 Maximizing probability uzless Posers and Puzzles 17 Jan '08 17:17
1. 17 Jan '08 17:172 edits
There will be a draw held at the county fair. 10 tickets will be entered. 4 will be winning tickets. All tickets will be put into a box and randomly drawn. There will be 10 draws. After the first draw, the ticket will be discarded so there will only be 9 tickets remaining etc etc.

You must try to "predict" which draw will hold a winning ticket. Which draw would you bet on in order to maximize the probability that you will pick one of the draws that contain a winning ticket. You must pick a draw BEFORE any of the draws occur and you can only pick one draw.

Can you devise a strategy or do you think it makes no difference at all which draw you pick?
2. 17 Jan '08 17:35
Intuitively, i would bet that the first draw was the winning ticket. This has the highest probability to be a winning ticket.

The one with lowest probability must be the tenth ticket, with a P = 0.
3. 17 Jan '08 17:541 edit
Originally posted by FabianFnas
Intuitively, i would bet that the first draw was the winning ticket. This has the highest probability to be a winning ticket.

The one with lowest probability must be the tenth ticket, with a P = 0.
Actually, they should all be the same. The probability of a wining ticket being drawn on the first try is 4/10. The probability of a winning ticket being drawn on the second try is the sum of what happens if a winning ticket was drawn on the first try and if no winning ticket was drawn on the first try:

P(not drawn) = (6/10)*(4/9)
P(drawn) = (4/10)*(3/9)

P(total) = (6/10)*(4/9) + (4/10)*(3/9) = (12/90) + (24/90) = (36/90) = 4/10

This pattern continues with slight modifications as you work your way down the line. I also ran an Excel simulation to see if there was any bias for or against any particular position, and came up none. The winning tickets are equally likely to come up on any draw.

Having said that, I'm not 100% sure but I think there is an argument to made for strategy if you are required to pick the position of the first winning ticket drawn. I'll check it out.
4. 17 Jan '08 18:08
Yes, looking back it seems obvious, but there is a strategy for picking the first winning ticket, and FabianFnas has the right answer to this question: bet on the first ticket.

Using Excel notation, there are COMBIN(10,4) possible arrangements for the tickets, where COMBIN is the choose function from combinatorics. The number of arrangements where the first ticket is a winner is COMBIN(9,3), because once you've used one ticket in the first draw you have 9 spaces left to fill with 3 tickets. Following this pattern, the number of arrangements where the second ticket is the first winner is COMBIN(8,3), and so on down the line until you get to the seventh ticket being the first winner with the number of possible arrangements being COMBIN(3,3) = 1. Anything after seventh is not possible because you wouldn't use all 4 tickets.

The probability for any particular draw "n" having the first winning ticket is therefore P = COMBIN(10-n,3)/COMBIN(10,4). You can calculate all the values, but this is a decreasing function and indeed the first draw has the highest probability of hosting the first winning ticket, with P = 4/10.
5. 17 Jan '08 18:22
Originally posted by uzless
You must try to "predict" which draw will hold a winning ticket. Which draw would you bet on in order to maximize the probability that you will pick one of the draws that contain a winning ticket.
With the question posed like this ("one of the draws that contain a winning ticket"😉, then there is no difference.
6. 17 Jan '08 18:281 edit
Okay, but consider this assumption. Keep in mind the Let's make a deal problem about switching doors.

The first draw has a probability of 4/10. In other words, there is a 60% chance that the first draw will NOT be a winner. Why would you bet on the first draw if there is a 60% chance it will not win?

If you can assume the 1st draw will therefore not be a winner, then you can extrapolate that the odds of the 2nd draw will be 4/9. Much better odds. However, the odds in your favour are only increased to 44%...or a 56% chance a winner not be drawn. If you also assume the 2nd draw will not be a winner, the 3rd draw will be 4/8...or 50%.

I think the 3rd draw, based on the logic above, would be where I would place my money.

However, I'm not convinced of my logic so hack it up!
7. 17 Jan '08 19:13
4 winning tickets of 10, that makes 6 losing ticket, right?
So if the 1st, 2nd, 3rd, 4th, 5th, and 6th ticket was losing ones, then the 7th has to be a winning one, P=1, right?
Therefore there is no possibility for the 8th, 9th and 10th ticket to be winning at all, right?
Therefore, my conclusion is that there are different probabilities for the 7 first draws, and the same P=0 for the last 3 draws, right?
Yes, right!
8. 17 Jan '08 19:331 edit
Originally posted by FabianFnas
4 winning tickets of 10, that makes 6 losing ticket, right?
So if the 1st, 2nd, 3rd, 4th, 5th, and 6th ticket was losing ones, then the 7th has to be a winning one, P=1, right?
Therefore there is no possibility for the 8th, 9th and 10th ticket to be winning at all, right?
Therefore, my conclusion is that there are different probabilities for the 7 first draws, and the same P=0 for the last 3 draws, right?
Yes, right!
No, he says that there's more than one winning draw. So if 1 to 6 were losing, then with P=1 the winning draws would be 7,8,9 and 10.

Edit - "one of the draws that contain a winning ticket."
9. 17 Jan '08 19:34
Originally posted by uzless
Okay, but consider this assumption. Keep in mind the Let's make a deal problem about switching doors.

The first draw has a probability of 4/10. In other words, there is a 60% chance that the first draw will NOT be a winner. Why would you bet on the first draw if there is a 60% chance it will not win?

If you can assume the 1st draw will therefore not ...[text shortened]... uld be where I would place my money.

However, I'm not convinced of my logic so hack it up!
the difference is that in the monty hall problem he removes the doors that he knows are NOT the correct choice. so your choice at the end is essentially "did i pick the right door" or "did i pick the wrong door" and it's more likely you picked the wrong door at the start, so you should always switch.

in this problem, however, the fact that the first ticket has a 40% chance of being a winner should clue you in to the fact that it's the best choice at the start... imagine the first ticket had a 90% chance, but the 2nd ticket was a 100% chance of being a winner. your best choice would STILL be to pick the first one, because 90% of the time you're not even going to get to look at the 2nd ticket.

from the same logic, the first ticket has a 40% chance of being a winner, and though the 2nd ticket indeed is a 44% chance of being a winner, only 60% of all trials will get past the first ticket... so it's a (.44 x .6) = 26.4% chance of being the FIRST winner. as PBE6 noted, the function that tracks the first winner is based on combinatorics, and is a monotonically decreasing function, so the best strategy is picking the first ticket.

i haven't played around with it at all, but this might prove to be a more interesting problem if either the initial chance for a ticket to be a winner was smaller (e.g. 5 winners out of 100) or if the jackpot grew with each successive draw... and then a strategy to maximize your winnings? just some ideas. 🙂
10. 17 Jan '08 19:40
Originally posted by Palynka
No, he says that there's more than one winning draw. So if 1 to 6 were losing, then with P=1 the winning draws would be 7,8,9 and 10.

Edit - "one of the draws that contain a winning ticket."
yes, but 8,9, and 10 could never be a winning CHOICE for the player because it would never be the first winning ticket (since P=1 for draw 7). so it really depends on what you're defining P to mean - probability of being a "winning ticket" or probability of being a "winning choice for the player." getting these two ideas straight is the key to understanding why this logic isn't sound.
11. 17 Jan '08 19:46
Originally posted by Aetherael
yes, but 8,9, and 10 could never be a winning CHOICE for the player because it would never be the first winning ticket (since P=1 for draw 7). so it really depends on what you're defining P to mean - probability of being a "winning ticket" or probability of being a "winning choice for the player." getting these two ideas straight is the key to understanding why this logic isn't sound.
😞

Where does the problem say that it had to be the FIRST winning draw?

Which draw would you bet on in order to maximize the probability that you will pick one of the draws that contain a winning ticket.
12. 17 Jan '08 20:16
Originally posted by PBE6
Yes, looking back it seems obvious, but there is a strategy for picking the [b]first winning ticket, and FabianFnas has the right answer to this question: bet on the first ticket.
[/b]
Now when I know (or understood correctly) that it is not the first winning ticket that decides if 'my picked ticket' is the winning one or not. If I chose ticket #9 so could this one actually be the winning ticket, recardless if the other 3 twinning tickets was found or not.

So now, I change my view. Every chosen ticket have the same probability as any other.

Let's do this experiment: Take the 10 tickets up in a random order. Put these on the table in that particular order in the way that the information if it is a winning ticket or not is not shown. Now, does it matter if you read the tickets from the left or in the reversed order, from the right? Of course not. Therefore, each ticket has the same probability to be either a winning ticket or a loseing one.

Sorry PBE6, but I change my mind.
13. 17 Jan '08 21:01
Originally posted by FabianFnas
Now when I know (or understood correctly) that it is not the first winning ticket that decides if 'my picked ticket' is the winning one or not. If I chose ticket #9 so could this one actually be the winning ticket, recardless if the other 3 twinning tickets was found or not.

So now, I change my view. Every chosen ticket have the same probability as any ...[text shortened]... bability to be either a winning ticket or a loseing one.

Sorry PBE6, but I change my mind.
14. 17 Jan '08 23:04
I want to know what the prizes are before I start handing money over for tickets...
15. 18 Jan '08 01:55
Originally posted by Palynka
😞

Where does the problem say that it had to be the FIRST winning draw?

[b]Which draw would you bet on in order to maximize the probability that you will pick one of the draws that contain a winning ticket.
[/b]
oh man you got me misreading and assuming. haha i put together PBE6's and Fabian's discussions with the part about "picking a draw BEFORE any tickets are drawn", and took it all to mean you had to pick the FIRST winner.

I agree with Fabian's most recent analysis, and PBE6's first analysis in which all the tickets have the same probability of being a winner.

the difference between this and the "let's make a deal" problem is that there's no one removing draws that are already known to be non-winners. as PBE6 said, the probability of the second ticket being a winner is a sum of A:its probability of being a winner if the first ticket WAS a winner and B: its probability of being a winner if the first ticket was NOT a winner. and this works out to be 40%, as does Fabian's looking at the problem from a broader angle that you put 4 winners and 6 losers on the able and ask the probability of one being a winner, regardless of order.

EDIT: i forgot to post this 5 hours ago, lets see if someone has said the same thing lol