View Full Version : Scratcher Odds.

Let me know if my math is wrong, but I think it's fairly simple to calculate the odds (assuming Funzio hasn't rigged it)

There are 6 items that can show up: (Car, armor, melee, Gun, $$, RP), so in each slot you have a 1/6 chance to get a certain item. To get the top items, you then need 6 of the same, which is simply 1/6 ^ 6. or 1 in 46,656. Since there are 6 top items, your chances are 6 in 46,656 (1 in 7,776) to get a top reward.

Getting 5 of the same items is 6 x's (1/6)^5 x's (5/6)^1, etc

Summary:

6-items - 1 in 7,776

5-items - 1 in 259

4-items - 1 in 52

3-items - 1 in 10

2-items - 1 in 2

Luke 3457

06-15-2012, 09:03 AM

Hmm really crappy odds... I'll do my one a day... If I have 20g left from other purchases then yeah I'll chance it. But the common items don't help me much of at all. I'm sticking with crates to get weapons.

kykboxr

06-15-2012, 09:18 AM

Murf, your odds are slightly off, at least on the lower end as everyone is guaranteed to win something as long as they play, and unless the odds have been amended to not reflect what you stated, then you *should* be right. Based on probability, the odds reflect:

n! / (r!*(n-r)!)

Fredo

06-15-2012, 11:26 AM

Here is one probability of which we can all be sure: there is no chance Funzio will actually tell us the odds of getting the prizes.

Dr BoneCrusher

06-15-2012, 11:29 AM

The other thing you can be shore about is the odds will never better then they are now

Fredo

06-15-2012, 11:33 AM

And that scratchers will never be cheaper :)

GazLegend

06-15-2012, 02:58 PM

Let me know if my math is wrong, but I think it's fairly simple to calculate the odds (assuming Funzio hasn't rigged it)

There are 6 items that can show up: (Car, armor, melee, Gun, $$, RP), so in each slot you have a 1/6 chance to get a certain item. To get the top items, you then need 6 of the same, which is simply 1/6 ^ 6. or 1 in 46,656. Since there are 6 top items, your chances are 6 in 46,656 (1 in 7,776) to get a top reward.

Getting 5 of the same items is 6 x's (1/6)^5 x's (5/6)^1, etc

Summary:

6-items - 1 in 7,776

5-items - 1 in 259

4-items - 1 in 52

3-items - 1 in 10

2-items - 1 in 2

I play a lot of poker and this is basically a deck of cards with 6 suits

Math is wrong.

Maximum number of combinations 46,6566

If you want 6 out of 6 odds are:

Gun/gun/gun/gun/gun/gun

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6 = 1/46,656

There are 6 different (suits) giving 6/46,656 or 1 in 7,776

If you want 5 out of 6

Gun/gun/gun/gun/gun/any not gun

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 5/6 = 5/46,656

There are 6 different suits 30/46,656 or 1 in 1,555

If you want 4 out of 6

Gun/gun/gun/gun/ not gun/not gun

1/6 * 1/6 * 1/6 * 1/6 * 5/6 * 5/6 = 25/46,656

6 suits = 150/46,656 or 1 in 311 or for you % fans 0.32%

3 out of 6:

1/62 or 1.6%

2 out of 6

1/12.4 or 8%

I think

andymac106

06-15-2012, 03:30 PM

I play a lot of poker and this is basically a deck of cards with 6 suits

Math is wrong.

Maximum number of combinations 46,6566

If you want 6 out of 6 odds are:

Gun/gun/gun/gun/gun/gun

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6 = 1/46,656

There are 6 different (suits) giving 6/46,656 or 1 in 7,776

If you want 5 out of 6

Gun/gun/gun/gun/gun/any not gun

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 5/6 = 5/46,656

There are 6 different suits 30/46,656 or 1 in 1,555

If you want 4 out of 6

Gun/gun/gun/gun/ not gun/not gun

1/6 * 1/6 * 1/6 * 1/6 * 5/6 * 5/6 = 25/46,656

6 suits = 150/46,656 or 1 in 311 or for you % fans 0.32%

3 out of 6:

1/62 or 1.6%

2 out of 6

1/12.4 or 8%

I think

Im going to agree with you due to the 1/1555 fact for 5 not 1/259. If 259 were true the odds of winning would be so much better and pretty much everyone would be spending their gold on scrathers

I play a lot of poker and this is basically a deck of cards with 6 suits

Math is wrong.

Maximum number of combinations 46,6566

If you want 6 out of 6 odds are:

Gun/gun/gun/gun/gun/gun

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6 = 1/46,656

There are 6 different (suits) giving 6/46,656 or 1 in 7,776

If you want 5 out of 6

Gun/gun/gun/gun/gun/any not gun

1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 5/6 = 5/46,656

There are 6 different suits 30/46,656 or 1 in 1,555

If you want 4 out of 6

Gun/gun/gun/gun/ not gun/not gun

1/6 * 1/6 * 1/6 * 1/6 * 5/6 * 5/6 = 25/46,656

6 suits = 150/46,656 or 1 in 311 or for you % fans 0.32%

3 out of 6:

1/62 or 1.6%

2 out of 6

1/12.4 or 8%

I think

I may be wrong, but I don't think this is right...The odds only add up to about 9%

enahs1

06-15-2012, 03:54 PM

I think that is right, I believe to get the grand prize (6 matching) it would be 1/6^6 which equals a .002143347 percent chance.

G Wiz

06-15-2012, 03:55 PM

where is nicholost

Does anybody know how to get the scratcher back up? I downloaded the update, which launched the scratcher...and nerfed the event. Then the event fixed itself, but I don't see where you get the scratcher back...maybe I am missing something.

Anyway, first free scratch was trip respect for 1,000 respect...i'll take it.

enahs1

06-15-2012, 05:01 PM

Under the special tab

Nice thanks for the feedback.

I think that is right, I believe to get the grand prize (6 matching) it would be 1/6^6 which equals a .002143347 percent chance.

Yeah, that's the easy one...I think 2 out of 6 is the one we need to solve...kykboxr's formula looks right (I think it's the binomial theorem) but I forget how to apply it correctly.

dee17

06-15-2012, 11:38 PM

Easiest way is to use a spreadsheet to list all 46656 combinations then count the results:

1 item the same: 720/46656 = 1.54% or 1 in 64.8

2 items the same: 28800/46656 = 61.7% or 1 in 1.62

3 items the same: 14700/46656 = 31.5% or 1 in 3.17

4 items the same: 2250/46656 = 4.82% or 1 in 20.7

5 items the same: 180/46656 = 0.3858% or 1 in 259.2

6 items the same: 6/46656 = 0.0129% or 1 in 7776

If you're gonna do it manually, there's no simple formula for it. In fact, manually calculating for 2x, 3x, 4x items is a major pain even for a maths nerd like myself.

PawnXIIX

06-15-2012, 11:52 PM

Easiest way is to use a spreadsheet to list all 46656 combinations then count the results:

1 item the same: 720/46656 = 1.54% or 1 in 64.8

2 items the same: 28800/46656 = 61.7% or 1 in 1.62

3 items the same: 14700/46656 = 31.5% or 1 in 3.17

4 items the same: 2250/46656 = 4.82% or 1 in 20.7

5 items the same: 180/46656 = 0.3858% or 1 in 259.2

6 items the same: 6/46656 = 0.0129% or 1 in 7776

If you're gonna do it manually, there's no simple formula for it. In fact, manually calculating for 2x, 3x, 4x items is a major pain even for a maths nerd like myself.

*Raises hand*

Wouldn't the odds of getting 1/6 be 100%?

Unless Funzio is playing dirty xD Scratch off the card and get 6 trollfaces *facepalm*

dee17

06-16-2012, 12:42 AM

*Raises hand*

Wouldn't the odds of getting 1/6 be 100%?

Unless Funzio is playing dirty xD Scratch off the card and get 6 trollfaces *facepalm*

Getting AT LEAST 1/6 would be 100%. But getting EXACTLY 1x item max is not that easy since you need to get 6 different items.

This is probably more useful:

Get 1/6 or better: 100%

Get 2/6 or better: 98.46%

Get 3/6 or better: 36.73%

Get 4/6 or better: 5.22%

Get 5/6 or better: 0.3987%

Get 6/6: 0.0129%

Also check out Dreno33's thread listing his results; there's barely any 1x prizes; most are 2x/3x ones, which corroborate the odds here.

P.S. Don't mean to brag, but I just got the L4 vehicle prize (140/85!!) from my 2nd scratcher (both free). :cool:

Let me know if my math is wrong, but I think it's fairly simple to calculate the odds (assuming Funzio hasn't rigged it)

There are 6 items that can show up: (Car, armor, melee, Gun, $$, RP), so in each slot you have a 1/6 chance to get a certain item. To get the top items, you then need 6 of the same, which is simply 1/6 ^ 6. or 1 in 46,656. Since there are 6 top items, your chances are 6 in 46,656 (1 in 7,776) to get a top reward.

Getting 5 of the same items is 6 x's (1/6)^5 x's (5/6)^1, etc

Summary:

6-items - 1 in 7,776

5-items - 1 in 259

4-items - 1 in 52

3-items - 1 in 10

2-items - 1 in 2

Thanks for posting the odds up !

Save your money people unless your happy piss.sing it up the wall with those odds hahaa

Unbelievable, I got a better chance of winning a tenner on the lottery.

PawnXIIX

06-16-2012, 01:11 AM

Getting AT LEAST 1/6 would be 100%. But getting EXACTLY 1x item max is not that easy since you need to get 6 different items.

Now I get it :P

They should have made the reward for 1/6 more. Thinking of the odds those items are extremely difficult to obtain in comparison to the others.

McLovin

06-16-2012, 01:13 AM

dee17 is right. The total odds of all outcomes need to add up to 100% because every scratcher wins some kind of prize.

PawnXIIX - The odds of getting 'something' are 100% - which can be a prize for 1/6, 2/6...up to 6/6. As dee17 showed, there is only a 1.54% chance of getting a 1/6 prize because there are 720 possible combinations of winning a prize for 1/6 out of 46656 total possible combinations - basically a scratcher with 6 different items on it can be arranged 6 factorial=6!=6x5x4x3x2x1=720 different ways.

The odds of winning a specific prize eg. 2/6 vehicle would be 61.7/6=10.28% because there are six different possible 2/6 prizes.

So:

Odds of specific prizes types (vehicle, gun, etc):

Each 1/6 = 0.26%

Each 2/6 = 10.28%

Each 3/6 = 5.25%

Each 4/6 = 0.80%

Each 5/6 = 0.06%

Each 6/6 = 0.002%

You can also work out how often you should expect to get each prize level ie. how many days it should take to get each type of prize if you only use the once per day free scratchers, or indeed how many scratcher you should buy to expect to get each prize type:

For each Prize type:

1/6 Prize = Once every 64.94 days (or scratchers)

2/6 Prize = Once every 1.62 days (or scratchers)

3/6 Prize = Once every 3.17 days (or scratchers)

4/6 Prize = Once every 20.74 days (or scratchers)

5/6 Prize = Once every 259.20 days (or scratchers)

6/6 Prize = Once every 7751.94 days (or scratchers)

Also, to work out the chances of getting that sweet 3/6 level vehicle that you just multiply the expected number of days/scratchers by 6 because the chance of getting a vehicle is 1/6th the chance of getting any 3/6 level prize

ie. 3.17 * 6 = 19.04 days/scratchers

Good luck to anyone planning on playing for 20+ years for the 6/6 prizes, lol. Although this should mean that the 6/6 items remain rare enough for Funzio to maintain the 'balance' that they so covet. Unless you're a beta tester of course...

PawnXIIX

06-16-2012, 01:17 AM

dee17 is right. The total odds of all outcomes need to add up to 100% because every scratcher wins some kind of prize.

PawnXIIX - The odds of getting 'something' are 100% - which can be a prize for 1/6, 2/6...up to 6/6. As dee17 showed, there is only a 1.54% chance of getting a 1/6 prize because there are 720 possible combinations of winning a prize for 1/6 out of 46656 total possible combinations - basically a scratcher with 6 different items on it can be arranged 6 factorial=6!=6x5x4x3x2x1=720 different ways.

The odds of winning a specific prize eg. 2/6 vehicle would be 61.7/6=10.28% because there are six different possible 2/6 prizes.

Now I get it :P

They should have made the reward for 1/6 more. Thinking of the odds those items are extremely difficult to obtain in comparison to the others.

Yeah I know :P Don't worry xD

So to summize Funzio are advertising odds that dont tally with that table?

Shock horror

McLovin

06-16-2012, 01:24 AM

In case anyone is interested in how much you should expect to spend in Gold to get a 6/6 prize.

7752 scratchers would cost 7752 x 20 = 155 040 gold or 103.36 vaults = a bit over $10k RM.

Although you would get a lot of other items with the other 7751 scratchers.

Inzaghi

06-16-2012, 01:27 AM

In case anyone is interested in how much you should expect to spend in Gold to get a 6/6 prize.

7752 scratchers would cost 7752 x 20 = 155 040 gold or 103.36 vaults = a bit over $10k RM.

Although you would get a lot of other items with the other 7751 scratchers.

It's a pity if you spend 10K RM and get $100,000,000 :p

.Hold on a minute mclovin

Those odds only represent what you would need to spend to "guarantee" a 6/6 prize. That's not to say you don't get A 6/6 prize on the second try.

If you think about using your calculus a 7000 / 1 shot for a 10k rm prize isn't that bad after all.

dee17

06-16-2012, 02:02 AM

.Hold on a minute mclovin

Those odds only represent what you would need to spend to "guarantee" a 6/6 prize. That's not to say you don't get A 6/6 prize on the second try.

If you think about using your calculus a 7000 / 1 shot for a 10k rm prize isn't that bad after all.

Lol "guarantee" is definitely the wrong word to use in this case.

You scratch 7752 times and there's a (7751/7752)^7752 = 37% chance that you won't get a single grand prize.

Heck, do it 15000 times and there's 15% chance you still won't.

Probablities can be cruel like that. :)

Dipstik

06-16-2012, 08:30 AM

Good luck to anyone planning on playing for 20+ years for the 6/6 prizes, lol. Although this should mean that the 6/6 items remain rare enough for Funzio to maintain the 'balance' that they so covet. Unless you're a beta tester of course...

Yeah, I've noticed that all the beta testers who opened hundreds of these things still have the items. I suppose we should be grateful that the scratchers were tested so thoroughly.

WinkadInk

06-16-2012, 08:50 AM

You guys are all wrong on your math! Here is my math

6= hahaha no chance in heck, thx for wasting your money!

5= spend a few grand and we will think about it

4= 1 in 500,000

3= 1 in 499,999

2= everyone who tries, see guys this is how casinos get people who gamble hooked! Since everyone is fed up with our other events, we can make this the new event that pays out less but people will think they get more lol

Gi0rgi0s

06-16-2012, 10:21 AM

I give everyone here a lot of respect for trying to calculate the odds. I'm new to this game, so I'm not sure exactly how this event works.

But If you are trying to calculate the odds of getting any one specific item to match, you must delve into combinatorics.

GazLegend your technique is close but you disregarded combinations.

so if you are looking for the probability of say 4 matches

1/6 * 1/6 * 1/6 * 1/6 * 5/6 * 5/6 isn't enough.

position plays a role here, so you need to consider also

5/6 * 5/6 * 1/6 * 1/6 * 1/6 * 1/6 +

5/6 * 1/6 * 5/6 * 1/6 * 1/6 * 1/6 +

1/6 * 5/6 * 1/6 * 5/6 * 1/6 * 1/6 + etc.....

so you have to take the probability (5/6 * 5/6 * 1/6...) * the number of sequences (or combinations)

luckily there's a formula for this called "from n choose r" or nCr which kykboxr touched on.

In this case n = 6, and r changes...

so the formulas for probability are:

6: (1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 1/6) * 6 choose 6 = 0.0021%

5: (1/6 * 1/6 * 1/6 * 1/6 * 1/6 * 5/6) * 6 choose 5 = 0.0643%

4: (1/6 * 1/6 * 1/6 * 1/6 * 5/6 * 5/6) * 6 choose 4 = 0.8037%

3: (1/6 * 1/6 * 1/6 * 5/6 * 5/6 * 5/6) * 6 choose 3 = 5.3584%

2: (1/6 * 1/6 * 5/6 * 5/6 * 5/6 * 5/6) * 6 choose 2 = 20.094%

1: (1/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6) * 6 choose 1 = 40.188%

0: (5/6 * 5/6 * 5/6 * 5/6 * 5/6 * 5/6) * 6 choose 0 = 33.490%

Now, if you are looking for ANY item to match, the math is slightly different, let me know if you are interested. Hopes this helps :)

Ramshutu

06-16-2012, 01:03 PM

Everyone here is right. Ish. but the stats do not depend on combinatorics, or straight odds, but how the underlying system works.

From n Choose x works only in situations where there is a memory, like in cards or a lottery where the chosen item is "taken out" of the pool, so picking a heart from the deck is 13/52 but picking a second heart from the deck is 12/51.

If it is a straight dice role, in that rolling a 6 has odds 1/6 regardless of whether you have previously rolled 0 or 99 sixes.

In reality, the latter is marginally easier to implement, but both are trivial. The only way of proving which is which is to note down your own experiences and see whether they match 1 or 2.

This is, of course, assuming that the odds are not stacked as with events; which will throw everything off and make it impossible to determine the stats other than simply building up a large data set and comparing odds.

Gi0rgi0s

06-16-2012, 02:20 PM

Everyone here is right. Ish. but the stats do not depend on combinatorics, or straight odds, but how the underlying system works.

From n Choose x works only in situations where there is a memory, like in cards or a lottery where the chosen item is "taken out" of the pool, so picking a heart from the deck is 13/52 but picking a second heart from the deck is 12/51.

If it is a straight dice role, in that rolling a 6 has odds 1/6 regardless of whether you have previously rolled 0 or 99 sixes.

In reality, the latter is marginally easier to implement, but both are trivial. The only way of proving which is which is to note down your own experiences and see whether they match 1 or 2.

This is, of course, assuming that the odds are not stacked as with events; which will throw everything off and make it impossible to determine the stats other than simply building up a large data set and comparing odds.

You are wrong in saying that n choose x only works in situations where there is memory. I am assuming each event is an independent (or is a memoryless) Bernoulli trial with a binomial distribution. I would suggest looking up the binomial distribution on wiki.

According to what you are saying, the probability of rolling a 6 and a 5 with dice is equal to that of rolling a 6 and a 6 (1/6 * 1/6), which it is not. That's because position is important, and that's why you need to use nCr. In the former case your odds are (1/6) chance of rolling a 5 and (1/6) chance of rolling a 6 with 2 choices of position for 1 die, which yields 1/6 * 1/6 * 2 choose 1 = 2 / 36.

Assuming that each "roll" is dependent on (or remembers) the last, the solution becomes obfuscated. OP made an assumption that each event is independent so that's the question we are all trying to answer.

Luciferianism

06-16-2012, 02:30 PM

Lmao, whenever a maths problem crops up everyone turns into a math professor and comes up with different conclusions from each other.

Ramshutu

06-16-2012, 02:56 PM

You are wrong in saying that n choose x only works in situations where there is memory. I am assuming each event is an independent (or is a memoryless) Bernoulli trial with a binomial distribution. I would suggest looking up the binomial distribution on wiki.

According to what you are saying, the probability of rolling a 6 and a 5 with dice is equal to that of rolling a 6 and a 6 (1/6 * 1/6), which it is not. That's because position is important, and that's why you need to use nCr. In the former case your odds are (1/6) chance of rolling a 5 and (1/6) chance of rolling a 6 with 2 choices of position for 1 die, which yields 1/6 * 1/6 * 2 choose 1 = 2 / 36.

Assuming that each "roll" is dependent on (or remembers) the last, the solution becomes obfuscated. OP made an assumption that each event is independent so that's the question we are all trying to answer.

Bah, I was thinking of the nPr ... the other one. My bad.

Eyelusion

06-16-2012, 03:00 PM

...an independent (or is a memoryless) Bernoulli trial with a binomial distribution.

I was just going to say that... :rolleyes:

Nudie

06-16-2012, 03:02 PM

Odds are very good that with the scratchers you get something. Odds of getting the best items - not so good.:p

Amber_

06-16-2012, 04:02 PM

Haha you guys are incredible! I guess some people really DO like math.. It's interesting to read all your theories. Anyone can understand that the odds are small as hell though. It's like, a better chance to actually win 1mil at a real lottery than to win some crappy imaginary car in this one :S

PawnXIIX

06-16-2012, 07:57 PM

According to what you are saying, the probability of rolling a 6 and a 5 with dice is equal to that of rolling a 6 and a 6 (1/6 * 1/6), which it is not. That's because position is important, and that's why you need to use nCr. In the former case your odds are (1/6) chance of rolling a 5 and (1/6) chance of rolling a 6 with 2 choices of position for 1 die, which yields 1/6 * 1/6 * 2 choose 1 = 2 / 36.

I reach the same conclusion but the way I remember being taught this particular example was that when there are 2 separate numbers, take for example a 5 and a 6 as you said, that the formula looks like this:

(2/6) * (1/6) = (2/36)

Because the first die can be either a 5 or a 6, and then the next die must be the complementary pair to make the set. It's the same thing as (1/6) * (1/6) * 2 but I just learned it a different way and I want to feel smart because I still remember this :)

McLovin

06-16-2012, 09:47 PM

Ramshutu has a good point, there is always an element that Funzio can control so that these scratchers aren't truly 'random' and a good example is the box events where the chances of getting an 'item' with a successfully opened box decrease the more 'items' you already have (as stated by CCMark, in the original post for the first event I think).

QPR, I didn't mean to imply that you are 'guaranteed' to get any prize because that is not true, it's just the expected number of scratchers that it would take to get one of that level, it is the way any similar gambling game or lottery works. We all know that you can buy a lottery ticket and hit the jackpot on your first try but the odds that you won't and in fact will die or stop buying tickets before you ever do far outweigh that which is how they make money.

I guess unless we hear anything from the devs the true odds will remain a mystery, as is likely the intention.

ShawnBB

06-16-2012, 10:44 PM

Late for the math and statistics party...... I'm so sorry about it.

Binomial distribution and Hypergeometric distribution.

As Ramshutu mentioned about whether or not there is a replacement after each scratch.(good point!)

B distribution is applied on a fixed probability of success which means it is with replacement.

H distribution is applied on situation without replacement.

We don't know the actual mechanism yet, but one thing we are sure is that 6 same item is extremely low in B d and even lower in H d.

------An important thing about probability calculating.(using B distribution here)

6 identical,5 identical,4 identical can be perfectly fit into B distribution formula. (multiply by 6 for the final probability) Because the success and failure is strictly 1:5.

But,the chance of hitting 3 identical and 2 identical are a lot more higher, since the rest 3 attempts and 4 attempts are not defined as failure. They can hit 3 identical and 2 identical as well to make it success.

We can't apply normal B distribution fomula here! A lot of folks get trapped at this point actually.

And meanwhile the chance of hitting 6 different is very low(5!/6^5) , funzio actually stupid enough by putting such poor prize on hitting 6 diverse icon,and there should be only one prize for hitting that because it only has one way representing hit which is 6 different! LOL

Inzaghi

06-16-2012, 10:51 PM

Late for the math and statistics party...... I'm so sorry about it.

And meanwhile the chance of hitting 6 different is very low(5!/6^5) , funzio actually stupid enough by putting such poor prize on hitting 6 diverse icon,and there should be only one prize for hitting that because it only has one way representing hit which is 6 different! LOL

Right! I also found that when I calculated the probability. Just imagine A K Q J 10 > A A K Q J

PawnXIIX

06-17-2012, 12:48 AM

Right! I also found that when I calculated the probability. Just imagine A K Q J 10 > A A K Q J

I'm implying you're trying to make a case that the odds of five unique cards are better than the odds of 3 unique and and pair, and is also better in terms of the winning. There's 311,875,200 ways to make 5 unique cards, and I'm going to assume order doesn't matter in this case, so therefore the cards can go A K Q J 10 or 10 K J Q A. So the odds are:

100 * ((20 * 16 * 12 * 8 * 4) / 311,875,200) = .039%

The odds of a pair and 3 unique cards are:

100 * ((52 * 3 * 48 * 44 * 40) / 311,875,200) = 4.226%

I wrote all this out and I forget my point. But the math was fun so. Yeah.

ShawnBB

06-17-2012, 06:55 AM

It is well known that the odds in poker is sorted like this (descending)

High card

One pair

Two pairs

Three of a kind

Straight

Flush

Full house

Four of a kind

STFL

They judge the hand value according to the probability,so it's pretty fair.

---The chance of hitting a pair in 5 out of 13 without replacement is way lower than the chance of hitting a pair in 6 out of 6 with replacement,

Inzaghi, I would say they are two slightly different cases.:)

Wally64

02-13-2013, 11:46 AM

wewt, played this game for like a week, camping at low level (8-9)...

then i got a lucky scratcher, 5x$... got me 10,000,000...

so got weapons, armor, building loft atm etc, pretty crazy. now i read the chances are:

"5-items - 1 in 259" :D

i looked up on how much this would cost you to buy in the game: like 700$!

(add me: 213-017-768)

cditti

02-13-2013, 12:08 PM

Build that loft and soon as it is done buy another and upgrade the first asap

Wally64

02-13-2013, 05:39 PM

Build that loft and soon as it is done buy another and upgrade the first asap

yea i will do this right away, only it takes 100 hours to build a loft...

i dont even want to know how long it takes to upgrade :P

by the way i spent my money on: 20x "Steel Garrote", 20x "Blast Guard Helmet" and an Audi of 900,000 LOL (rest is in the bank). i reckon that should keep me safe for a while, but still i get attacks, ur never safe!

Dipstik

02-13-2013, 07:31 PM

Smells like noob in here. I'm curious, Wally... I can understand the excitement of winning $10 million in game. I mean, when I won that I wanted to put my ipad through a wall because cash is utterly worthless, but I'm sure you're new enough not to know that. That said, what on earth made you think "Gosh, I think I'll go make a forum account just to tell people... but I better bump up an 8 month old thread to do it"?

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