[superprismatic]

Circular hopscotch is played on six

squares arranged in a circle. A coin

is tossed and a player starting on

square 1 jumps, advancing either 1 or

2 squares according as the coin falls

head or tail. When he lands on square

6, he is 'out'. Each player gets a

score equal to the number of squares

stood on or passed over during his

turn. (Thus the only possible scores

are multiples of 6.) What is the

average number of points a player may

expect to score?

SUPERPRISMATIC CLARIFICATION:

Each advancement ( 1 or 2 squares)

is preceded by a separate coin toss.

That is to say, a single coin toss

does not control all the jumps a

player makes.

[bushindo]

Circular hopscotch is played on six

squares arranged in a circle. A coin

is tossed and a player starting on

square 1 jumps, advancing either 1 or

2 squares according as the coin falls

head or tail. When he lands on square

6, he is 'out'. Each player gets a

score equal to the number of squares

stood on or passed over during his

turn. (Thus the only possible scores

are multiples of 6.) What is the

average number of points a player may

expect to score?

SUPERPRISMATIC CLARIFICATION:

Each advancement ( 1 or 2 squares)

is preceded by a separate coin toss.

That is to say, a single coin toss

does not control all the jumps a

player makes.

Is it weird of me to say that I start to experience withdrawal symptoms if I go too long without seeing a new Walter Penney puzzle?

average number of points is 9.142857. This is a geometric distribution, or distribution of time to first success (a success being defined in this case as getting a 6). This game can be discretized into rounds of trying to get to 6 from 1. If you pass by the 6 without landing on it, then you must be at square 1 and another round starts. The number of points for every round is 6. A system of recursive equations shows that the probability of successfully getting to 6 from 1 is always 0.6562500. The expected number of round before getting a 6 is 1/0.6562500 = 1.523810. Multiplying that by 6 gives the expected score of 9.142857.

Edited August 23, 2009 by bushindo

[pdqkemp]

I understood this problem a little differently. I figured you just keep going around and around and around until 1 of 2 things happens. Either you are on 'square 4' and you flip "heads" to move 2 spaces (or whichever of heads/tails was chosen to be 2), OR you are on 'square 5' and you flip "tails" to move only 1 square.

Well...maybe I understood the question the same, but this is what I got:

There is a 1/2 shot that you will get exactly 6 points because the first time you go around, you've got 50/50 that you'll land on 'square 6'. You've got a 1/4 shot that you will get exactly 12, landing on 'square 6' on exactly the 2nd round. 1/8 says you'll land on square 6 exactly the 3rd time around, giving you 18 points. 1/16 that you get 24 points, etc.

So...the formula in my mind is (1/2 * 6) + (1/4 * 12) + (1/8 * 18) + (1/16 * 24) + (1/32 * 30) + (1/64 * 36) + (1/128 * 42) ...... + [1/(2?n) * (6*n)]

Unfortunately, I got a "D-" in Calculus about a million years ago in high school, so I can't be positive; but after manually going out to about 12 powers of '2', it seems that the number is approaching 12.

So, 12 is my guess.

I've still yet to be right at one of these, so if I'm wrong AGAIN, I've always appreciated someone pointing out where I went wrong!

[Guest]

Every time around, you either end on 6 or restart on 1.

You get to or pass 6 in 3 flips 4 times, in 4 flips 7 times, and in 5 flips 2 times (13 combos)

That means your odds of getting to or passing 6 in 3 flip are 4/2³, in 4 flip 7/2⁴ and 5 flips 2/2⁵

Then 3 of the three flip get you to six, 4 of the four flip turns get you to six and 1 of the 2 five flips.

So your overall odds, when starting at 1 of getting to 6 are 3/2³ + 4/2⁴ + 1/2⁵ = 21/32. So your expected number of turns around the wheel is 32/21. 6 points per turn is 64/7 points.

[bushindo]

I understood this problem a little differently. I figured you just keep going around and around and around until 1 of 2 things happens. Either you are on 'square 4' and you flip "heads" to move 2 spaces (or whichever of heads/tails was chosen to be 2), OR you are on 'square 5' and you flip "tails" to move only 1 square.

Well...maybe I understood the question the same, but this is what I got:

There is a 1/2 shot that you will get exactly 6 points because the first time you go around, you've got 50/50 that you'll land on 'square 6'. You've got a 1/4 shot that you will get exactly 12, landing on 'square 6' on exactly the 2nd round. 1/8 says you'll land on square 6 exactly the 3rd time around, giving you 18 points. 1/16 that you get 24 points, etc.

So...the formula in my mind is (1/2 * 6) + (1/4 * 12) + (1/8 * 18) + (1/16 * 24) + (1/32 * 30) + (1/64 * 36) + (1/128 * 42) ...... + [1/(2?n) * (6*n)]

Unfortunately, I got a "D-" in Calculus about a million years ago in high school, so I can't be positive; but after manually going out to about 12 powers of '2', it seems that the number is approaching 12.

So, 12 is my guess.

I've still yet to be right at one of these, so if I'm wrong AGAIN, I've always appreciated someone pointing out where I went wrong!

You have the right idea, except that you started with the wrong value for the change of landing on a 6, which isn't 1/2. The post above described one way to get that value. Here's another way to get that value

The chance of landing on a 6 changes depending on what square you are presently on. If you are on square 5, your chance of landing on a 6 is 1/2. If you are on square 4, your chance of landing on a 6 is 3/4. You want to compute the chance of landing on a 6 from square 1. Let P(N ) be the chance of landing on square 6, given that you are on square N, N != 6. The following system of equation becomes apparent.

P(1) = (1/2) P( 2 ) + (1/2) P(3)

P(2) = (1/2) P( 3 ) + (1/2) P(4)

P(3) = (1/2) P( 4 ) + (1/2) P(5)

P(4) = (1/2) P( 5 ) + (1/2)

P(5) = (1/2)

If you start from P(4) and solves upwards, you'll get P(1), which you can plug into your existing equation to give the right answer

[pdqkemp]

bushindo,

Thanks for helping my wee brain understand this one!

I've got just enough brain cells to understand these puzzlers, but not quite enough to solve one!

[Guest]

do you count the first square as a point? So, if you flip a 'head' you move to square two for a total of two points? Also, if you flip a 'tail' do you only count the point for the square jumped or do you count 2 points for 1 jumped and one landed on?

sorry it's at the end of the day and my head is mushy.

Edited August 25, 2009 by badudi