A little puzzle

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Charlie Reams
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A little puzzle

Post by Charlie Reams » Sat Apr 12, 2008 6:50 pm

Fred has no friends and his only companion is a coin. He likes to toss the coin until both sides have come up at least once. How many turns of this game does he get to play on average?

Fred eventually gets bored of the coin and starts to play with a die, rolling it until all 6 faces have come up at least once. How long does this game last on average?

[The first part is a warm-up which might give a hint on how to solve the second, but neither case is that difficult if you approach it in the right way. I made it up while I was sitting here so it could be thoroughly embarrassing if I got the answer wrong.]

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Re: A little puzzle

Post by Gavin Chipper » Sat Apr 12, 2008 7:14 pm

Charlie Reams wrote:Fred has no friends and his only companion is a coin. He likes to toss the coin until both sides have come up at least once. How many turns of this game does he get to play on average?

Fred eventually gets bored of the coin and starts to play with a die, rolling it until all 6 faces have come up at least once. How long does this game last on average?

[The first part is a warm-up which might give a hint on how to solve the second, but neither case is that difficult if you approach it in the right way. I made it up while I was sitting here so it could be thoroughly embarrassing if I got the answer wrong.]
I won't answer this because I think it came up fairly recently on the Yahoo Group and I answered it then (the dice one).

Edit - http://tv.groups.yahoo.com/group/c4coun ... sage/16385 by the way.
Last edited by Gavin Chipper on Sat Apr 12, 2008 7:29 pm, edited 1 time in total.

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Re: A little puzzle

Post by Joseph Bolas » Sat Apr 12, 2008 7:27 pm

Charlie Reams wrote:He likes to toss the coin until both sides have come up at least once. How many turns of this game does he get to play on average?
The minimum number of tosses is 2, if you were lucky to get H and T straight away. If you did 3 tosses then you have to focus on all the possible outcomes of 3 tosses and also which ones have a H and T in them. The possible outcomes of 3 coins would be HHH, HHT, HTH, HTT, THH, THT, TTH, TTT, where 6 out of 8 have one of each side so you have a 75% chance of doing the challenge with 3 tosses.

So if correct, its just a number of working out set of outcomes depending on the number of tosses and then finding the one with the higher percentage (if that makes sense).

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Re: A little puzzle

Post by Charlie Reams » Sat Apr 12, 2008 9:47 pm

Joseph Bolas wrote:
Charlie Reams wrote:He likes to toss the coin until both sides have come up at least once. How many turns of this game does he get to play on average?
The minimum number of tosses is 2, if you were lucky to get H and T straight away. If you did 3 tosses then you have to focus on all the possible outcomes of 3 tosses and also which ones have a H and T in them. The possible outcomes of 3 coins would be HHH, HHT, HTH, HTT, THH, THT, TTH, TTT, where 6 out of 8 have one of each side so you have a 75% chance of doing the challenge with 3 tosses.

So if correct, its just a number of working out set of outcomes depending on the number of tosses and then finding the one with the higher percentage (if that makes sense).
It makes sense, but it's not the right solution. The most likely outcome is not the same as the average outcome. For example, the average outcome when rolling a dice is 3.5, but the most likely outcome is... well, they're all equally likely.

Anyway, Gevin has spoilt my fun, because his answer on the forum is correct. I don't remember reading that post but maybe it inspired me subconciously. Can you generalise the method to a, say, an N-faced die?

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Re: A little puzzle

Post by Joseph Bolas » Sat Apr 12, 2008 11:15 pm

Charlie Reams wrote:Anyway, Gevin has spoilt my fun, because his answer on the forum is correct. I don't remember reading that post but maybe it inspired me subconciously. Can you generalise the method to a, say, an N-faced die?
Yeah, I've just seen that Gevin's edit was just after I posted my message, so I missed that. Sorry :oops:.

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Re: A little puzzle

Post by Jason Larsen » Sun Apr 13, 2008 2:09 pm

Two, because 2 x 1 = 2

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Re: A little puzzle

Post by M. George Quinn » Sun Apr 13, 2008 10:41 pm

Jason. That is brilliant.

This topic should be locked as it has nothing to do with either cheese or mushrooms.

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Re: A little puzzle

Post by Jason Larsen » Sun Apr 13, 2008 10:54 pm

Thank you.

I will accept what you say, George.

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Re: A little puzzle

Post by Gavin Chipper » Wed Apr 16, 2008 9:36 pm

Anyway, for an n-sided dice: n/n (or 1) + n/(n-1) + n/(n-2) ... + n/(n-(n-1)) (or n)

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Re: A little puzzle

Post by Jason Larsen » Wed Apr 16, 2008 9:49 pm

How did you come up with that?

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Re: A little puzzle

Post by Gavin Chipper » Thu Apr 17, 2008 8:21 pm

Jason Larsen wrote:How did you come up with that?
Well, to get all the sides of the dice to come up, you first have to get one of them - there is a 100% chance of achieving this on your first go, and the average number of goes is one. Then you need to get the second number. Each time you roll, there is one chance in n of failing (the chance of rolling the number you got the first time). You have an (n-1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n-2)/n chance, until you get down to the last side left, where the chance is 1/n and the average number of goes is n. So you just add up all the averages to get the total average. I know this reads like a ramble but I can't be bothered to make it any clearer.

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Re: A little puzzle

Post by Joseph Bolas » Thu Apr 17, 2008 8:32 pm

Gevin-Gavin wrote:Well, to get all the sides of the dice to come up, you first have to get one of them - there is a 100% chance of achieving this on your first go, and the average number of goes is one. Then you need to get the second number. Each time you roll, there is one chance in n of failing (the chance of rolling the number you got the first time). You have an (n-1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n-2)/n chance, until you get down to the last side left, where the chance is 1/n and the average number of goes is n. So you just add up all the averages to get the total average. I know this reads like a ramble but I can't be bothered to make it any clearer.
So then for an 8-sided die for example it would be then:

8/8 + 7/8 + 6/8 + 5/8 + 4/8 + 3/8 + 2/8 + 1/8?

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Re: A little puzzle

Post by Gavin Chipper » Thu Apr 17, 2008 8:38 pm

Joseph Bolas wrote:
Gevin-Gavin wrote:Well, to get all the sides of the dice to come up, you first have to get one of them - there is a 100% chance of achieving this on your first go, and the average number of goes is one. Then you need to get the second number. Each time you roll, there is one chance in n of failing (the chance of rolling the number you got the first time). You have an (n-1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n-2)/n chance, until you get down to the last side left, where the chance is 1/n and the average number of goes is n. So you just add up all the averages to get the total average. I know this reads like a ramble but I can't be bothered to make it any clearer.
So then for an 8-sided die for example it would be then:

8/8 + 7/8 + 6/8 + 5/8 + 4/8 + 3/8 + 2/8 + 1/8?
No - it would be 8/8 + 8/7 + 8/6 + 8/5 + 8/4 + 8/3 + 8/2 + 8/1.

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Re: A little puzzle

Post by Jason Larsen » Fri Apr 18, 2008 2:05 am

If you were all having fun with this, I bet you would all shout, "Yahtzee!"

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Re: A little puzzle

Post by Jon Corby » Fri Apr 18, 2008 8:12 am

I'll take that bet.

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Re: A little puzzle

Post by Jason Larsen » Fri Apr 18, 2008 1:26 pm

You know what Yahtzee is, don't you, Jon?

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Re: A little puzzle

Post by Joseph Bolas » Fri Apr 18, 2008 1:36 pm

Gevin-Gavin wrote:
Joseph Bolas wrote:
Gevin-Gavin wrote:Well, to get all the sides of the dice to come up, you first have to get one of them - there is a 100% chance of achieving this on your first go, and the average number of goes is one. Then you need to get the second number. Each time you roll, there is one chance in n of failing (the chance of rolling the number you got the first time). You have an (n-1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n-2)/n chance, until you get down to the last side left, where the chance is 1/n and the average number of goes is n. So you just add up all the averages to get the total average. I know this reads like a ramble but I can't be bothered to make it any clearer.
So then for an 8-sided die for example it would be then:

8/8 + 7/8 + 6/8 + 5/8 + 4/8 + 3/8 + 2/8 + 1/8?
No - it would be 8/8 + 8/7 + 8/6 + 8/5 + 8/4 + 8/3 + 8/2 + 8/1.
My bad, I got confused by what side of the '/' you put (n-1) :oops:.

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Re: A little puzzle

Post by Jon Corby » Fri Apr 18, 2008 3:49 pm

Jason Larsen wrote:You know what Yahtzee is, don't you, Jon?
It's a river in China, isn't it?

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Re: A little puzzle

Post by Jason Larsen » Fri Apr 18, 2008 3:52 pm

No, that's Yangtze. Yahtzee is a dice game similar to poker with a cup.

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Re: A little puzzle

Post by Joseph Bolas » Fri Apr 18, 2008 3:56 pm

Corby wrote:
Jason Larsen wrote:You know what Yahtzee is, don't you, Jon?
It's a river in China, isn't it?
You love sarcasm don't you :lol:.

EDIT: I was asking this to Corby.
Last edited by Joseph Bolas on Fri Apr 18, 2008 4:17 pm, edited 1 time in total.

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Re: A little puzzle

Post by Jason Larsen » Fri Apr 18, 2008 4:09 pm

It's pronounced differently, whether you have an accent or not.

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Richard Brittain
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Re: A little puzzle

Post by Richard Brittain » Fri Apr 18, 2008 4:22 pm

How can you be sure of that Jason? There could easily be some little town in Wales where people pronounce 'Yahtzee' and 'Yangtze' the same.

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Re: A little puzzle

Post by Dinos Sfyris » Fri Apr 18, 2008 4:30 pm

Stop Yangtzing Jason's chain!

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Re: A little puzzle

Post by Ben Wilson » Fri Apr 18, 2008 4:41 pm

Jason Larsen wrote:No, that's Yangtze. Yahtzee is a dice game similar to poker with a cup.
Throw in 2 girls and that's the type of game I want to play. ;)

Back on topic, did the puzzle ever get solved?

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Re: A little puzzle

Post by Joseph Bolas » Fri Apr 18, 2008 4:44 pm

Ben Wilson wrote:Back on topic, did the puzzle ever get solved?
If you are on about the question "Can you generalise the method to a, say, an N-faced die?", then yes, Gevin-Gavin solved it.
Gevin-Gavin wrote:Anyway, for an n-sided dice: n/n (or 1) + n/(n-1) + n/(n-2) ... + n/(n-(n-1)) (or n)

To get all the sides of the dice to come up, you first have to get one of them - there is a 100% chance of achieving this on your first go, and the average number of goes is one. Then you need to get the second number. Each time you roll, there is one chance in n of failing (the chance of rolling the number you got the first time). You have an (n-1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n-2)/n chance, until you get down to the last side left, where the chance is 1/n and the average number of goes is n. So you just add up all the averages to get the total average. I know this reads like a ramble but I can't be bothered to make it any clearer.

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Re: A little puzzle

Post by Jason Larsen » Sat Apr 19, 2008 2:16 am

The game is pronounced "Yahtzee," with the "ah" sound meaning, "Open wide for the doctor!"

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Re: A little puzzle

Post by Richard Brittain » Sat Apr 19, 2008 7:51 pm

Jason, are you trying to replicate the crazy British sense of humour?

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Re: A little puzzle

Post by Jason Larsen » Sat Apr 19, 2008 9:12 pm

No, that's just the way it's pronounced, Yahtzee.

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