Now, imagine the same game played with a die that has an infinite number of sides. What is the expected number of rolls to end the game?

Closed-form solutions only please.

1) If 2,5,5, game ends, total 3 rolls

2) If 4 ,4, game ends, total 2 rolls

3) If 2, 4 ,5, 6, game ends, total 4 rolls

4)If 2 ,3, 5, 2, game ends, total 4 rolls

For a six sided die game, average rolls = 2.1614 ?

For N sided die, average rolls = 1 + 1/N * ∑[(N+1)/N]^(N-n), n = 2 to N

Simplify further, average rolls =[(N+1)/N]

^{(N – 1)}

I suspect average rolls = e = Euler's number, when N = infinity !

I tend to agree using this logic.Quote:ssho88...I suspect average rolls = e = Euler's number, when N = infinity !

Rather than assuming the dice is N sided assume we are picking random numbers between 0 and 1 (technically depending on the number of faces the choices would be 1/N 2/N 3/N etc, but as N tends to infinity it's just every possible number).

After n numbers have been rolled the chances they are in ascending order, and you can roll again, is 1/n! Thus to get the next roll the chances are 1/n! (I might be 1 out, but who cares, this is proof by induction).

You'll always have the first roll and last roll, so this corresponds to 1 + 1/1!.

The chances of having an extra/third roll is that the second number is bigger than the first number, the chances of that are 1/2!.

Hence the total IS 1 + 1/1! + 1/2! etc = e.

Quote:charliepatrickI tend to agree using this logic.

Rather than assuming the dice is N sided assume we are picking random numbers between 0 and 1 (technically depending on the number of faces the choices would be 1/N 2/N 3/N etc, but as N tends to infinity it's just every possible number).

After n numbers have been rolled the chances they are in ascending order, and you can roll again, is 1/n! Thus to get the next roll the chances are 1/n! (I might be 1 out, but who cares, this is proof by induction).

You'll always have the first roll and last roll, so this corresponds to 1 + 1/1!.

The chances of having an extra/third roll is that the second number is bigger than the first number, the chances of that are 1/2!.

Hence the total IS 1 + 1/1! + 1/2! etc = e.

I am using Markov Chain to find the average rolls, from there, part of the formula can be simply with Geometric series formula, then finally it is Euler's number when N = infinity !

Those are incorrect answers for a six-sided die and for an infinite-sided dieQuote:ssho88My assumptions for 6 sided die :-

1) If 2,5,5, game ends, total 3 rolls

2) If 4 ,4, game ends, total 2 rolls

3) If 2, 4 ,5, 6, game ends, total 4 rolls

4)If 2 ,3, 5, 2, game ends, total 4 rolls

For a six sided die game, average rolls = 2.1614 ?

For N sided die, average rolls = 1 + 1/N * ∑[(N+1)/N]^(N-n), n = 2 to N

Simplify further, average rolls =[(N+1)/N]^{(N – 1)}

I suspect average rolls = e = Euler's number, when N = infinity !

Quote:Ace2Those are incorrect answers for a six-sided die and for an infinite-sided dieQuote:ssho88My assumptions for 6 sided die :-

1) If 2,5,5, game ends, total 3 rolls

2) If 4 ,4, game ends, total 2 rolls

3) If 2, 4 ,5, 6, game ends, total 4 rolls

4)If 2 ,3, 5, 2, game ends, total 4 rolls

For a six sided die game, average rolls = 2.1614 ?

For N sided die, average rolls = 1 + 1/N * ∑[(N+1)/N]^(N-n), n = 2 to N

Simplify further, average rolls =[(N+1)/N]^{(N – 1)}

I suspect average rolls = e = Euler's number, when N = infinity !

My assumptions for 6 sided die correct ?

My both algebraic calculations and simulations give same answer.

six sided die, rolls = 2.1614

Please double check, Assumptions 3 correct ?

Only increasing rolls are counted. So 2455, 1342, and 236 are all counted as 3 rolls.

It’s not difficult to see what the expected value converges to as the number of sides goes up. But the proof of the answer is clever in my opinion.

ssho88, you are correct for the six-sided die EV for the originally stated problem (when all rolls are counted). Sorry about that

Quote:Ace2My apologies, I misstated the problem.

Only increasing rolls are counted. So 2455, 1342, and 236 are all counted as 3 rolls.

It’s not difficult to see what the expected value converges to as the number of sides goes up. But the proof of the answer is clever in my opinion.

ssho88, you are correct for the six-sided die EV for the originally stated problem (when all rolls are counted). Sorry about that

Thanks for the clarification. I get 1.52163 for a 6-sided die. By varying the number of sides, I see that for an infinite-sided die, the answer would be near 1.718. (Like sscho88, I first thought the answer would have been infinity.)

Please show us your technique to get the exact answer.