Probability of Raffle with 100 total tickets, 16 different games, 5 drawings,

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Rimi

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So, my friends and I are making a gaming tournament. We are going to have 5 drawings out of a pool of 16 games. In total, our group will split 100 tickets to allocate among games we want higher odds of playing, and then do the 5 drawings. All games must have at least 1 ticket in for them.

Example Possible Ticket Allocation: Game 1: 20 G2: 10 G3: 10 G4: 10 G5: 5 G6: 5 G7: 5 G8: 5 G9: 5 G10: 5 G11: 5 G12: 5 G13: 3 G14: 3 G15: 3 G16: 1

How do we calculate the percentage for each game to be in the tournament after the 5 drawings?

Initially I thought binomial expressions, but I am not getting accurate numbers with the binomial expressions I've tried. Then I tried, for the first game, (20/100) * (20/99) * (20/98) * (20/97) * (20/96), ie. Odds of the first game being drawn at least once out of 5 rounds (or so I thought was the way to calculate it). This also gives a weird number (0.0003541977). This formula also doesn't account for the drawings to draw repeats of OTHER games, which would be thrown out and not counted. Ie. Drawing 1 is game 4, drawing 2 draws game 4 again so that ticket is thrown out and drawing 2 draws again, thus increasing the odds for all the other games. Instinctually, I believe the percentage chance would be around 83-87% for game 1 with 20 tickets to be drawn out of 100 possible tickets after 5 drawings, but I don't know the formula to get to that.

Very simplified version of my problem: let's say a raffle has 2 drawings an item, of which there are two of. Person A buys 99 tickets and Person B buys 1 ticket and there are 100 in total. Person A has a 1% chance to not win the first drawing. Let's assume that happens. Drawing 1 happens and Person B wins (the odds were 99/100 for Person A to win). Next, drawing 2 happens. The remaining 99 tickets are owned by Person A, so his chances are 100%. The math seems like it would be (99/100) * (99/99), so 99% chance to win the first drawing and 100% chance to win the second drawing, but (99/100) * (99/99) equals 99%, not 100%. This is where my instinct breaks down with the mathematical formula I thought would be correct. Obviously Person A has a 100% of winning, but why doesn't the math reflect that?
 
Rimi said:
How do we calculate the percentage for each game to be in the tournament after the 5 drawings?
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What does the result of the five drawings mean? Is it supposed to be only one winner, i.e., the majority of the five draws? Or multiple winners? It looks like you're describing is a hypergeometric distribution (sampling without replacement) rather than a binomial distribution.
 
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BigBeachBananas said:
What does the result of the five drawings mean? Is it supposed to be only one winner, i.e., the majority of the five draws? Or multiple winners? It looks like you're describing is a hypergeometric distribution (sampling without replacement) rather than a binomial distribution.
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Multiple winners. First 5 games to be chosen will be in the tournament.
 
Rimi said:
Multiple winners. First 5 games to be chosen will be in the tournament.
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I think we're getting closer to the answer with hypergeometric distribution (sampling without replacement)! What N, k, n, and x be? N would probably be 100 as that's the total. Unsure of what k, n, and x would be. Here is the question again: from a total of 100 tickets of which there are X for each game (where X is the number of tickets allocated for each game; there are 16 possible games), what is the probability game Y will be drawn at least once?
 
BigBeachBananas said:
What if you draw the same game five times? Then you'll be only playing one game?
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Ah my apologies, I forgot to include that situation in my post! If a game has already been drawn, we're going to throw away that ticket and continue drawing in that round until we draw a game that hasn't been drawn, so 5 different games are guaranteed.
 
Rimi said:
Ah my apologies, I forgot to include that situation in my post! If a game has already been drawn, we're going to throw away that ticket and continue drawing in that round until we draw a game that hasn't been drawn, so 5 different games are guaranteed.
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Which tickets are you throwing away? For example, your 1st draw is Game1, then what?
Option 1: do you throw out all of the Game1 tickets from the raffle? i.e it's not possible to draw Game1 again from the raffle; OR
Option 2: the Game1 tickets are still part of the raffle, but if you draw it again, you won't count it and continue to draw until a different Game is picked?
 
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BigBeachBananas said:
Which tickets are you throwing away? For example, your 1st draw is Game1, then what?
Option 1: do you throw out all of the Game1 tickets from the raffle? i.e it's not possible to draw Game1 again from the raffle; OR
Option 2: the Game1 tickets are still part of the raffle, but if you draw it again, you won't count it and continue to draw until a different Game is picked?
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Option 2
 
The scenario you're describing doesn't fit into any distribution, at least I'm aware of, because of 2 reasons.
First, the number of times you draw isn't fixed. There's a possibility you have to draw more than five times to get five unique games.
Second, your probability of drawing each Game isn't fixed as you're throwing out the repeated tickets.
For instance, you picked Game1 on your first draw (20/100 chances), then you picked Game1 again(19/99), then picked Game2(10/98)
In another instance, you picked Game1 on your first draw (20/100 chances), then you picked Game2(10/99).
As you can see, the probability of picking Game2 is heavily dependent on how the previous tickets are picked; thus, it changes in every scenario.
Note that I've only shown you two simple cases that have only drawn two games so far instead of 5 games out of 16. There isn't a "shortcut" that I know of to accommodate your raffle scenario besides what I call the "brute force" method, where you'll have to account for all of the possibilities of picking repeated tickets and in different orders. This amounts to tremendous combinations and is computationally expensive.
 
Hello, it's nice to see an interesting approach to a game. There are many games around for children to learn math[/URL].
However, your game does not suit any of the possible answers that I can think of. I will agree with the comment above that you can use the brute force method for the solution to this.
This approach is inefficient, in the sense that it takes a long time. The rationale for this is that if we have a full graph, K-N, with N vertices, we will have (N-1)! circuits to list, compute the weight, and then choose the least.
Brute Force Algorithms are exactly what they sound like: simple ways of addressing a problem that relies on brute processing power and exhaustive testing rather than complex strategies to increase efficiency.
Consider a little padlock with four numbers, each ranging from 0-9. You can't remember your combination and don't want to buy a new padlock. You'll have to employ a brute force approach to open the lock because you can't recall any of the digits. So you reset all the numbers to 0 and try them one by one till it opens: 0001, 0002, 0003, and so on. In the worst-case situation, finding your combination would take 104, or 10,000, trials.
The temporal complexity of brute force is O(mn), which is represented as O(n*m) in some cases. So, if we used brute force to find a string of "n" characters in a string of "m" characters, it would take n * m attempts.
 
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