Here is the scenario. There are 25 individuals. These individuals are rolling a 100 sided die to win a prize. If 10 of these individuals are allowed to add +10 to their roll...
1. What is the percent chance that one of the 10 individuals allowed to add +10 to their roll will win the prize?
2. What is the percent chance that the other 15 individuals rolling 1-100 will have to win the prize?
3. If there are 12 prizes, what is the percent chance of one of the 10 individuals allowed to add +10 winning a prize?
4. If there are 12 prizes, what is the percent chance of one of the 15 individuals rolling 1-100 winning a prize?
5. At the end of 12 rolls what is the projected distribution of these prizes between the two groups?
6. Extra question. What number would you have to add to the 10 individuals roll to have the distribution be 2 prizes for the 10 people every one prize for the 15 people?
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Thank you so much if you are able to help out. If you can show any work that would be amazing.
Zipper, that's still UNCLEAR.
Say 1st two to roll are ones who get +10, and both roll 100 for 110:
so the 15 who don't get +10 are eliminated right away;
BUT these 2 with 110 cannot re-roll right away: they need to wait
to see if one or more of the other 8 will also roll 100, right?
As Wilmer noted, we still don't know enough. My questions:
Will the +10 people still add 10 for subsequent tie-breaker rolls? Does question #5 refer to 12 prizes or 1 prize? If 12 prizes, are we concerned about who gets 1st, who gets 2nd, etc., or is it simply that a person either wins or loses? For question #6, we need to know if it's exactly 2 prizes for every 1 prize, or at least. Also we need to know whether the amount we add has to be an integer. Also how many prizes are there total in regards to question #6. It's not really possible to answer your questions without this kind of clarity.
My general approach would be
Fix one of the players, say the name is Zach.
Let A = event that Zach wins
Let B = event that Zach is one of the 10 who adds +10
Now we have
Then you can do stuff like, probability that Zach rolls 100 * probability that everyone else rolls less than 100, (also handle ties), plus probability Zach rolls 99 * probability that everyone rolls less than 99, etc.
I would probably sketch some stuff out on paper and then write a computer program.
Does this give you enough to fill the details?
CorrectSay 1st two to roll are ones who get +10, and both roll 100 for 110:
so the 15 who don't get +10 are eliminated right away;
BUT these 2 with 110 cannot re-roll right away: they need to wait
to see if one or more of the other 8 will also roll 100, right?
-The +10 people will still add 10 to subsequent rolls.Will the +10 people still add 10 for subsequent tie-breaker rolls? Does question #5 refer to 12 prizes or 1 prize? If 12 prizes, are we concerned about who gets 1st, who gets 2nd, etc., or is it simply that a person either wins or loses? For question #6, we need to know if it's exactly 2 prizes for every 1 prize, or at least. Also we need to know whether the amount we add has to be an integer. Also how many prizes are there total in regards to question #6. It's not really possible to answer your questions without this kind of clarity.
-Question #5 refers to 12 prizes. To clarify as soon as someone gets a prize they may no longer roll. Each prize is to be rolled for separately. As an example All individuals roll for the first prize, The winner may no longer roll eliminating him from the pool.
-For the amount to add it has to be an integer for question 6 and the distribution would be at least 2 prizes for every 1.
Hopefully this helps. And thank you for your patience.
I think that pretty much clears up the meaning, unless there's some detail I'm overlooking (which is entirely possible). However, this is not a forum to do people's homework for them, but rather to help out. Please let us know how far you got, where you got stuck, etc. I already provided a sketch for a solution method.
Actually this isn't a homework assignment. Its to figure out a reward distribution method on a team and to put its effectiveness into perspective. Basically the goal of adding 10 to a roll was to ensure that those individuals get at least 2 prizes for every one prize won for the other group [non +10]. I do not believe the 10 is close to doing that and wanted to try to get some experienced eyes and proof behind it.
I actually do not know how to make use of the formulas and was looking to get the sharper eye of someone that is good with math and probability. I was looking more for percentages. If this is possible that would be amazing, if not I appreciate the look at the question and I will revise it and try to forward it to a math major or my old prof.
Hmm, well that does change the situation, although it's slightly complicated since theoretically tiebreakers could continue indefinitely; so in the computer program I'm planning in my head, I will need to check when the probability becomes less than some value like 10^-10 and discontinue such branches.. anyway, I'm not putting an extremely high priority on this project but I'll give it a shot, I'd like to have an answer to you within two days. (Although I recently failed to keep a time estimate given on another thread... I'll try though.) If someone else wants to give it a go, by all means do so.
Well, I think the way problem is presented makes it appear more complicated than it is.
You are trying to find out in a 15:10 group how much the score of the "10" needs to be
increased by so that the "10" ends up winning twice as often as the "15".
NO NEED to worry about ties within a group: we're looking at the 2 groups as individuals.
A tie really ONLY happens if top score in "15" group = top score in "10" group.
The "numbers" in your problem can be reduced by dividing by 5; so re-wording problem:
Here is the scenario. There are 5 individuals. These individuals are rolling a 20 sided die to win a prize.
2 of these individuals are allowed to add +x to their roll.
If the intent is for these 2 (G2) to win twice as often as the other 3 (G3), what must x be?
With x=1, 1 million tries at random (rounded):
Ties: 70,000
G3 : 492,000
G2 : 438,000
With x=5, 1 million tries at random (rounded):
Ties: 55,000
G3 : 312,000
G2 : 633,000
For the "purposes" of WHY you need this, then adding 5 is good enough:
no need to worry about "exactness", since you'll end up with stuff like
x = 4.567 to 5.123...
Again, I would not worry about tie-breakers; simply add them to the 2 groups
on a ratio basis; in this case 312:633, to get:
With x=5, 1 million tries at random (rounded):
G3 : 330,000
G2 : 670,000
Just my opinion !!
I'm not sure what Wilmer meant exactly; I wrote a straightforward simulation (hopefully bug free) in Java
Playing around with the variable named add, I find that adding 5 gives the advantaged people about 1.90 advantage, and adding 6 gives them about 2.25 advantage (I discarded ties altogether), so I would go with 6.Code:import java.util.Random; public class Dice100Group25Sim { static Random g = new Random(); static int add = 6, trials = 10000000, groupAdv = 0, groupDisadv = 0; public static void main(String[] args) { monteCarlo(); } static void monteCarlo() { int i = 0, j, highestAdv = 0, highestDisadv = 0; while(i < trials) { for(j = 0; j < 10; j++) highestAdv = Math.max(highestAdv, g.nextInt(100) + add); for(j = 0; j < 15; j++) highestDisadv = Math.max(highestDisadv, g.nextInt(100)); if(highestAdv != highestDisadv) { if(highestAdv > highestDisadv) groupAdv++; else groupDisadv++; i++; } highestAdv = 0; highestDisadv = 0; } System.out.println("Advantaged person won: " + groupAdv + " times"); System.out.println("Disadvantaged person won: " + groupDisadv + " times"); System.out.println("Ratio Adv/Disadv = " + (groupAdv/(double)(groupDisadv))); } }
Here is sample output for add = 5
and add = 6Code:Advantaged person won: 6543911 times Disadvantaged person won: 3456089 times Ratio Adv/Disadv = 1.8934440056375863
Code:Advantaged person won: 6921674 times Disadvantaged person won: 3078326 times Ratio Adv/Disadv = 2.248518837835889
Undefined, I'm unfamiliar with Java (Ubasic is my LOVE!); is this correct:
> add = 6
> for(j = 0; j < 10; j++)
> highestAdv = Math.max(highestAdv, g.nextInt(100) + add);
This picks 10 numbers at random from 1 to 100, adds 6 to the numbers picked,
and returns the highest of the 10 resulting numbers.
Example: 47, 7, 83, 18, 11, 34, 11, 65, 11, 34: returned is 83+6 = 89
NOTE: if above returned number > 100, this may be skipped:
> for(j = 0; j < 15; j++)
> highestDisadv = Math.max(highestDisadv, g.nextInt(100));
Similarly to above; 15 numbers picked, highest returned;
Example: 1,1,1,1,1,1,1,1,1,1,7,7,7,7,7: returned is 7
> if(highestAdv != highestDisadv) {
skip if the 2 returned numbers are the same
> if(highestAdv > highestDisadv) groupAdv++;
> else groupDisadv++;
update the proper group
I agree with your results.
Ran 10 million cases each for add = 3 to 7; rounded (add: [ties], 15group, 10group):
3: [49],434,566
4: [44],387,613
5: [40],346,654*
6: [36],308,692*
7: [33],275,725
* quite close to yours!!