- 20 Oct '05 20:08 / 1 editA group of n players are rolling to see who goes first in a Monopoly game by throwing 2 evenly weighted 6-sided dice. Which roll (total score) is most likely to be the winning roll if there are:

(a) 4 players?

(b) 10 players?

(c) 25 players?

(For clarification, if you ran the game 1,000,000 times, which roll would be expected to appear as the winner the most number of times? HINT: The bigger the roll, the more likely it is to win, but the less likely it is to come up in the first place.) - 20 Oct '05 20:22

Seven (for all), this is because the two dices are more likely to add up to seven:*Originally posted by PBE6***A group of n players are rolling to see who goes first in a Monopoly game by throwing 2 evenly weighted 6-sided dice. Which roll (total score) is most likely to be the winning roll if there are:**

(a) 4 players?

(b) 10 players?

(c) 25 players?

(For clarification, if you ran the game 1,000,000 times, which roll would be expected to appear as the winner th ...[text shortened]... he roll, the more likely it is to win, but the less likely it is to come up in the first place.)

1 + 6

2 + 5

3 + 4

and the reverse. - 20 Oct '05 21:34

(a) 9*Originally posted by PBE6***A group of n players are rolling to see who goes first in a Monopoly game by throwing 2 evenly weighted 6-sided dice. Which roll (total score) is most likely to be the winning roll if there are:**

(a) 4 players?

(b) 10 players?

(c) 25 players?

(For clarification, if you ran the game 1,000,000 times, which roll would be expected to appear as the winner th ...[text shortened]... he roll, the more likely it is to win, but the less likely it is to come up in the first place.)

(b) 11

(c) 12 - 21 Oct '05 06:39

is it 0.0255. I get 0.0277.*Originally posted by XanthosNZ***With 4: 10**

With 10: 11

With 25: 12

I wrote some quick matlab code to do all the calculations.

Basically chance of a number winning is:

Chance of number being rolled by player X * Chance of number less than that being rolled ^ (n-1)

So for 12 it's 0.0255 with n=4.

well....if there is no tie breaker....my results change.

for 4 ppl - roll of 9 should win

for 10 ppl - roll of 10 should win

for 25 ppl - there'll be 2 ppl rolling 12. Hence, no clear winner.

Dont u have the answer? - 21 Oct '05 14:31

My numbers agree more closely with XanthosNZ, as does my calculation. Here's what I did:*Originally posted by gaurav2711***is it 0.0255. I get 0.0277.**

well....if there is no tie breaker....my results change.

for 4 ppl - roll of 9 should win

for 10 ppl - roll of 10 should win

for 25 ppl - there'll be 2 ppl rolling 12. Hence, no clear winner.

Dont u have the answer?

1. Calculate the probability of each roll coming up.

2. Calculate the area under the probability distribution graph left of the number in question (in Excel, I approximated this using the trapezoid rule).

3. Calculate the probability that all opponents' rolls will be less than a given roll (P(my roll=highest) = (area to left of roll)^(# opponents)).

4. Calculate the most likely winning roll (likelihood = P(roll)*P(my roll=highest)).

5. Find the most likely winning roll.

Depending on how you approximate the area under the graph, the numbers will be slightly different, but here's what I got:

(a) 4 players - most likely winner is 10 at 22.7%

(b) 10 players - most likely winner is 11 at 34.2%

(c) 25 players - most likely winner is 12 at 53.3% - 21 Oct '05 21:47

Why are you approximating?*Originally posted by PBE6***My numbers agree more closely with XanthosNZ, as does my calculation. Here's what I did:**

1. Calculate the probability of each roll coming up.

2. Calculate the area under the probability distribution graph left of the number in question (in Excel, I approximated this using the trapezoid rule).

3. Calculate the probability that all opponents' rolls will b ...[text shortened]... players - most likely winner is 11 at 34.2%

(c) 25 players - most likely winner is 12 at 53.3%

Say you are working out the chance of 7 winning.

It's 1/6 (chance of rolling a 7) * (15/36)^(n-1)

The 15/36 is the chance of every number less than 7 summed together. For anyone that is interested the code to test this is quite short:

clear;

clc;

n=25;

Z=1/36;

A=1;

for I=12:-1:7

C= Z * (A-Z)^(n-1);

A=A-Z;

Z=Z+1/36;

I

C

end

This outputs the number and then the probability of that number being the winning roll one after each other. The probabilities don't add to 1 as the remainder will be tied rolls which are ignored for the purpose of this test. - 22 Oct '05 06:07 / 3 edits

I think your formula is not quite right. It should be a binomial distribution. If we have n players, and only one can win, but any one can win, then the formula for 7 is:*Originally posted by XanthosNZ***Why are you approximating?**

Say you are working out the chance of 7 winning.

It's 1/6 (chance of rolling a 7) * (15/36)^(n-1)

The 15/36 is the chance of every number less than 7 summed together. For anyone that is interested the code to test this is quite short:

clear;

clc;

n=25;

Z=1/36;

A=1;

for I=12:-1:7

C= Z * (A-Z)^(n-1); ...[text shortened]... 't add to 1 as the remainder will be tied rolls which are ignored for the purpose of this test.

C(n, 1) * (1/6) *(15/36)^(n-1) = n * (1/6)*(15/36)^(n-1).