September 19th, 2014, 01:28 PM
When buying a lottery ticket* you can chose your own numbers or take a lucky dip and have the machine give you random numbers.
My question; Is there a difference in the probability of the two methods? i.e. is there more chance of a win by selecting numbers or using random ones?
I have read up on probability but I don't quite get it enough to work this out myself.
*I only buy a ticket once in a blue moon - it's a cheap way to have a nice daydream !
OB
Simplify the problem. Assume that the lottery was just one number between 1 and 6. What would the probabilities of winning be in that case?
agreed. Probability does not change unless the game is rigged.Originally Posted by MagiMaster
The odds of last weeks powerball lottery had just as much odds of coming up as 1, 2, 3, 4, 5, 6.
flip a quarter a million times & you have the odds of heads coming up every single time as you do any other random combination.
Here in the UK ( and it still is the UK ) the national lottery tells us that no matter how your numbers are selected, your chances of winning the jackpot is 14 million to one. Very long odds indead, but a winning ticket always defies the odds, so go for it dude.
Surely if you choose the exact combination of numbers that have already previously won then you are even less likely to win than if you picka combination of numbers that haven't ever come up, would this not be changing the probability by not only having to factor in the chance of a particular combination of numbers coming out but also the probability for that same combination coming out twice, which would surely be far less likely.
Incorrect. Previous actions do not effect future actions when it comes to random probability.
As others have stated, simplify it. Throw a dice or flip a coin. You have just as many odds of getting a string of all 4's or a string of all tails as any other combination.
The odds of the same number coming around again are the same odds as any other combination of numbers coming up. This applies to nearly all gambling games including craps, roulette, poker, slots, lottery, scrach off tickets, ect. Your odds do not differ based on previous actions.
I think I understand that, it seems easier with like single numbers for say a dice. The concept did seem harder to grasp for combinations of numbers though. The idea that the same combination of lottery numbers coming up again and again but at the same probability of any other combination does just somehow seem counter intuitive, yet it also does make sense that they shouldn't be affected by previous actions.
No, that would be the gamblers fallacy; Gambler's fallacy - Wikipedia, the free encyclopedia
Different trials are independent.
Edit; Oop, ninja'd!
Right. Once you grasp how it all works with a single die, you can move on to, say, rolling 3 dice.
There's a slight kink in that most lotteries are more like drawing 3 cards than rolling 3 dice, but that makes for a good next step. (Just remember to shuffle the cards between draws.)
Consider this, your odds of actually winning the lottery regardless of what number you played are exactly the same as last weeks numbers coming up again next week & its the same odds as those numbers being 1,2,3,4,5,6(or however many numbers are played)
Which is why its not even worth a buck to me.
Why? Assuming the cards are sufficiently randomized the first time around, shuffling them again will in no way affect the probability outcome.
Lets pretend you are attempting to draw 4 aces in a row. after the first draw (which for humor sake is an ace) leaves you with 51 cards in the deck, shuffling or not shuffling will still leave you with a 1:51 odds of drawing another ace. The ace might be the next card or shuffling the deck might being an ace to the top, it doesn't matter as your odds have not changed.
1st Ace 1:13Why? Assuming the cards are sufficiently randomized the first time around, shuffling them again will in no way affect the probability outcome.
Lets pretend you are attempting to draw 4 aces in a row. after the first draw (which for humor sake is an ace) leaves you with 51 cards in the deck, shuffling or not shuffling will still leave you with a 1:51 odds of drawing another ace. The ace might be the next card or shuffling the deck might being an ace to the top, it doesn't matter as your odds have not changed.
2nd Ace 1:17
3rd Ace 1:25
4th Ace 1:49
Total chance of drawing four aces in a row on a single trial of four draws is I think 1:270725
However, the point of reshuffling the deck is if the drawn card has been replaced so that single trials remain independent. When rolling multiple dice each die is independent. Getting a number on one die doesn't exhaust that number for additional dice.
If you wanted to calculate the probability of drawing a particular poker hand that would be somewhat complicated involving a bit of combinatorics.
Your right, my bad. Its late & forgot there are 4 aces in a deck. Was for some reason assuming there was just 1 ace in the deck (hints the 1:52 odds) XD The simple crap escapes me.
I disagree about the odds. Drawing 4 in a row if that is what you are shooting for is mathematically higher odds, however each card draw will be independent from each other. Thus, even with replacing the card in the deck. you will still have a 1:13 odds of drawing an ace each go around. My point I was attempting to make is shuffling the cards after each draw will not change the outcome of the probabilities (on the whiteboard)Total chance of drawing four aces in a row on a single trial of four draws is I think 1:270725
However, the point of reshuffling the deck is if the drawn card has been replaced so that single trials remain independent. When rolling multiple dice each die is independent. Getting a number on one die doesn't exhaust that number for additional dice.
If you wanted to calculate the probability of drawing a particular poker hand that would be somewhat complicated involving a bit of combinatorics.
You are correct, though I am speaking on a thought experiment more than a practical one. Spherical cows mate
Still think 1:270725 is way too high, I have gotten 4 of a kind quite a few times in games & fairly positive I have not played 270k hands in my lifetime.
13*17*25*49=270725. That spread is based on a single trial of four cards drawn without replacement. If we draw four cards independently with replacement we would have odds of four Aces at 13^4. Also there are 13 ways to get any four of a kind, and if five cards are drawn in a trial then the odds of any particular permutation of four cards goes up.
Just a side note of possible interest:
There are 80,658,175,170,943,878,571,660,636,856,403,766,975 ,289,505,440,883,277,824,000,000,000,000 different ways a pack of cards can be sorted.
That be a mighty big number.
While it is true that the odds of the same number coming around again are the same odds as any other combination of numbers coming up, there most certainly is a bias in the various logic above due to the predictable humans involved.
When humans pick numbers, they are almost by definition, pick in a predictable manner. They will pick birthdays, sequences (eg, 1,3,5,7 or 12,12,12,12 or etc), famous dates (1,7,7,6), etc. If a somewhat predictable person, select the numbers to play, the person will have a higher chance (than with computer randomized) of selecting numbers that someone else has predicted, and multiple winners will dilute your winnings.
When the machine picks random numbers, the sequence is more likely to be truly random, and so, the winnings are less diluted.
A similar effect occurs when the jackpot has grown to absurdly huge amounts. More people — including people who "never play the lottery" — will play, and thus, dilute the winnings. The best day to play the lottery is the day after the absurdly huge jackpot is won, and many fewer people care to play for a variety of reasons: they spent all their pocket money on the big jackpot, they're despondent over not winning the huge jackpot, the small jackpot doesn't look worth playing, etc.
So, you get the biggest bang for your buck if you play the day after the huge jackpot is won and allow the machine to pick t he numbers.
52!=8*10^67
It is in fact such a big number that if you thoroughly shuffle a deck of cards that deck is almost certainly now in an order that no other deck of cards has every been in.Just a side note of possible interest:
There are 80,658,175,170,943,878,571,660,636,856,403,766,975 ,289,505,440,883,277,824,000,000,000,000 different ways a pack of cards can be sorted.
That be a mighty big number.
If your goal is drawing an ace, there's a 1:13 chance on the first draw. If you place the drawn card on the bottom of the deck afterwards, your chances of drawing an ace on the second draw are no longer 1:13 since you will necessarily be drawing from the remaining 53 cards (you know one card you definitely won't be drawing), which might have 3 or 4 aces in it depending on your first draw. Shuffling the cards removes this dependency.
Edit: If you could return the drawn card to a truly random position in the deck after each draw, then reshuffling the entire deck probably wouldn't be necessary, but humans are terrible random number generators and the usual approach of somewhere-near-the-middle isn't very random.
Biggest bang for your buck will be spending the money on damn near anything else besides a lotto ticket
You got that right ! But like I say, once in a while it's a cheap daydream - thinking what you would spend the money on.Biggest bang for your buck will be spending the money on damn near anything else besides a lotto ticket
Thanks for the replies to my question.
OB
I believe the odds of buying the winning ticket is about the same as finding the winning ticket in the gutter. So I spend a lot of time looking in gutters and save the cost of a ticket.You got that right ! But like I say, once in a while it's a cheap daydream - thinking what you would spend the money on.Biggest bang for your buck will be spending the money on damn near anything else besides a lotto ticket
Thanks for the replies to my question.
OB
Refresh my math memory. "52!" is 52 x 51 x 50 x 49.......all the way to one...right?
Yep that's right Mac.
If you wanted you could forego calculating the one, as it is the product identity element.Refresh my math memory. "52!" is 52 x 51 x 50 x 49.......all the way to one...right?
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