[Spaceman Spiff]

Here is how my seat assignment progression has gone:

1. Cards and suits method
2. Tags pulled from a cup
3. Excel "random seat generator" that I wrote
4. Tournament Director

Coincidentally, those are also in order from roughly longest time to shortest time to accomplish.

I don't redraw for the button. Seat 1 at each table is the initial button and whoever gets assigned there gets the button. Random enough for our purposes and also quickest.


As for the Monty Hall paradox. I forget the reasoning exactly. The way I remember it is to extend the situation out. Say there are 100 doors. You pick 1. Monty then eliminates 98 of the other doors. You should switch your pick because your initial pick was a 1 in 100 shot and now you're a 1 in 2 shot to guess right.

[Himewad]

Here's an idea ... those pieces of paper with numbers on them that you shuffle up ... we have those ... they're called playing cards!

I use two decks to assign seats. I use the ace through 10 of spades from each deck (my table seats 10 players). The cards from the first deck are placed face-up at each seat on the table. The cards from the second deck are placed face-down on another table. As players come in, they take a face-down card, and they are assigned a seat automatically.

[whitepotatoe]

Originally Posted by R Deckard

I believe the answer is D. The reasoning is similar. Your first pick probabilities are: hit 0.2, miss 0.8. The second pick, by switching doors, your hit probability goes to 0.8/3 = 0.2667. The third pick, by switching doors, hit probability goes to 0.36667.

The answer is not D! (this is picked by most people who know the answer to the regular Monte Hall problem) But you are 100% correct that you will increase your chances to 36.67% by switching both times. But there is an even better method.

Originally Posted by Spiff

As for the Monty Hall paradox. I forget the reasoning exactly. The way I remember it is to extend the situation out. Say there are 100 doors. You pick 1. Monty then eliminates 98 of the other doors. You should switch your pick because your initial pick was a 1 in 100 shot and now you're a 1 in 2 shot to guess right.

You should switch your pick because the other door will be correct 99% of the time and your initial pick is only right 1% of the time. Your chances of beign right and wrong must add up to 100%.

[R Deckard]

Originally Posted by whitepotatoe

The answer is not D! (this is picked by most people who know the answer to the regular Monte Hall problem) But you are 100% correct that you will increase your chances to 36.67% by switching both times. But there is an even better method.

Duh, I should have thought out the other options--B is the correct answer. The extreme case would be to let Monty continue to pick losers until there is only one unchosen door left, then to switch. Math: After Monty eliminates two losers, there are only three doors left. Staying still is only 0.2. Switching is 0.8/2 = 0.4!

[JM]

LOL, I'm glad I went to bed after my last post or I would have been up as late as spacemonkey!

[JM]

Originally Posted by R Deckard

Your definition of "best chance" has changed the original question, then, and you've actually added additional ones as well:

The question originally was: "If five people draw out of the hat, who has the best chance of drawing the Ace?" Answer: "Nobody, they all have a 1 in 5 chance of getting it."

Your new Question #1 is: "Who has the best chance of getting the Ace on the first draw?" Answer: "The person who makes the first draw (of course)."

Question #2: If the first person to draw doesn't get the Ace, what are the chances of the others getting it? Answer: "Each of the others have a 1 in 4 chance of getting it."

It is a rather subtle point I'm making, and so is yours. But we are talking about two completely different problems.

I guess I didn't word it as well as I could have, so your answer was absolutely correct. I guess I meant that the person making the first draw has the advantage in that he is the only one with a 100% shot of getting the ace from the start. And if the ace determines both seat number as well as button position, then that's a double advantage. His odds of picking the ace are of course no different than any other card in the hat.

I think I will go with the 2 stage system, one draw for seat assignments and a second draw for button, so seat #1 doesn't have the advantage of both playing on the button for the first hand and getting to see several hands for free.

As for Monty's 5 doors, I'm torn between B & C. I'm going to go with C on that one. Thanks for that link wp, I'm going to check out that software.

[vanquish]

Originally Posted by whitepotatoe

B. Switch doors. You increase your chance of being right from 33% to 67%. When you initialy pick, your chances of being right are 33%. When one door is removed, you still have a 33% chance about being right with your initial pick, so that means the other door is the winner 67% of the time.

I went and ran through this myself when I found myself saying "that doesn't quite look right" at this answer. Really this answer just confused me since I'm a visual learner. So for the other visual learners out there, look at it this way-

Here are the three doors:

edit: this stupid bb truncates all of my spaces to one space, so the figures will look a little funny.

|---| |---| |---|
| 1 | | 2 | | 3 |
|---| |---| |---|

One with the Prize, two and three with crap:

|---| |---| |---|
| P | | 2 | | 3 |
|---| |---| |---|

We pick one door:

|---| |---| |---|
| P | | X | | 3 |
|---| |---| |---|


Monty nixes a gag door:

|---| |---|
| P | | X |
|---| |---|

What action, if taken, would get us the prize? Switch doors.

Okay, now what would happen if we had started by picking a different door?

|---| |---| |---|
| P | | 2 | | X |
|---| |---| |---|


Monty eliminates door 2:

|---| |---|
| P | | X |
|---| |---|

What action, if taken, gets us the prize? Switch doors.

And the last possibility?

|---| |---| |---|
| X | | 2 | | 3 |
|---| |---| |---|

Monty eliminates a junk door.

|---| |---|
| X | | 2 |
|---| |---|

What action, if taken, gets us the prize? Stay.

----------

So switching gets you the prize two out of three times. It will be two out of three times for any placement of the prize, so the net probability is that ~67% of the time switching doors get you the new car.

Okay, I know many of you probably understood this anyway, but I figured I'd help some of the straglers with and explanation that made more sense to me as a visual learner.

I'm out!

[vanquish]

Originally Posted by JM

I guess I didn't word it as well as I could have, so your answer was absolutely correct. I guess I meant that the person making the first draw has the advantage in that he is the only one with a 100% shot of getting the ace from the start. And if the ace determines both seat number as well as button position, then that's a double advantage. His odds of picking the ace are of course no different than any other card in the hat.

I think I will go with the 2 stage system, one draw for seat assignments and a second draw for button, so seat #1 doesn't have the advantage of both playing on the button for the first hand and getting to see several hands for free.

As for Monty's 5 doors, I'm torn between B & C. I'm going to go with C on that one. Thanks for that link wp, I'm going to check out that software.

I think the point that he's trying to make is that there is really and truly absolutely no advantage to going first. A two stage system is definitely the way to go, since that is the cardroom standard, but it is a little beside the point.

What Deckard said:

Originally Posted by R Deckard

Your new Question #1 is: "Who has the best chance of getting the Ace on the first draw?" Answer: "The person who makes the first draw (of course)."

This is the important part. Let me rewrite this: "Who has the best chance of getting the Ace on the first draw?" The answer isn't quite "the person who makes the first draw (of course)." The answer is the person who makes the first draw is the only person who can draw an Ace on the first draw. There is no advantage here. All this line of reasoning proves (really) is that the guy that picks first is the only guy that can pick first.

You can continue to "update" the odds of who will get the Ace as you watch people NOT get the ace, but that is pointless. What, for all practical purposes, is the difference between five guys drawing a card in succession and five guys drawing a card at the same time? None. Everyone has a 1 in 5 chance of getting the coveted Ace. If you let all five guys draw their cards before anyone turns a card over then you will eliminate this percieved advantage the first drawer has. Or just draw for the button also, which is what you're doing anyway. Problem solved.

[JM]

Originally Posted by vanquish

So switching gets you the prize two out of three times. It will be two out of three times for any placement of the prize, so the net probability is that ~67% of the time switching doors get you the new car.

Okay, I know many of you probably understood this anyway, but I figured I'd help some of the straglers with and explanation that made more sense to me as a visual learner.

I'm out!

The downside of this, the new car is a 1976 Chevy Cavalier station wagon! (reruns, get it!)

[vanquish]

You actually made me laugh. I like it! Good job.