A little puzzle
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 Charlie Reams
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A little puzzle
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 warmup 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.]
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 warmup 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
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).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 warmup 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.]
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.
 Joseph Bolas
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Re: A little puzzle
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.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?
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).
 Charlie Reams
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Re: A little puzzle
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.Joseph Bolas wrote: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.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?
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).
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 Nfaced die?
 Joseph Bolas
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Re: A little puzzle
Yeah, I've just seen that Gevin's edit was just after I posted my message, so I missed that. Sorry .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 Nfaced die?
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Re: A little puzzle
Two, because 2 x 1 = 2
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Re: A little puzzle
Jason. That is brilliant.
This topic should be locked as it has nothing to do with either cheese or mushrooms.
This topic should be locked as it has nothing to do with either cheese or mushrooms.
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Re: A little puzzle
Thank you.
I will accept what you say, George.
I will accept what you say, George.

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Re: A little puzzle
Anyway, for an nsided dice: n/n (or 1) + n/(n1) + n/(n2) ... + n/(n(n1)) (or n)
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Re: A little puzzle
How did you come up with that?

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Re: A little puzzle
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 (n1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n2)/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.Jason Larsen wrote:How did you come up with that?
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Re: A little puzzle
So then for an 8sided die for example it would be then:GevinGavin 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 (n1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n2)/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.
8/8 + 7/8 + 6/8 + 5/8 + 4/8 + 3/8 + 2/8 + 1/8?

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Re: A little puzzle
No  it would be 8/8 + 8/7 + 8/6 + 8/5 + 8/4 + 8/3 + 8/2 + 8/1.Joseph Bolas wrote:So then for an 8sided die for example it would be then:GevinGavin 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 (n1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n2)/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.
8/8 + 7/8 + 6/8 + 5/8 + 4/8 + 3/8 + 2/8 + 1/8?
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Re: A little puzzle
If you were all having fun with this, I bet you would all shout, "Yahtzee!"
Re: A little puzzle
I'll take that bet.
 Jason Larsen
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Re: A little puzzle
You know what Yahtzee is, don't you, Jon?
 Joseph Bolas
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Re: A little puzzle
My bad, I got confused by what side of the '/' you put (n1) .GevinGavin wrote:No  it would be 8/8 + 8/7 + 8/6 + 8/5 + 8/4 + 8/3 + 8/2 + 8/1.Joseph Bolas wrote:So then for an 8sided die for example it would be then:GevinGavin 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 (n1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n2)/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.
8/8 + 7/8 + 6/8 + 5/8 + 4/8 + 3/8 + 2/8 + 1/8?
Re: A little puzzle
It's a river in China, isn't it?Jason Larsen wrote:You know what Yahtzee is, don't you, Jon?
 Jason Larsen
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Re: A little puzzle
No, that's Yangtze. Yahtzee is a dice game similar to poker with a cup.
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Re: A little puzzle
You love sarcasm don't you .Corby wrote:It's a river in China, isn't it?Jason Larsen wrote:You know what Yahtzee is, don't you, Jon?
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
It's pronounced differently, whether you have an accent or not.
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Re: A little puzzle
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
Stop Yangtzing Jason's chain!
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Re: A little puzzle
Throw in 2 girls and that's the type of game I want to play.Jason Larsen wrote:No, that's Yangtze. Yahtzee is a dice game similar to poker with a cup.
Back on topic, did the puzzle ever get solved?
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Re: A little puzzle
If you are on about the question "Can you generalise the method to a, say, an Nfaced die?", then yes, GevinGavin solved it.Ben Wilson wrote:Back on topic, did the puzzle ever get solved?
GevinGavin wrote:Anyway, for an nsided dice: n/n (or 1) + n/(n1) + n/(n2) ... + n/(n(n1)) (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 (n1)/n chance of succeeding, and the average number of goes is the reciprocal of this. For the next one you have an (n2)/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.
 Jason Larsen
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Re: A little puzzle
The game is pronounced "Yahtzee," with the "ah" sound meaning, "Open wide for the doctor!"
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Re: A little puzzle
Jason, are you trying to replicate the crazy British sense of humour?
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Re: A little puzzle
No, that's just the way it's pronounced, Yahtzee.
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