Statistics Puzzle

Nick McClick

This might only be a puzzle for me, but that's because stats is one of my weaknesses. I've been noodling on this for a while and finally just figured it out tonight;

Consider there are 30 daily raffle drawings. Each day, 1 ticket is chosen randomly to receive a prize. I give you 1 ticket before every raffle. You can either enter that ticket into the day's raffle, or you can save up all the tickets to enter into the last day's raffle.

Assume that each raffle has the same number of other entries. Which has a better payout, saving up your tickets for the last raffle, or entering 1 ticket every day, or are they equal? Show proof.

VonSzarvas

As we cannot calculate for the statistical influence of the number of participants involved, particularly <> 30, and also for the number of players holding on until the last draw, then I think the answer might be simply based on your rule:

"Assume that each raffle has the same number of other entries"

To enter all the draws would give the best odds as each draw has the same odds of winning (as each draw has same number of entrants). Thus the old adage of "you gotta be in it to win it" holds true. The more you enter, the more chances to win. If you only enter one draw, even with multiple entries, you only have one chance to win.

The odds of winning any draw are far improved from actually entering the draw! Additional entries do not increase your chance of winning by a greater factor than the initial first entry, unless your entries count for 100% of the ticket holders!

Depends on the number of entrants, but we might say, the first entry gives you 10% "chance" to win. As you had 0% chance before entering, that is a big statistical leap. Additional tickets in the same draw might increase your odds by only 0.1% each, so not a great difference! Usually a good reason not to buy multiple tickets for the state lotto, just buy 1 !

Phil Pilgrim (PhiPi)

Your expected winnings from entering 30 raffles, each with n other entries, plus 1 (you), and a purse P is:

30 raffles * P / (n + 1)

Your expected winnings from entering 1 raffle with n other entries, plus 30 (you), and a purse P is:

30 entries * P / (n + 30)

Clearly, one entry in each of 30 raffles, beats 30 entries in one raffle. This advantage diminishes, however, as n becomes very large.

-Phil

WBA Consulting

I would save up then enter the second to last drawing........

Statistically though, enter each drawing.

Nick McClick

I wish Phil could pretend to struggle on that one!

Originally, I thought they were equal, but then I tested it with a population of 1 (if you were the only person entering the drawing), and figured it out.

wjsteele

I wish I could figure this stuff out... I'm only average at stats.

I need a median... you know what I mean, one of those psyhics to help me out on this stuff!

Bill

Phil Pilgrim (PhiPi)

Even the probability experts struggle and stub their toes from time to time. 'Remember the Monty Hall puzzle and the firestorm of controversy it ignited? Here it is, in case it's new to some of you. It was first popularized by Marilyn Vos Savant in Parade Magazine twenty years ago:

On Let's Make a Deal, Monty Hall offers a contestant the chance to choose from among three curtains. Behind one is a brand new car; behind the other two, donkeys. The contestant chooses a curtain. But Monty, who knows which curtain conceals the car, doesn't open the chosen curtain but opens one of the other two to reveal a donkey. He then offers the contestant a chance to pick the remaining curtain instead of her first choice. Should she switch?

If you are familiar with the puzzle and know the answer, please refrain from posting until others have a chance to weigh in.

-Phil

ElectricAye

Phil Pilgrim (PhiPi) wrote: »

...
On Let's Make a Deal, Monty Hall offers a contestant the chance to choose from among three donkeys...

Quick: which of the worst US presidents said it best? "Liars don't statistify, but statistifiers sometimes do."


If you want to read about the donkey probem in excruciating detail, check it out:
http://en.wikipedia.org/wiki/Monty_Hall_problem

bobo0330

me too, stats is one of my weaknesses

Bobb Fwed

I say stay with the curtain you have chosen. With (a little) logic, you would assume that because he has knowledge of you choosing the right one, he would give you an additional chance to your change your answer. If you had chosen the wrong one, he would just reveal that, rather than giving you a chance to change their answer to the correct curtain.

This assumes that he doesn't always give you the option to change.

This also assumes that he doesn't already think you have thought about it and won't reverse the logic.

"...so I can clearly not choose the wine in front of me!" - Vizzini

Phil Pilgrim (PhiPi)

He always reveals a donkey behind an unchosen curtain and always gives you the option to change. There's no deliberate subterfuge.

-Phil

Bobb Fwed

Phil Pilgrim (PhiPi) wrote: »

He always reveals a donkey and always gives you the option to change. There's no deliberate subterfuge.

-Phil

In which case, I say it's all the same. It goes from a 33% chance to a 50% chance, but without any new information, thus, your first answer is just as good as second guessing and changing your answer.

This is only an answer based on (possibly flawed) logic, no statisticians on this side of keyboard. But if it caused a "firestorm of controversy" I somehow doubt it is as simple as my mind thinks it is.

mctrivia

Bobb Fwed think of it this way.

You pick a door. You have 33% chance you are right. There is 67% chance that the car is behind one of the other 2 doors. The host shows you 1 of the 2 other doors has a donkey. There is still 67% chance that the car is behind one of those 2 doors so if you switch your odds will double.

W9GFO

From the Wiki article that ElectricAye posted;

Increasing the number of doors
It may be easier to appreciate the solution by considering the same problem with 1,000,000 doors instead of just three (vos Savant 1990). In this case there are 999,999 doors with goats behind them and one door with a prize. The player picks a door. The game host then opens 999,998 of the other doors revealing 999,998 goats—imagine the host starting with the first door and going down a line of 1,000,000 doors, opening each one, skipping over only the player's door and one other door. The host then offers the player the chance to switch to the only other unopened door. On average, in 999,999 out of 1,000,000 times the other door will contain the prize, as 999,999 out of 1,000,000 times the player first picked a door with a goat. A rational player should switch. Intuitively speaking, the player should ask how likely is it, that given a million doors, he or she managed to pick the right one. The example can be used to show how the likelihood of success by switching is equal to (1 minus the likelihood of picking correctly the first time) for any given number of doors. It is important to remember, however, that this is based on the assumption that the host knows where the prize is and must not open a door that contains that prize, randomly selecting which other door to leave closed if the contestant manages to select the prize door initially.

Rich H

Gadgetman

The critical fact is that the host does not always have a choice (whether random or not) between the two remaining doors. He always chooses a door that he knows hides a goat after the contestant has made their choice. He always can do this, since he knows the location of the car in advance. If the host chooses completely at random when he has a choice, and if the car is initially likely to be behind any of the three doors, it turns out that the host's choice does not affect the probability that the car is behind the contestant's door. But even if the host has a bias to one door or another when he has a choice, the probability that the car is behind the contestant's door, so it turns out, can never exceed 1/2, again as long as initially all doors are equally likely. It is never unfavourable to switch, while on average it definitely does pay off to switch. The contestant will be asked if he wants to switch, and there was a 1 in 3 chance that his original choice hides a car and a 2 in 3 chance that his original choice hides a goat. By opening another door and revealing a goat the host has removed one of the two other doors, and the 1 in 3 chance of the contestant's initial door hiding a car means there is a 2 in 3 chance that the other closed door will turn out to hide a car.

I call Bovine Fecal matter on this.

Sure, you had a 1:3chance of picking the right door, and a 2:3 that the car was behind the other two doors, but as soon as the third door is opened to reveal a goat, you CANNOT still work with thirds.
The statistical 'math' they throw up is all about thirds, but there are no longer three possibilities.

I've done some rather suspicios maths in my time(lots of 'Big O' notation for calculating effiiency of algorithms comes to mind), but this is...

There are lies,
Damn Lies,
And statistics...

W9GFO

How can you argue with a truth table?



Rich H

mctrivia

The car does not move. So your odds of being correct will not change from 1/3 to 1/2 if you stick with the first door. The only way you will win if you stick is of you guessed correctly the first time. If you sitch then if either of the other 2 was correct then you will won since the one with the goat is shown to you.

Gadgetman

The truth table is wrong...

for one reason oranother, they think that showing both door 2 and 3 is important, but it isn't.

It SHOULD show TWO columns, 'Door A' and 'Remaining door' only.

you can only switch to the 'Remaining door', not 'Door 2' or 'Door 3'.

mctrivia

if you pick door 1

Door 1,2,3 |Host show |
C,G,G |2 or 3 |
G,C,G |3 |
G,G,C |2 |


Leaving you with the option
C,G,X
G,C,X
G,X,C
where x is shown car

you can keep with door 1 or you can switch if you switch you will get the car in 2 out of 3 of the options. Same holds true if you picked any door to start with.

W9GFO

The truth table is setup so that your first choice is door #1. It shows the end result if you stick with door number one or switch to one of the other doors for each possible location of the car.

You are twice as likely to get the car if you switch.

Rich H

Gadgetman

mctrivia wrote: »

if you pick door 1

Door 1,2,3 |Host show |
C,G,G |2 or 3 |
G,C,G |3 |
G,G,C |2 |


Leaving you with the option
C,G,X
G,C,X
G,X,C
where x is shown car

Actually, you need to EXPAND the first table, specifically, the first row must be changed to TWO rows:
C,G,G|2 |
C,G,G|3 |

Everyone thinks this problem has to do with thirds, but it's about quarters and halves, and that's not apparent until you have the correct tables.

mctrivia

No it does not need to be split. The fact the host has a choice between the 2 doors does not constitute a permutation. What matters is what you and the car do.

Gadgetman

Sure it does!

Otherwise, the last two lines should also be reduced to one row.
Because THERE IS NO DIFFERENCE between those two.

mctrivia

No. The first is 1 row because the car is in the same place and the option is a meaningless one that the boat has.

The second and third are 2 sperate rows because the car is in different posititions. The Host has no choice with these ones as he must pick the door that has a donkey in it.

Gadgetman

That's exactly my point...

On the last rows, the donkey is behind door 1, and the car is behind the 'remaining door'!
Ignore 'Door ' and the 'Door 3' labels, At this point in the game they don't mean anything any more.

The host has no choice, and the player has only one choice, whether to switch or not.
That means the two last rows boils down to whether or not to switch. Whether he switches with door 2 or 3 is irrelevant.

mctrivia

Here is a list of every possible and not possible option. At first glance it may look like 50/50 odds but because the hosts choice always results in the same result or an invalid result then each block must be considered as 1 result. This is a fundamental statistical rule. As a result there is 67% odds of winning if you pick the donkey.

If you disagree then lets play this proposed game. We each put down $500,000. We will put down the full $1,000,000 behind 1 door out of 1,000,000. You can pick 1 door. I will then open up 999,998 of the doors showing them all empty. By your reasoning you have a 50/50 odds of winning now and you will stay with the door you have. Take a guess who is really going home with the money.

mctrivia

Gadgetman maybe you are like me and don't like 50/50 odds. Let me sweeten the deal for you.

I will make a program to simulate 2^64 doors. You may chose 999 doors. My simulation will then reveal all but 1 door of the remainders. Now by your reasoning you have 99.9% odds of winning. I will match any amount of money you want that the money will be behind the door my sim did not expose. A good bet right.

If you are right you have 99.9% odds of winning. If I am right though you have only 5.416x10^-15% odds of winning. Are you still sure you are correct?

wjsteele

I love this puzzle because it confuses people quite easily due to their "intuition."

Here's a very simple explanation.

If I have 3 doors... each door had 1/3 of a chance of being the car.

If I pick one door, the other 2 doors have 2/3rd chance of containing the car.

If Monty reveals one of the other two doors, the remaining door still has a 2/3rds chance of being the car. (The odds do NOT change, but YOU have more information!!!)

The kicker... remeber that the odds do NOT change... meaining my door still maintains its original 1/3 chance of being the car.

So, switching to the other door gives me a 2/3rds chance of winning, which is 50% greater chance of winning than if I kept my old door.

If you run this through a billion simulations, you will find that if you switch, you have a greater chance of winning the car by two to one.

Bill

Bobb Fwed

I thought I'd put together a simple simulator. And before I finished it, I realized they are correct. There is double the chance to win if you switch your choice. Attached is the simulator I wrote (in PHP). I was going to have it change it's answer after one of the "doors" were removed, but I realized, before the answer changes, you can determine if they get the car or not (so there is a little extra, unneeded code).

So here is the logic that makes more sense to me (and hopefully simpler than what has been said -- or maybe exactly the same). When randomly choosing one of the three doors you have a 67% chance of choosing wrong. Once they remove one of the wrong answers, switching your answer gives you a 67% chance of choosing correctly. Yeah, so that sounds exactly like what others have been saying.

Well, look at the simulation program, it may allow you to see the light as I have.

Nick McClick

The way I think about it is by extending the values; imagine there were a million doors, you chose one, then monty eliminated all but 2 remaining doors; one with the goat, one with the car.