c4countdown

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.]

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?

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).

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?

Jason Larsen wrote:How did you come up with that?

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.

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?

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.

Jason Larsen wrote:You know what Yahtzee is, don't you, Jon?

Corby wrote:

Jason Larsen wrote:You know what Yahtzee is, don't you, Jon?

It's a river in China, isn't it?

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

Ben Wilson wrote:Back on topic, did the puzzle ever get solved?

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.