- 26 Sep '08 18:32 / 6 edits10 pieces of paper numbered 1 to 10 are placed into a bucket. 10 people draw one piece of of paper from the bucket at a time. You hold the bucket and can pick a piece of paper at anytime. Your objective is to pick as close to number 1 as possible.

If the first person picks a 10, should you pick next or wait?

If the first person picks a 1, should you pick next or wait?

EDIT: To clarify, this is for a sports draft where you want number 1 because you get to have the number 1 draft pick. So you obviously want your number to be closer to one so you can get a better player. - 26 Sep '08 21:06Suppose it's with three papers and only one paper means the winning of a grand prize:

The first to draw has a 1/3 chance to pick the right one. (1/3)

The second to draw has 2/3 chance that the first one hasn't picked the right one and 1/2 chance that he picks the right one, which makes his chance to win 1/3 as well. (2/3 * 1/2)

And well, if the second drawer fails, that automatically means the third wins, so his chance to win the grand prize is 1/3 as well (2/3 * 1/2 * 1). - 27 Sep '08 01:55I don't think there is a pat answer to this sort of question. The reason is because until the next draw occurs, either of your two options gives you the same expected value.

The nature of the shift in average value is that it tends to change slightly, but erraticly in terms of direction. There is, therefore no predictability, and no real useful pattern to aid your decision. - 28 Sep '08 02:23 / 5 edits

grr, this isn't helping.*Originally posted by geepamoogle***I don't think there is a pat answer to this sort of question. The reason is because until the next draw occurs, either of your two options gives you the same expected value.**

The nature of the shift in average value is that it tends to change slightly, but erraticly in terms of direction. There is, therefore no predictability, and no real useful pattern to aid your decision.

We'll be having our draw soon. 10 guys. I'll be holding the bucket and can draw a number at any time.

Here's my thinking. I don't have to get 1...just get close to one and i'll be happy. So if the first number comes up an 8 9 or 10, the odds are in my favour that the next one would be lower than an 8 9 or 10 (if you take 8 9 10 as a "group" ) It's more likely that numbers from another "group" will be picked next ...ie 2 3 4 or 5 6 7.

If a 2 comes up, then it's more likely that the next "group" picked will be from the 5 6 7 or 8 9 10 "group" so i'd be better off to not pick next.

If the first number is either a 5 6 or 7, it does me no good because the next number is equally likely to come from 8 9 10 or 2 3 4. In this case i guess i'd have to wait for the next number to be picked to get more info and then make the decision at that point.

Any flaws to that thinking?? - 28 Sep '08 13:43

Well, as far as I can see it works as well as any other thought process. However, it does sound like Gambler's Fallacy has creeped in a bit.*Originally posted by uzless***Here's my thinking. I don't have to get 1...just get close to one and i'll be happy. So if the first number comes up an 8 9 or 10, the odds are in my favour that the next one would be lower than an 8 9 or 10 (if you take 8 9 10 as a "group" ) It's more likely that numbers from another "group" will be picked next ...ie 2 3 4 or 5 6 7.**

If a 2 comes up ...[text shortened]... re info and then make the decision at that point.

Any flaws to that thinking??

The difference between this and a set of coin tosses or roulette spins is that these draws aren't independent events. Early on, though, it still functions almost as independent draws in a pragmatic sense, or at least so I think.

But, as I said, to draw now or to wait at any point have the same expected value at that point, so it's not harmful to adopt this tactic in the least. Best of luck. - 28 Sep '08 15:38If you hold the bucket and can draw at any time...

Just rig it and draw first.

If not write out the entirety of expected draws.

Start with - 1st draw you have a 4/10 chance of getting a 'good' number.

if the 1st draw is a 8-10 your chance becomes 4/9

if 5-7, 4/9

if 1-4, 3/9

etc. and then compare highest values. To be honest 40% is actually fairly decent afaik unless there is a long string of 5-10s, which chance debates. - 30 Sep '08 15:20 / 1 editIt doesn't matter if you pick or wait. Not even if you enter considerations like the one you mention above.

For example, imagine that there are 3 amazing players that you're nearly indifferent between, 1 good player and 7 very crap players. Then 1-3 would give you (say) 10 regardless of which, 4 would give you 5 and 5-10 would give you 0.

Imagine the first person gets 1. You now have 2 "10s", 1 "5" and 6 "0s".

E(picking) = 2/9*10+1/9*5 = 2.77(7)

E(waiting) = 2/9*(1/8*10+1/8*5)+1/9*2/8*10+6/9*(2/8*10+1/8*5)=2.77(7) - 30 Sep '08 15:41

It is nonsense.*Originally posted by uzless***grr, this isn't helping.**

We'll be having our draw soon. 10 guys. I'll be holding the bucket and can draw a number at any time.

Here's my thinking. I don't have to get 1...just get close to one and i'll be happy. So if the first number comes up an 8 9 or 10, the odds are in my favour that the next one would be lower than an 8 9 or 10 (if you take 8 ...[text shortened]... info and then make the decision at that point.

Any flaws to that thinking??

You say there is more chace to draw a 5,6,7,8,9 or 10 after drawing a 2.And what if you count 2,3,4,5,6 and 7,8,9,10 as groups, do you have less chance of drawing a 6 or 5 then? - 30 Sep '08 16:21 / 5 edits

No because you've grouped it into 5 numbers vs 4 numbers. It's not even. Mine have 3 numbers in each. Your example leaves the odds at 50-50 after you draw a 2.*Originally posted by Thomaster***It is nonsense.**

You say there is more chace to draw a 5,6,7,8,9 or 10 after drawing a 2.And what if you count 2,3,4,5,6 and 7,8,9,10 as groups, do you have less chance of drawing a 6 or 5 then?

So, consider this.

I have 3 groups of letters. I want to pick an X

xxx

yyy

zzz

Someone else draws a letter from a bucket. It is an X.

We now have left in the bucket,

xx

yyy

zzz

Would you say it is now less likely that an X will be picked on the next pick? - 30 Sep '08 16:51

Hindsight is 20-20 vision.*Originally posted by uzless***No because you've grouped it into 5 numbers vs 4 numbers. It's not even. Mine have 3 numbers in each. Your example leaves the odds at 50-50 after you draw a 2.**

So, consider this.

I have 3 groups of letters. I want to pick an X

xxx

yyy

zzz

Someone else draws a letter from a bucket. It is an X.

We now have left in the bucket,

xx

yyy

zzz

Would you say it is now less likely that an X will be picked on the next pick?

It is as likely to get an x if you pick or wait. This is true for all picks. The reason is that the probability gain in case another doesn't pick an x is exactly cancelled by the probability he does pick it.

n/N*(n-1)/(N-1)+(N-n)/N*n/(N-1) = (same denominator, collect terms) [n*(n-1)+(N-n)*n] / [N*(N-1)] = (collect numerator terms in n) = [n*(N-1+n-n)]/ [N*(N-1)] = n/N - 30 Sep '08 17:19

Look, i'm the first one to admit I dont' have the math skills. But it seems pretty clear that if you have the following letters in a bucket,*Originally posted by Palynka***Hindsight is 20-20 vision.**

It is as likely to get an x if you pick or wait. This is true for all picks. The reason is that the probability gain in case another doesn't pick an x is exactly cancelled by the probability he does pick it.

n/N*(n-1)/(N-1)+(N-n)/N*n/(N-1) = (same denominator, collect terms) [n*(n-1)+(N-n)*n] / [N*(N-1)] = (collect numerator terms in n) = [n*(N-1+n-n)]/ [N*(N-1)] = n/N

x

yyy

zzz

That if you were to reach in and grab a letter, it's unlikely you will get an X. In fact, the odds are 1/7. More to the point, the odds are 6/7 that you will NOT get an X. So you are better off waiting, hoping that something else will be picked so you can better your odds from 1/7 to 1/6.

I am basically trying to employ the strategy used by Blackjack players where you count cards. Specifically you are counting how many cards above and below 10 have been played in order to be able to increase the chances that the next card played will be below or above 10 depending on what actual cards have already been played. (If the dealer deals 20 jacks and kings in a row, you can be pretty sure some lower cards are coming!) - 30 Sep '08 18:10

OK lets use your example*Originally posted by uzless***Look, i'm the first one to admit I dont' have the math skills. But it seems pretty clear that if you have the following letters in a bucket,**

x

yyy

zzz

That if you were to reach in and grab a letter, it's unlikely you will get an X. In fact, the odds are 1/7. More to the point, the odds are 6/7 that you will NOT get an X. So you are better off ...[text shortened]... ealer deals 20 jacks and kings in a row, you can be pretty sure some lower cards are coming!)

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.