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#26 Re: Puzzles and Games » dice puzzle » 2007-01-28 19:14:30
The question as asked, the answer is indeed 6 - as far as I can tell.
However, if it was a normal numbered dice covered by six letters - by my calculations the number of possible combinations between numbers and letters is 720! 
#27 Re: Puzzles and Games » The Roulette Conundrum » 2007-01-28 18:37:27
Interesting observations Pi Man.
You said the average might be 18 or 19 non-zero spins before zero comes up.
I should point out immediately, the question was "how many non-zero spins are most likely to occur?" That is not the same as asking "on average, how many non-zero spins will
occur?". The answer to both might or might not be the same. However, I submit that 18 or 19 is the wrong answer in terms of being either the average or the most likely number of spins.
Sorry!
Just to illustrate the difference: a practical test would be to do exactly as you suggested - sample as many wheels as possible. Suppose you tried 38 wheels, just as an example. On 37 wheels you found it took a different number of spins ranging from none to 36. On the 37th table you found it took 30 spins. That means 30 non-zero spins occured twice. The rest in the none-to-36-range occured once. Although the average was 18 spins, the indications so far suggest that 30 spins might occur more often than any other number of spins.
With a large enough sample - say 10,000 wheels - that is what you would be looking for. You would want the number of spins that occurs the most frequently.
In other words, in any representative sample, the most likely number of non-zero spins to occur is always the Mode rather than the Mean. But feel free to offer the Mean as well.
There will probably be less debate about that.
Simon
#28 Re: Puzzles and Games » The Roulette Conundrum » 2007-01-28 15:50:32
Ok that's cool!
And your answer is?
#29 Puzzles and Games » The Roulette Conundrum » 2007-01-28 14:59:30
- Simon
- Replies: 10
Here is a curious probability puzzle, the answer to which has been hotly disputed elsewhere.
Suppose there is an unbiased roulette wheel with the usual 36 regular numbers plus a single Zero.
A Zero has just come up:
"How many non-zero spins are most likely to occur before you hit zero again?"
Simon
#30 Re: Puzzles and Games » The Six Card Trick - an advanced variation on Monty Hall » 2007-01-28 14:27:10
Apologies for throwing you all in the deep end!
I'm going to make it much simpler.
Forget six cards. Let's do it with four, and only have one card eliminated.
By the way, this is not a trick. Just a probability exercise!
There are four cards face down - two Kings and two Queens. There is a a host on hand who knows what the cards are.
1) You choose a card.
2) You decide who is going to eliminate one of the three remaining cards you or the host.
a) If it's you, then simply turn over one of the cards.
b) If it's the host, you choose whether he reveals a King or Queen. He will use his knowledge to comply.
Based on what is revealed, what are the odds that your chosen card is a King or Queen?
Do the other two cards have the same odds?
Does it make a difference who performed the elimination?
As an example, imagine two games were played and the same card was shown each time.
In game 1, a Queen was revealed. You turned over the card youself.
In game 2, a Queen was revealed. You told the host to reveal a Queen.
In both cases you were left with two Kings and one Queen.
In each game:
Where is the Queen most likely to be?
Are the odds the same for each card?
Does it matter who performed the elimination?
Hope that's more straightforward!
Simon
#31 Puzzles and Games » The Six Card Trick - an advanced variation on Monty Hall » 2007-01-27 00:28:27
- Simon
- Replies: 7
The following should be solvable by everyone, but will be of particular interest to those familiar with Monty Hall.
There are six cards face down - three are Kings, three are Queens. There is a host present who knows what each card is. Naturally, you dont.
1. You choose one of the cards.
2. From the five remaining cards, four cards must now be revealed and eliminated. The rule is that three of a kind must not be shown. No game can ever be voided.
3. You have three choices as to how this is done.
a) You can instruct the host to reveal all four cards. He will use his knowledge to reveal 2 Kings and 2 Queens in any order you wish.
b) You can turn over the first two cards yourself. and the host will turn over the opposite.
c) You can turn over the first and third card yourself and the host will turn over the opposite.
Whichever method is used, exactly two Kings and two Queens are always removed. You are left with a King and a Queen. Your chosen card is one of them.
Which is your card more likely to be? Are the odds 50/50?
Does it make a difference who revealed what or is the probabilty constant?
Consider the following three games that were played using each method.
In game a), you instructed the host to remove two Kings and then two Queens.
In game b) you turned over two Kings and the host showed two Queens
In game c) you alternated with the host in a sequence of King, Queen, King, Queen.
In each game there is a King and Queen left. Your chosen card could be either. Let's say you want the Queen.
Should you stick or swap? Does it make no difference? Do the odds vary, depending on the method of elimination?
Have fun!
Simon
#32 Re: Puzzles and Games » Variation on the Monty Hall conundrum. » 2006-06-20 08:19:46
OK, so before we proceed to the two host scenario, I need to convince you that in any game where a host doesn't know where the car is, but nevertheless avoids revealing it, the odds are no longer 1/3 that you originally picked the car. They are 1/2.
There is no contradiction. The best way to illustrate it is to show the six possible outcomes.
So let's look at six games where those outcomes get played out.
Imagine three doors. The door the player selects we call X. The one the host opens is Y. The remaining door we call Z. Behind each could be Goat 1, Goat 2 or the Car.
X Y Z
-------------
1. G1 G2 C
2. G2 G1 C
3. G1 C G2 voided
4. G2 C G1 voided
5. C G1 G2
6. C G2 G1
The above is simply the most probable average set of outcomes. You see what is most likely to happen? Outcomes 3 & 4 get eliminated.
In those six games you will most likely originally choose 4 goats and 2 cars. However, because the host doesn't know where the car was, he is most likely to nullify two out of the six games. On average, you will originally pick the car one third of the time. On average, the random host will prematurely reveal the car one third of the time. This cancels your advantage. A non-random host will never reveal the car. Therein lies the difference
So two games have been voided. Of the remaining non-voided games you picked 2 goats and 2 cars. If you swapped in all four of those games, you would still end up with 2 goats and 2 cars. The two games voided by the random host were the extra games you would have won by swapping if the host had known where the car was and avoided revealing it.
So in any single game where the host randomly opens a door, but nevertheless avoids revealing the car, the odds are 1/2 for the two doors left. Why? Because you know there was always a chance he might have revealed the prize, even though he didn't. That's why the host's state of knowledge is crucial.
Have tried this ready made simulator?
http://math.ucsd.edu/~anistat/chi-an/MonteHallParadox.html
Ok are we agreed on that yet?
If so we can move on to the two host game!
#33 Re: Puzzles and Games » Variation on the Monty Hall conundrum. » 2006-06-20 03:53:23
Actually no Ricky. In the Monty Hall debate it's already established that the 2/3 advantage in swapping only occurs when the single Host knows where the car is and will avoid revealing it.
If you doubt this, try runnning a one host simulation where he may randomly open either door. Then discount all cames where the host prematurely revealed the car. These are obviously voided games because all has been revealed and you the player have no decision to make.
You will find that among the non-voided games, the odds are now 50/50 whether you stick or swap.
So strangely enough, when you get a single game where the host has revealed a goat but had no knowledge where the car was, the odds are 1/2 whatever you do.
Once you're convinced of that, see what you think about my amended version of the two host game.
#34 Puzzles and Games » Variation on the Monty Hall conundrum. » 2006-06-17 15:56:20
- Simon
- Replies: 20
I hope you find this a challenging variation of the conundrum.
Before I begin, let's make sure we're all familiar with the old three door Monty Hall concept. Here is a brief refresher.
"Contestants are given three doors to choose from. Behind two doors are goats, and behind the other door is a car. The host knows what's behind each door. The contestant gets to pick one door and the host opens one of the doors the contestant did not pick to reveal a goat. The host then gives the contestant a chance to change his mind.
The question is, is it more likely for the contestant to win the car if he sticks with his original door, or if he changes doors, or are the odds equal either way?"
Answer: as long as the host knows where the car is (and so never cancels a game by revealing it) the contestant's chance of winning if he sticks is 1/3 and if he swaps is 2/3.
For a full discussion of this solution which many find hard to beleive, see:
http://www.mathsisfun.com/forum/viewtopic.php?id=3106
And for those who still can't beleive the answer, here are a couple of excellent online simulators to test it.
http://www.curiouser.co.uk/monty/montygame.htm
http://math.ucsd.edu/~anistat/chi-an/MonteHallParadox.html
The Monty Hall debate has already established the following:
As long as the host knows where the car is - and so never cancels a game by revealing it - the contestant's chance of winning if he sticks is 1/3 and if he swaps is 2/3.
However, when the host doesn't know where the car is but reveals a goat randomly - the odds whether you stick or swap become 50/50.
The following variation is a simple, but perhaps more mysterious, version of the original. It is aimed at those who already agree on the above. Hope you're ready for this:
There are three doors. Behind one is a car. Behind the others are goats. You select one door.
Enter Host 1, who does not know where the car is. From the two doors you didnt pick, he randomly reveals a goat. He then closes that door and leaves, but you remember what's behind it. (If a car was revealed the game is void. Shuffle the doors and start again. Keep going until you get a game where the host has randomly revealed a goat).
Because the host's elimination was random, we're agreed at this point that your chances of finding the car are now 50/50 whether you stick to your original selection or swap to the door that wasn't opened.
However, before you make your decision:
Enter Host 2, who did not witness the above but knows where the car is and will not reveal it. As it happens, he opens the same door as Host 1. You are shown nothing new.
What are the odds now?
Are they the same as when Host 1 opened this door? Or should your estimate be different?
(If Host 2 opens a different door to Host 1, once again the game is void. As before, 'shuffle' the doors and keep going until you get a complete game where Host 1 randomly opens a door with a goat and Host 2 opens the same door.)
Simon