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IGOR

My math is rusty. What's the probability of guessing one number and having it come up in the Florida lottery?

I found a PDF explaining how to calculate the odds of getting X matches out of 6, but that (naturally) assumes you're picking six numbers.
 
If 6 balls are picked out of 45 without replacement, then the odds of guessing one of them are somewhat pretty close to 8.5 : 1.

If he has to do four successful and consecutive trials as described, he'll be facing something around 5220 : 1 odds.
 
Hmmm..?

Protocol says six balls are drawn out of a pool of 53 balls.

Chance of getting one right is 1/53 + 1/52 + 1/51 + 1/50 + 1/49 + 1/48 = 0.1189...

Doing it on four separate attempts = (0.1189...) ^ 4 = 0.0002... = about 1/5,000

Doing it on seven separate attempts = (0.1132...) ^ 7 = 0.00000034... = about 1/3,000,000



Moose: You seem to have gotten the odds right, but with a different number of balls. Did you mean 53 or 45?
 
Oh, was it 53? Sorry, I got confused with the other lottery protocol. Igor's I think, where there's talk of playing powerball. Powerball's based on a pool of 45, isn't it?

Yeah, your numbers are right.

[Edit: Ah hell, the powerball protocol was an earlier proposal, I think. I'm so confused. I'll get myself sorted out in a bit. *chuckle*]
 
Timothy said:
.

Chance of getting one right is 1/53 + 1/52 + 1/51 + 1/50 + 1/49 + 1/48 = 0.1189...
Click to expand...
That series sums to greater than one. Which means if 53 balls were drawn, his probability of winning would be, according to this formula, greater than one. Which shows the equation is incorrect.

His chances of guessing 1 out of 6 in this case should be 6/53.
 
roger said:
That series sums to greater than one. Which means if 53 balls were drawn, his probability of winning would be, according to this formula, greater than one. Which shows the equation is incorrect.

His chances of guessing 1 out of 6 in this case should be 6/53.
Click to expand...

sorry, but nope....
the problem is getting 1 right out of 6 draws. first draw, there are 53 different balls. yours is 1 of them. 1/53.
if not, the odd go up now--in your favor. There are only 52 balls left, and yours is one of them. 1/52 for this draw
repeat for 51, 50, etc. 1/51, 1/50, etc. Add 'em up to get the probability for a single ball from 6 draws. (53+52+51+50+49+48)/(53*52*51*50*49*48)
You are right that the series sums to >1. IFF you draw 50 times, it is a certainty that you will get one of them right, with the odds increasing in your favor each draw.

This assumes that the drawn ball is removed from the pool, of course.

Roger
 
How can you be guaranteed to pick the number if only 50 are drawn? You may have chosen one of the 3 balls that weren't drawn.


Also, (53+52+51+50+49+48)/(53*52*51*50*49*48) is around 10^-8, a very small number. Worse, it gets smaller the more numbers that are drawn. It can't be right.


Change this to a more intuitive, yet equivalent domain.

There are 10 doors in front of you. 2 contain lions. Pick a door. What are the chances you get attacked by a tiger?

2/10, or 1/5. It doesn't matter what order the lions were put in the rooms.

Now, there are 53 doors, and 6 tigers. Pick a single door. What are the chances you get attacked by a tiger? 6/53. Again, the order the rooms were filled by tigers doesn't matter.

I argue this is exactly the same as the lotto drawing. 6 out of 53 balls are picked. You name a number between 1 and 53. 6/53 of those will be a hit, and 47/53 will be a miss.

I know its been awhile since I took statistics, but this seems right to me....
 
The 6/53 number is correct.

The chance of the first ball drawn NOT matching a single selected number is 52/53. Given that this happens, the chance that the next ball drawn does not match the number is 51/52. So on until the chance of six drawn numbers not matching the single chosen number is (52*51*50*49*48*47)/(53*52*51*50*49*48). Everything cancels here but 47/53. So there's a 47/53 chance of NOT matching the number and a 6/53 chance of matching it.

To look at it the positive way, you still need to work the conditional probabilities in. The chance of matching the first ball is 1/53. The chance of matching the second ball is not 1/52, but (52/53)*(1/52) - GIVEN that you missed the first one, it's 1/52, but you have to account for the probability that you missed the first one when calculating the total probability. Note that (52/53)*(1/52) = 1/53. So doing it this way, you're going to wind up adding 1/53 six times and again get 6/53.
 
Silly?

Despite the debate here between statisticians amateur and pro (and I know this is about as redundant as I can possibly be), if Igor can really do what he claims to be able to do...

WHY is he coming to us for a paltry million when he can just wait for the next $26M Florida Lotto and collect 26x more?

These lottery psychics are truly a breed unto themselves. It's a unique species of Silly.

I wish I was a statistician, too, so I could have some fun with this claim.
 
Re: Silly?

KRAMER said:
Despite the debate here between statisticians amateur and pro (and I know this is about as redundant as I can possibly be), if Igor can really do what he claims to be able to do...

WHY is he coming to us for a paltry million when he can just wait for the next $26M Florida Lotto and collect 26x more?

These lottery psychics are truly a breed unto themselves. It's a unique species of Silly.

I wish I was a statistician, too, so I could have some fun with this claim.
Click to expand...

If he can only get one number at a time, I guess the JREF million is the best he can hope for.;)
 
The 6/53 number is correct.
Click to expand...

No, it's not. Apparently posting a link to the thread where we already discussed it isn't clear enough....

6 balls are chosen and their values range from 1 to 53.
So, the probability of getting exactly one ball correct=
6/53 (the chance of getting the first ball to match)
*
47/52 (chance of not matching the next ball, since there are 52 possible values for your remaining 5 balls to match)
*
46/51 (51 possible numbers left, 4 balls left to match)
*
45/50
*
44/49
*
43/48

so, we have (6/53)*(47/52)*(46/51)*(45/50)*(44/49)*(43/48) = 0.06
BUT, there are six possible ways to do this, since the order in which it happens is irrelevant. Hence, multiply this by 6 and we get 0.40, or 40%
i.e. 2 out of 5 times you will get exactly one ball correct.


To get the probability of hitting the first two balls is much the same:

= (6/53)*(5/52)*(47/51)*(46/50)*(45/49)*(44/48) = 0.0078

So to get any two balls, we multiply by 15 (combinations of ordering two hits and four misses) to get 0.117, or about 12%

You can continue to work the rest out in much the same way.


I really didn't want to type all this out, but it drive me crazy when people post things like "I don't know what I'm talking about, but it looks right to me".
 
The applicant states they will select a SINGLE number, not six numbers. The probability of that single number coming up in one drawing is certainly not 40%. It's 6/53.

If the applicant picks 6 balls (like a regular lottery play), the probability of hitting none of them is:

6C0 * 47C6 / 53C6

where xCy is the number of combinations that exist when selecting y numbers out of a pool of x. That number is 46.8%, so the probability of hitting more than zero (at least 1) is 53.2%. The probability of hitting EXACTLY one is 40%, as you say. But the claim states that ONE number will be selected, and for this claim the 6/53 number is correct for one drawing.
 
No need to be rude, spacefluffer.


Note that the applicant is only specifying 1 number, not six. He's not picking 6 numbers, and trying to get 1 match. He's specifying 1 number. There's no way you can hit that 40% of the time.

edit: opps, I was beaten to it.
 
choosing one

IGOR is specifiying one number, but the lottery drawing will show six in total. IGOR has to get one of these six.
By taking the six all from one bin (whereby each number can come up only once), if 53 balls are in the starting mix, then as each individual ball pops out, the remaining balls would have differing mathematical computations as to the chances of any other number to appear next.

I really don't know what I'm talking about, but I think that if someone offered me a million dollars for this 'guessing game" then it wouldn't be too much bother.

Here I go for practice:

Maryland Lottery May 14th Drawing: 11:07PM local time
(1 hour, 17 minutes from now)


My psychic number for tonight:

36


(((( ETA: Jeff from Tasmania, Australia noted:
"The Applicant clearly does not understand anything about statistics or number theory." ))))))
 
missed it by THAT much

Number 35 came up. Missed by one.

I must have been experiencing some weird Vortex in my psychic powers, that caused a shift in the time-space continuum by a slight degree and caused me to guess- er- predict 36 instead of 35. Oh well.

Onwards to the testing of IGOR... I wish him all the best.
 
roger said:
Change this to a more intuitive, yet equivalent domain.

There are 10 doors in front of you. 2 contain lions. Pick a door. What are the chances you get attacked by a tiger?
Click to expand...

erm, zero?


assuming there are no tigers behind the other 8 doors, of course

:p
 
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