Probability Question

[BlueEyedBoy]

Trying to work out the probability of each contenstant in a game:

There are four players that have to race round a race track.

Lane 1 has 12 squares
Lane 2 has 18 squares
Lane 3 has 24 squares
Lane 4 has 30 squares

Each player throws a dice in turn.

What is the probability of Player 1, 2, 3, 4 of winning? And how do you work it out, I haven't done this for years, and have an answer, just want to check its right.

[CosmicMatt]

Do you want us to really do your homework for you?

[grumbledoak]

Not really enough info about how the game works, though I can guess. I'm also guessing you are asking us to do your homework. No dice.

[g4ry13]

How do you win in this game and what does the squares have to do with it?

What answer do you have anyway?

[BlueEyedBoy]

It is homework, not mine, but my 10 year old son's, I just want to check I have explained it correctly.

[BlueEyedBoy]

g4ry13 said:

How do you win in this game and what does the squares have to do with it?

What answer do you have anyway?

First person round the board wins. Think of it has a oval track, lane four has more distance, so we have added more squares.

Edited by BlueEyedBoy on Saturday 11th October 17:12

[sstein]

g4ry13 said:

How do you win in this game and what does the squares have to do with it?

What answer do you have anyway?

I think he means the players take turns rolling a dice and the number on the dice dictates how many squares to advance, the player wins when they advance all their squares.

-

Stuart

[grumbledoak]

sstein said:

I think he means the players take turns rolling a dice and the number on the dice dictates how many squares to advance, the player wins when they advance all their squares.

At least, that was my guess.

[g4ry13]

What answer do you have? Are we talking about a one single race or over numerous races. Assuming the die is fair any number has the same chance (1/6) but a contestant may roll above the average expected value and others may roll below average distorting the true probability. So for a single race, you'll probably have the result affected by variance and possibly not have the outcome with the highest probability winning.

So do they want a fractional answer or something or am I thinking too much about this and thinking it's slightly more complex than the stuff I did as a 10 year old.

[BlueEyedBoy]

g4ry13 said:

What answer do you have? Are we talking about a one single race or over numerous races. Assuming the die is fair any number has the same chance (1/6) but a contestant may roll above the average expected value and others may roll below average distorting the true probability. So for a single race, you'll probably have the result affected by variance and possibly not have the outcome with the highest probability winning.

So do they want a fractional answer or something or am I thinking too much about this and thinking it's slightly more complex than the stuff I did as a 10 year old.

I am killing his for coming up with this! But now I want to be able to make sure I can answer it!! Yes I think its far more complex than stuff I did at 10, or probably at 16.

Yes, talking about 1 single race and a 1 in 20 type answer, so yes fractional.

Edited by BlueEyedBoy on Saturday 11th October 17:29

[g4ry13]

Fair die (1/6) average roll will be 3. Rolling the die 1000 times and adding them all up the answer should be very close to 3 (assuming the first sentence.)

So the results "should" be:

Player 1: 4 throws, Player 2: 6 throws etc.

This probably seems very obvious, but for a 10 year old it must be the sort of thing they're looking for

[BlueEyedBoy]

g4ry13 said:

Fair die (1/6) average roll will be 3. Rolling the die 1000 times and adding them all up the answer should be very close to 3 (assuming the first sentence.)

So the results "should" be:

Player 1: 4 throws, Player 2: 6 throws etc.

This probably seems very obvious, but for a 10 year old it must be the sort of thing they're looking for

Would the average not be 3.5 though?

[carmonk]

BlueEyedBoy said:

Trying to work out the probability of each contenstant in a game:

There are four players that have to race round a race track.

Lane 1 has 12 squares
Lane 2 has 18 squares
Lane 3 has 24 squares
Lane 4 has 30 squares

Each player throws a dice in turn.

What is the probability of Player 1, 2, 3, 4 of winning? And how do you work it out, I haven't done this for years, and have an answer, just want to check its right.

LOL! The answer's there, written down for you!

12:18:24:30

or

2:3:4:5

I thought from the title we were talking quantum

[BlueEyedBoy]

g4ry13 said:

BlueEyedBoy said:

g4ry13 said:

Fair die (1/6) average roll will be 3. Rolling the die 1000 times and adding them all up the answer should be very close to 3 (assuming the first sentence.)

So the results "should" be:

Player 1: 4 throws, Player 2: 6 throws etc.

This probably seems very obvious, but for a 10 year old it must be the sort of thing they're looking for

Would the average not be 3.5 though?

Why? Try rolling it a few thousand times and let me know

(1+2+3+4+5+6)/6

[g4ry13]

BlueEyedBoy said:

g4ry13 said:

Fair die (1/6) average roll will be 3. Rolling the die 1000 times and adding them all up the answer should be very close to 3 (assuming the first sentence.)

So the results "should" be:

Player 1: 4 throws, Player 2: 6 throws etc.

This probably seems very obvious, but for a 10 year old it must be the sort of thing they're looking for

Would the average not be 3.5 though?

That is a point thinking about it Now I look like a bit of a muppet

edit: Damn, you're quick. I deleted that after thinking for about 10 seconds and you already replied.

Edited by g4ry13 on Saturday 11th October 17:42

[BlueEyedBoy]

g4ry13 said:

BlueEyedBoy said:

g4ry13 said:

Fair die (1/6) average roll will be 3. Rolling the die 1000 times and adding them all up the answer should be very close to 3 (assuming the first sentence.)

So the results "should" be:

Player 1: 4 throws, Player 2: 6 throws etc.

This probably seems very obvious, but for a 10 year old it must be the sort of thing they're looking for

Would the average not be 3.5 though?

That is a point thinking about it Now I look like a bit of a muppet

Thankgod, getting sick of the jibes myself

[grumbledoak]

Jibes aside, then!

Dude, if the 'die' is fair, it doesn't matter at all; it'll just be down to the number of squares each player has to move. Probability != one game; it's the average over all of the possible games.

Edited by grumbledoak on Saturday 11th October 17:51

[BlueEyedBoy]

Using the figure of 3.5 as the average, (I have also worked it out with the die been a constant 6, just picking one number), I get about a 97% chance of track 1 winning, i.e. 12 squares, about 3% chance of track 2 and the other 2 pretty remote chances.


Track 1 = 4 throws = 1/6^4 = 1/1296 = 8029 / 7776 = 96.8
Track 2 = 6 throws = 1/6^6 = 1/46656 = 8029 / 216 = 2.7
Track 3 = 7 throws = 1/6^7 = 1/279936 = 8029 / 36 = 0.45
Track 4 = 9 throws = 1/6^9 = 1/10077696 = 8029 / 1 = 0.1

Edited by BlueEyedBoy on Saturday 11th October 18:07

[grumbledoak]

Dude, forget the dice. Really.

[sstein]

This is for a ten year old ?