random draws

You can divide the numbers however you wish. The fact remains each number has a differing value.

2 and 4 may both be 'good', but 2 is better than 4..

It seems reasonable to look at expected outcome, which is obtained by assigning a value to each outcome and doing an average of the remaining outcomes.

I used the position number as the value, but even if you assign values differently, there remains one fact which renders the question moot.

For any given circumstance and any set of values, your expected outcome for drawing at that moment matches exactly the expected outcome for waiting for the next person to draw, because you do not know what will be picked until it is too late to decide whether you wanted to draw beforehand or not.

Now, after the next draw is made, the expected outcome will most likely shift somewhat, but this doesn't help you BEFOREHAND, because it could have just as easily went the other way.

uzless, everyone here is correct: Your method is very good at determining what you SHOULD HAVE DONE on the previous choice, but there is nothing that can help you decide to wait or pick if the odds are truly even.

Originally posted by wolfgang59
OK lets use your example
we have xyyyzzz and we want to choose an x.

If we choose right away we have a 1/7 chance of getting it.
So (according to you) we wait.

If our opppponent draws the x (1/7) we are doomed. If he doesnt draw x (6/7) we have a 1/6 (yippee) chance of the x.

So if we wait our chances of an x are 6/7 * 1/6 = 1/7
Exactly the same as if we had NOT waited.

Originally posted by uzless
ok, i'll go with this one.

It just seems crazy that you can't improve you odds by going with the odds of waiting on the first draw. There will be 6 bad numbers and 4 good ones so it would seem a worthwhile gamble to not pick first hoping a bad one will be picked thus increasing your odds on the 2nd draw.

I understand that by looking at BOTH draws (fir ...[text shortened]... otal odds are the same. I'd just rather not leave things to chance. Oh well.

Thanks all.