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How do I calculate the odds of winning a lottery when there are multiple prize pools?
Sorry if this is the wrong sub, I need help solving this problem and don't know where to start. I am trying to calculate raffle odds for if someone has multiple tickets to a lottery. Here are the constraints:
There are three prize categories of thirty prizes each (90 total prizes)
30 participants are given 1 ticket from each category. (3 total tickets)
6 prizes are selected at random
A participant could win multiple prizes.
How do you calculate the odds of a participant winning a prize? Surely it isn't 6/30 since a participant could win a prize from more than one category. I don't think it would be 6/90 since each person has 3 tickets, but I could be wrong.
How do I calculate the odds of winning a lottery when there are multiple prize pools?
Sorry if this is the wrong sub, I need help solving this problem and don't know where to start. I am trying to calculate raffle odds for if someone has multiple tickets to a lottery. Here are the constraints:
There are three prize categories of thirty prizes each (90 total prizes)
30 participants are given 1 ticket from each category. (3 total tickets)
6 prizes are selected at random
A participant could win multiple prizes.
How do you calculate the odds of a participant winning a prize? Surely it isn't 6/30 since a participant could win a prize from more than one category. I don't think it would be 6/90 since each person has 3 tickets, but I could be wrong.
[Q] How do I calculate the odds of winning a lottery when there are multiple prize pools?
Sorry if this is the wrong sub, I need help solving this problem and don't know where to start. I am trying to calculate raffle odds for if someone has multiple tickets to a lottery. Here are the constraints:
There are three prize categories of thirty prizes each (90 total prizes)
30 participants are given 1 ticket from each category. (3 total tickets)
6 prizes are selected at random
A participant could win multiple prizes.
How do you calculate the odds of a participant winning a prize? Surely it isn't 6/30 since a participant could win a prize from more than one category. I don't think it would be 6/90 since each person has 3 tickets, but I could be wrong.
Let's say like Logic, "If A then B, if B then C" type thing, but with percentages. So if you are trying to figure out the odds of C happening based on pre-existing odds, how would you if: A has a 5% chance of occurring. B has a 40% chance of occurring (only if A happens) C has a 1% chance of occurring (only if A and B happens) So C obviously has less than a 1% objective chance of happening, since it is 1% of 40% of 5%, is it straight break down? .0002? Bonus question: If there are say 10 things that each have a 1% chance of happening, does the probability still set technically at "1%" of anything happening? Is there some increase in the odds of SOMETHING occurring in mathematical consideration?
How do I calculate the odds of me receiving a rare drop from killing multiple monsters in an RPG?
Let's say I'm killing a monster that drops a rare item 1 in 300 kills, and I plan on killing them for two hours which, according to my previous experience, means I'll get around 100 kills. What's the formula for calculating my odds of success in things like this where the system has no memory?
Calculating the odds of successs with multiple attempts.
In poker there are 5 community cards that get played on the board. I'm trying to figure out how often one of those five will be a 7 AND a 2. The odds of either a 7 or a 2 being one of those cards is 13/1, the odds of them both coming, if there were only two community cards played is 149/1, however, there's 5 community cards. How do you calculate the odds 2 out of 5 successes at 13/1 odds? How often should a board of five cards produce both a 7 and a 2?
How do I calculate the odds of something happening when there are multiple (but limited) chances of it happening, at a fixed rate?
I've been playing a tower defense game that involves attacking enemy players' bases. During most attacks, certain parts of the route (monuments) can be destroyed for a chance at a chest dropping.
There are generally 8 monuments that can be destroyed.
There is a 20% chance that destroying a monument will yield a chest.
A monument will never yield more than 1 chest.
There is no apparent limit to how many monuments can yield a chest in a given attack.
How do I calculate the overall odds of getting a chest from a single attack if I destroy all 8 monuments every time? This is a formula I've been wondering about for a while (not good at math as I haven't had to study it in a decade), but I'm terrible at wording it for online search purposes.
Counting Outs vs. Binomial Coefficient for calculating odds over multiple cards.
SOLVED: by MaxineZJohnson -- because I'm a terrible proof reader So everyone knows that when you have four-to-a-flush on the Flop (two in the hole, two on the board), your odds of hitting your flush by the River are roughly 35%. I wanted to figure out the exact calculation, so I did some research. On this page (under the "Doing the Math" heading) calculates it as follows:
With 9 hearts remaining there would be 36 combinations of getting 2 hearts and making your flush with 5 hearts. This is calculated as follows: (9 x 8 / 2 x 1) = (72 / 2) ≈ 36. This is the probability of 2 running hearts when you only need 1 but this has to be figured. Of the 47 unknown remaining cards, 38 of them can combine with any of the 9 remaining hearts: 9 x 38 ≈ 342. Now we know there are 342 combinations of any non heart/heart combination. So we then add the two combinations that can make you your flush: 36 + 342 ≈ 380. The total number of turn and river combos is 1081 which is calculated as follows: (47 x 46 / 2 x 1) = (2162 / 2) ≈ 1081. Now you take the 380 possible ways to make it and divide by the 1081 total possible outcomes: 380 / 1081 = 35.18518%
However, I tried to calculate it differently, by using binomial coefficient to determine the chance that neither card helps you, with the remainder thus being your chance to hit the flush. That would be expressed as: (38c2)/(47c2), Or: 1 - ((38!/(2!x36!)) / (47!/(2!x45!))) Which equals 0.34967622571, or 34.967622571% Obviously the difference is so small as to be irrelevant in practice, but it is bugging the hell out of me trying to figure out which one is technically correct (the best kind of correct). I'm hoping some of you math wizards can help me out. Cheers! EDIT: had to use "x" instead of an asterisk to denote multiplication because reddit formatting makes it italics and I'm too dumb to figure out how to stop it.
Looking for Formula: Calculating odds across multiple variable tests.
Just found this subreddit. If this question is incorrectly placed, please let me know. I'm trying to determine the proper formula for a series of tests with differing odds, looking for a single success. For example: if I have a test that starts at a 5% success rate, then increases 5% with each successive test, what is the chance I'll get 1 success over three tests where I will stop as soon as it is achieved. so 1/20, 1/10, and 3/20 but if the first is a success, the following are not even tested. I'm wanting to apply this to a few different combinations, so I'm more focused on the formula than the specifics above (bot won't reject them as an example case).
Simple odds calculator to calculate drop chances for multiple runs
A while back I built a simple odds calculator for Warframe as a way to give back to the community. I got lots of great feedback and now an updated version is ready to use. The simple odds calculator now calculates the chances of dropping at least once in 10/20/30 runs or 4 runs (equals one rad share run). It also displays how many runs are needed to have a 90% or 99% percent chance of dropping at least once. You can also share calculations by copying and pasting of the URL. Example calculations for 4.17% drop chance: http://wfsoc.larshaendler.com/?percent=4.17
[no clue discipline level, video game math/statistics] How are odds of something happening calculated when one statistic is rolled multiple times?
This is a pokemon question, i know, silly, but for a shiny pokemon (simple aesthetic difference) the base likelyhood of encountering one is 1/4096. through certain methods this can be raised to 8/4096. and i have seen it said that this is not EXACTLY the same as 1/512, why is that, and how can the likelyhood be calculated based on multiple rolls of the same (or even different) odds?
Trying to figure out/find a calculator for multiple percentage odds.
I feel like there's a specific name for this but I can't for the life of me find it in a search term, and that I also previously knew how to do this back in highschool, but I can't remember anymore. Basically I'm looking to figure out how to find out which probability is a better bet, given 2 or more sets of probability. A calculator would be great since I'd need to do the somewhat frequently (unless it's actually really simple and I'm just being dumb). In case what I'm looking for doesn't make sense, an example issue could be something like ten 5% attempts vs three 20% attempts, which would more likely result in at least one success.
Calculating multiple events, odds per turn in a TCG.
So I’m aware of hypegeometric distribution and the calculator available online. However, in order to check the odds and possible progression of draws and decks, I’m wondering how to calculate draw odds for multiple events. Is anyone familiar with these sorts of deck building probability maths? Looking up how to do such calculations online, but needed some guidance in the right direction. Edit: it seems if I’m dealing with drawing specific NONoverlapping card draw events then it may be calculation of the probability of dependent events as well... not even sure where to start. Thanks all.
Just trying to work something out, how can i calculate multiple odds? For example if i roll a dice 3 times, what are the odds that i will get 3 of the same number? The specific equation i needed was 1/15, 1/10, 1/35 then 1/3. What i tried was (1/15) * (1/10) * (1/35) * (1/3) but i just got 0.33...
Hello all, I've been redirect here from math (where my problem was solved as 1-(7/8)3), but I'd like to get a better understanding of the solution as there wasn't much of an explanation or where to continue learning from there. Original post below: Problem: I have a 1 in 8 chance of something occuring, there are 3 events with this 1/8 chance. How would I calculate the chance of at least one of the Events occuring? Currently I'm calculating this inside an Excel spreadsheet, and using the formula 1/8 * 3, which seemed to do the trick until those three Events turned into eight Events and the formula shows 1/8 * 8=100%. As none of them are garunteed to happen, I know you can't reach 100%... it's obvious now that my math skills are terrible and I need to seek help. Is there any names for this kind of problem besides just "Probabilities" and where would be the best place to learn about how to calculate these? Thank you :)
Calculating the odds for a raffle with multiple prizes and a person holding multiple raffle tickets
I can't wrap my mind around this one. If a raffle had one prize and 500 tickets were given out to 500 individuals, the odds of winning would be 1 in 500. But what if a raffle had 200 prizes and 500 tickets were given out to 500 individuals? Would the odds be 2 in 5? Now what happens if one individual has four raffle tickets. So 200 prizes, 503 raffle tickets given out. How do I figure this out? The odds of a person with four raffle tickets. Real world situation so I'm very curious. My current guess would be (Sum of): 4/503 + 4/502 + 4/501 ... 4/(503-199) but that would be greater than 1.
I came across a very odd situation. I was playing cards and within 2 hours there were 3 people who had identical hands. Not that unique except for the fact that all 3 were four of a kind (4's) with a jack kicker. The individual odds of any quads are 224,848 to 1. How would I calculate the odds of three instances of this exact hand? How would I then add in the variable of the secondary card (jack)? Lastly there were only 50 people in the room, how would this affect it's rarity?
[General dice probabilities] How do I calculate the odds of rolling certain results with multiple D6 dice?
So on a D6 or six-sided die, I understand that there's a 3/6 chance of rolling a result of 4 or higher. At this point it's also my assumption that the chance of rolling a 4+ with 2 D6 is 3/4. But going a little more complex I can't figure it out in my head, how would I find the odds of rolling a 4+ (as in greater than or equal to 4) on at least2 D6 if I rolled more than 2 D6? Like 3D6, 4D6, etc. Thank you!
How do I calculate the total odds of an event occurring once, given the chance of this event occurring multiple times?
For example, assume I am female and that I am trying to get pregnant. I will try exactly once each month for one year (12 total times). Suppose further, that my odds of getting pregnant each time I try are 10% (So 10% chance of getting pregnant per month). After trying to get pregnant for one year, what is the overall chance of getting pregnant at least once, given these parameters? Is it still 10% no matter how many tries? Or does the probability or odds increase according to the number of tries? Why or why not? I'm looking for a formula/procedure and an explanation. Thanks!
Host or participate in Heists - Min. Level 1 / to participate; Min. Level 12 / or own a high-end apartment to host
I'm a millionaire already, just give me a grind:
CEO Import/Export Vehicle Work by Psychko - Currently most reliable and profitable money grind with high end only method - Min. Level 1 / if you already own a CEO office / if you do not own a CEO office
Do an I/E sourcing mission, then do VIP work (Headhunter or Sightseer are a breeze with a Buzzard) in the cool down, then do an I/E delivery mission. Rinse and repeat!
Earn 2000-3000RP per source delivery, 5000RP per sale delivery. Buy 1 crate and sell immediately for maximum RP since the same RP is given whether you source/sell 1 crate or multiple.
Earn 200-600RP bonuses when you stay near the CEO/VIP's location
Help with a probability/counting problem: you roll a fair 6-sided dice 8 times, what is the probability you obtain at least 2 odd numbers and at least 3 even numbers?
From this, I believe I have correctly calculated the probabilities of each criteria occurring independently. At least 2 odd numbers has probability 1 - 9(0.5)^8, and at least 3 even has probability of 1 - 37(0.5)^8. The above were calculated using binomial theory. I am now stuck with combining the probabilities. I don't think you can simply multiply them because they are not mutually exclusive. And I would use the probability addition rule to find P(A n B) (where A and B are the two required criteria), but then I do not know how to find P(A u B). The other way I thought about doing it is by finding the number of way to arrange 2 odd numbers and 3 even numbers in a set of 8 number. I thought this would be 32 C(8,2) * 33 C(6,3) * 63 where the first term is all the ways of arranging any two odd numbers (3 possible numbers, 1,3,5), the second term is the number of ways of arranging 3 even numbers in the remaining 6 positions, and the final term is the number of ways of arranging any number in the remaining 3 positions. But this cannot be correct as it yields a different value if you decide to assign the 3 even numbers before the 2 odds (i.e. change the combinatorics to C(8,3) and C(6,2))
Host or participate in Heists - Min. Level 1 / to participate; Min. Level 12 / or own a high-end apartment to host
I'm a millionaire already, just give me a grind:
CEO Import/Export Vehicle Work by Psychko - Currently most reliable and profitable money grind with high end only method - Min. Level 1 / if you already own a CEO office / if you do not own a CEO office
Do an I/E sourcing mission, then do VIP work (Headhunter or Sightseer are a breeze with a Buzzard) in the cool down, then do an I/E delivery mission. Rinse and repeat!
Earn 2000-3000RP per source delivery, 5000RP per sale delivery. Buy 1 crate and sell immediately for maximum RP since the same RP is given whether you source/sell 1 crate or multiple.
Earn 200-600RP bonuses when you stay near the CEO/VIP's location
Some bookmakers, however, now offer "Same Game Multis", allowing users to combine multiple markets from the same event. The odds for Same Game Multis are calculated differently and are different to the Multi Calculator on this website. Some bookmakers now also allow bettors to "cash out" before all legs of their multibet have resulted. Probability Theory: Definition, Misconceptions, and Importance - Guide Authored by Corin B. Arenas, published on September 24, 2019 . Ever thought about your chances of winning the lottery? The following steps will enable you to find the probability through our Experimental Probability Calculator. Click on the "Single" button for finding a single event probability or "Multiple" button if you want to find the probability of multiple events. Now, enter the number of possible outcomes in our Statistics Probability Calculator. When you input the odds, your stake, the desired market type and the outcome of the game, the Asian Handicap calculator will display the result, payout and profit of the bet. This amazing betting tool is extremely convenient and helps you avoid risky bets, as well as determine what is the most profitable option for yourself. This Bet Calculator allows bettors to calculate the potential Payout for any single bet and has a simple ‘Add Odds’ function to calculate the Payout for a multiple bet. It might be one of the most basic calculations in betting but using the Bet Calculator above will help bettors enhance their understanding of what betting odds represent and what it means for their bet. The calculator may also ask you to either enter the odds in the decimal or fraction format, but we will touch upon the key issues related to that further on in this article. You will then repeat that process however many times is necessary – for example, if you have placed multiple bets – and then may also be able to add information related to Dead Heats or any situation when Rule 4 may ... Howtobet4free’s easy to use Bet Calculator shows you the profit, returns and chance of winning on single, multiple, spread and accumulator bets. Simply input the odds to get started – and leave the mental maths in the classroom! Multi bets and our calculator. A multi bet combines a series of single bets into one wager.Our multi bet calculator allows you to calculate the return for your multi bet with any number of selections. You can choose to place an each-way multi bet by selecting each-way from the settings menu. Betting Calculator Before making any bet, it helps to know what you're risking for the expected payout. Enter Your 'Bet Amount' - that's what you're risking, along with the American, fractional or decimal odds. American Odds are the default odds at American sportsbooks. These odds are based on winning $100 for a given bet. Betting a Favorite: The odds for favorites will have a minus (-) sign, and represent the money you need to risk to win $100. So if you're betting on the Packers at -140 against the Vikings, that means Green Bay is a slight favorite.
Interpreting Odds Ratio with Two Independent Variables in ...
What is the probability of two events occurring together? First determine if the events and independant or dependant on eachother. Does replacement occur? ... This video will show you how to calculate a Linear Regression using the Casio fx-911ms. It uses an example to show you step by step. It will show you how t... How-To :: SingaporePools App - Multiple Bet Calculator probability of intersection of multiple events http://bit.ly/1zvFp4s Calculate your accumulator/parlays quickly and easily. Add and remove by simply selecting which odds to include. Ideal for horse racing... This video illustrates how to calculate the Probability of Multiple Events. For part 2 of this video, including examples 3, 4 and 5, as well as many more ins... This video demonstrates how to interpret the odds ratio (exponentiated beta) in a binary logistic regression using SPSS with two independent variables. A bin...
