paper and 9 pilots left to draw (and all 9 pilots missed the first opportunity to draw, so now have to work with what is left (1, plus 8 of the other 9 remaining non-

... non-"1" pieces of paper. That is why the "probability map" seems applicable. http://www.onlinemathlearning.com/pr...-diagrams.html the problem is, I can't find an exact probability map / tree for the exact situation we are looking at here at the pilots meeting that took place.