# Yet another probability question **URL:** \ **Category:** Factual Questions\ **Created:** [February 4, 2008, 9:13pm UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271 "2008-02-04T21:13:14Z")\ **Posts on this page:** 9\ **Page:** 1
**Author:** ![Intelligently\_Designed](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/intelligently_designed/32/20957_2.png) [@Intelligently\_Designed](https://boards.straightdope.com/u/Intelligently_Designed)\ **Post date:** [February 4, 2008, 9:13pm UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271/1 "2008-02-04T21:13:14Z")
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](http://diffusion.loto-quebec.com/sw3/res/asp/index.asp?j=30&m=1&a=2008&pRequest=1&submit=s?pRequest=0&cProduit=0&l=1) , 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! ---
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [February 4, 2008, 9:28pm UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271/2 "2008-02-04T21:28:44Z")
Assuming that this isn’t a homework problem, the answer is (6!/2!) : (49!/45!), or 15:211876. High odds, but not astronomical. Stranger ---
**Author:** ![Intelligently\_Designed](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/intelligently_designed/32/20957_2.png) [@Intelligently\_Designed](https://boards.straightdope.com/u/Intelligently_Designed)\ **Post date:** [February 4, 2008, 9:39pm UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271/3 "2008-02-04T21:39:42Z")
No, **Stranger** , it’s not for a homework. Just out of curiosity, and trying to remember my own old math courses. Thanks. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [February 4, 2008, 11:15pm UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271/4 "2008-02-04T23:15:00Z")
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. ---
**Author:** ![Stranger\_On\_A\_Train](https://avatars.discourse-cdn.com/v4/letter/s/13edae/32.png) [@Stranger\_On\_A\_Train](https://boards.straightdope.com/u/Stranger_On_A_Train)\ **Post date:** [February 5, 2008, 2:11am UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271/5 "2008-02-05T02:11:08Z")
[QUOTE=Usram] 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. [/QUOTE] 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 ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [February 5, 2008, 6:48pm UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271/6 "2008-02-05T18:48:25Z")
[QUOTE=Stranger On A Train] 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. [/QUOTE] I don’t see how that’s different from a lottery player getting four numbers right. ---
**Author:** ![pmwgreen](https://avatars.discourse-cdn.com/v4/letter/p/edb3f5/32.png) [@pmwgreen](https://boards.straightdope.com/u/pmwgreen)\ **Post date:** [February 5, 2008, 7:47pm UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271/7 "2008-02-05T19:47:04Z")
[QUOTE=Usram] I don’t see how that’s different from a lottery player getting four numbers right. [/QUOTE] I’ve got to second that, although I calculate the answer as 1032.4 : 1. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [February 5, 2008, 8:20pm UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271/8 "2008-02-05T20:20:12Z")
1 _in_ 1032.4. That’s 1031.4 to 1 against. ---
**Author:** ![Ximenean](https://avatars.discourse-cdn.com/v4/letter/x/aca169/32.png) [@Ximenean](https://boards.straightdope.com/u/Ximenean)\ **Post date:** [February 5, 2008, 8:45pm UTC](https://boards.straightdope.com/t/yet-another-probability-question/436271/9 "2008-02-05T20:45:39Z")
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