Last week in Quebec’s public lotteries, two 6/49-type drawings (49 balls, 6 are drawn + 1 complementary number) had 4 of the same numbers winning. Shown here , both Lotto 6-49 and Quebec 49 had the 1, 20, 21 and 40.
My simple question is: given these two lotteries, what is the probability, on any given week, of this happening?
Thanks for helping!
Assuming that this isn’t a homework problem, the answer is (6!/2!) : (49!/45!), or 15:211876. High odds, but not astronomical.
Stranger
No, Stranger, it’s not for a homework. Just out of curiosity, and trying to remember my own old math courses. Thanks.
Isn’t it the same as a punter guessing 4 of the 49 numbers correctly? A well established result of 1031.4:1 against. I don’t really follow Stranger’s reasoning, or arithmetic for that matter.
Nope, although on reflection my answer, which is getting four numbers out of a possible six options sequentially from a pool of forty-nine, is also wrong. The situation is actually picking a combination of four numbers that intersect a pool of six unique numbers (the original Lotto picks) in a possible six chances (the second Lotto picks) from a pool of forty-nine total unique options. I believe (although I’m doing this offhand and it’s been a while since I’ve fooled with combinatorics) that the correct odds are C[sup]4[/sup][sub]6[/sub] x C[sup]4[/sup][sub]6[/sub] : C[sup]6[/sup][sub]49[/sub] = (43!*6!*6!*6!) : (49!*4!*4!*2!*2!) = 75 : 4661272 ~ 0.00161%
Stranger
I don’t see how that’s different from a lottery player getting four numbers right.
I’ve got to second that, although I calculate the answer as 1032.4 : 1.
1 in 1032.4. That’s 1031.4 to 1 against.
Anyway, here’s the working:
Permutations of 6 numbers from 49 = 49 x 48 x 47 x 46 x 45 x 44 = 10,068,347,520
Since there’s no replacement of lottery balls, we can simplify things by counting combinations (unordered sets) rather than permutations, because all combinations are equally likely. Each combination can be ordered in 6! ways, so there are 10,068,347,520 / 6! = 13,983,816 possible sets of six numbers.
Take another set of 6 numbers, such as those from the other lottery. There are 6 x 5 x 4 x 3 / 4! = 15 ways that it can match the original set in 4 places, and for each of those ways there are 43 x 42 / 2! = 903 ways that the other 2 numbers can fail to match any of the original 6.
15 x 903 / 13,983,816 = 1/1032.4