Most likely winner

PBE6
Posers and Puzzles 20 Oct '05 20:08
1. PBE6
Bananarama
20 Oct '05 20:081 edit
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 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.)
2. 20 Oct '05 20:22
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.)
Seven (for all), this is because the two dices are more likely to add up to seven:
1 + 6
2 + 5
3 + 4
and the reverse.
3. PBE6
Bananarama
20 Oct '05 20:23
Originally posted by Bad wolf
Seven (for all), this is because the two dices are more likely to add up to seven:
1 + 6
2 + 5
3 + 4
and the reverse.
Nope! Seven is indeed the most likely roll, but if everyone rolls a seven then nobody won, did they?
4. XanthosNZ
Cancerous Bus Crash
20 Oct '05 21:22
What tiebreak procedures are in place?

I take it from you last post that there aren't any. So to win a player must be the highest roller alone.
5. gaurav2711
walking...
20 Oct '05 21:34
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.)
(a) 9
(b) 11
(c) 12
6. PBE6
Bananarama
20 Oct '05 22:30
Originally posted by gaurav2711
(a) 9
(b) 11
(c) 12
Close to what I got, but not quite. What was your reasoning/calculation?
7. PBE6
Bananarama
20 Oct '05 22:31
Originally posted by XanthosNZ
What tiebreak procedures are in place?

I take it from you last post that there aren't any. So to win a player must be the highest roller alone.
That's what I calculated for, yes. No tie-breakers, it has to be a clear win.
8. XanthosNZ
Cancerous Bus Crash
21 Oct '05 01:181 edit
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.
9. gaurav2711
walking...
21 Oct '05 06:39
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.
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?
10. PBE6
Bananarama
21 Oct '05 14:31
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?
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 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%
11. XanthosNZ
Cancerous Bus Crash
21 Oct '05 21:47
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%
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);
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.
12. PBE6
Bananarama
21 Oct '05 22:15
Originally posted by XanthosNZ
Why are you approximating?
Good point. I originally did this with a football pool, and the distribution was binomial so I had to approximate with Excel. But the dice distribution is linear (in 2 parts), which I totally missed.
13. 22 Oct '05 05:383 edits
Edit: Just read the tie rules.
14. 22 Oct '05 06:073 edits
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.
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:

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