the probability of winning 2nd prize in the lottery 6/51 by correctly picking 5 of the 6 #s?

Question

6/51 lottery means correctly pick 6 #s from 1-51

Answer 1

To all the people saying it's 1 in 2.3 million, that's assuming they ONLY have 5 numbers, and get them all correct, but in fact you have 6 numbers, and only need 5 of them to be correct, so the number is wrong.

There are six ways for the six numbers to include (exactly) five of the six numbers. The other number must be one of the other 45 numbers (I don't know if there is a bonus ball here, I'm assuming not).

There are 51c6 total ways, i.e. 18,009,460 total combinations

So, there are 6x45 = 270 ways to get exactly 5 from the six you choose, so the probability is 270/18,009,460, i.e. 27/1800946 = 0.001499212081%

so, odds wise: 66701.7 : 1

Answer 2

the probabilty of picking 5 out of 6 numbers in a 51 number lottery is 6/51 * 5/50 *4/49 * 3/48 * 2/47. that gives you 5 correct. and is .0000025542132 which is .000255%. that is odds of 391,512 to 1. each number chosen eliminates one from the next pick so that is why id doesnt stay at divisor 51. by the way the odds of picking all six is one in 18,094,962!

Answer 3

As we don't care what order the numbers are picked, this is a combination problem. Therefore you are asking to pick 5 things out of 51. This is written as C(51,5) = 51!/46!5!; where 51! = 51*50*49*...*1 and 5! = 5*4*3*2*1.

Using Excel's COMBIN function C(51,5) ~ 2.3 million, you have a 1 in 2.3 million chance of winning second place.

FYI C(51,6) ~ 18 million, which means you have a 1 in 18 million chance of winning first place...the big bucks.

Answer 4

Simple maths
51 choose 5
1 in 2349060 chance of it happening
I hope this is right cause I dont fully understand the lottery in America soo....