# Easy, easy probability question. **URL:** \ **Category:** Factual Questions\ **Created:** [November 7, 2000, 2:04pm UTC](https://boards.straightdope.com/t/easy-easy-probability-question/40107 "2000-11-07T14:04:10Z")\ **Posts on this page:** 8\ **Page:** 1
**Author:** ![Earl\_Snake-Hips\_Tucker](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/earl_snake-hips_tucker/32/3269_2.png) [@Earl\_Snake-Hips\_Tucker](https://boards.straightdope.com/u/Earl_Snake-Hips_Tucker)\ **Post date:** [November 7, 2000, 2:04pm UTC](https://boards.straightdope.com/t/easy-easy-probability-question/40107/1 "2000-11-07T14:04:10Z")
Let’s say the numbers 0-9 (or 1-10, it doesn’t matter). I want to select 4 numbers for a grand lottery, each of which can be used only once. What are the odds that I won’t get a single number correct? ---
**Author:** ![Friday](https://avatars.discourse-cdn.com/v4/letter/f/dec6dc/32.png) [@Friday](https://boards.straightdope.com/u/Friday)\ **Post date:** [November 7, 2000, 2:17pm UTC](https://boards.straightdope.com/t/easy-easy-probability-question/40107/2 "2000-11-07T14:17:11Z")
Each digit has 9/10 chance of being wrong. So (9/10)\*\*4 or .6561. Therefore, about 66% chance. ---
**Author:** ![kabbes](https://avatars.discourse-cdn.com/v4/letter/k/e99b99/32.png) [@kabbes](https://boards.straightdope.com/u/kabbes)\ **Post date:** [November 7, 2000, 2:25pm UTC](https://boards.straightdope.com/t/easy-easy-probability-question/40107/3 "2000-11-07T14:25:46Z")
5/9 x 4/8 x 3/7 x 2/6 = 120/3024 = 5/126 or 3.97% pan ---
**Author:** ![kabbes](https://avatars.discourse-cdn.com/v4/letter/k/e99b99/32.png) [@kabbes](https://boards.straightdope.com/u/kabbes)\ **Post date:** [November 7, 2000, 2:27pm UTC](https://boards.straightdope.com/t/easy-easy-probability-question/40107/4 "2000-11-07T14:27:44Z")
> [@](#): > > \*Originally posted by Friday \* > \*\*Each digit has 9/10 chance of being wrong. So (9/10)\*\*4 or .6561. Therefore, about 66% chance. \*\* I may be misunderstanding the way the lottery works. This would imply that the first number picked out of the hat must correspond to your first number, the second to your second and so on. Further, that although you can’t pick the same number more than once, the lottery can. Is this the way the question was supposed to have been interpreted? pan ---
**Author:** ![Cabbage](https://avatars.discourse-cdn.com/v4/letter/c/f07891/32.png) [@Cabbage](https://boards.straightdope.com/u/Cabbage)\ **Post date:** [November 7, 2000, 2:28pm UTC](https://boards.straightdope.com/t/easy-easy-probability-question/40107/5 "2000-11-07T14:28:28Z")
There are 210 different ways of picking four numbers out of 10. So, say you pick your four numbers. To get them all wrong, there are only six numbers that can occur in the drawing. There are 15 different ways to choose four numbers from 6. So the probability of getting them all wrong is 15/210, or about 7%. ---
**Author:** ![Earl\_Snake-Hips\_Tucker](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/earl_snake-hips_tucker/32/3269_2.png) [@Earl\_Snake-Hips\_Tucker](https://boards.straightdope.com/u/Earl_Snake-Hips_Tucker)\ **Post date:** [November 7, 2000, 2:34pm UTC](https://boards.straightdope.com/t/easy-easy-probability-question/40107/6 "2000-11-07T14:34:30Z")
If it makes any difference, the order is not important. ---
**Author:** ![Friday](https://avatars.discourse-cdn.com/v4/letter/f/dec6dc/32.png) [@Friday](https://boards.straightdope.com/u/Friday)\ **Post date:** [November 7, 2000, 2:45pm UTC](https://boards.straightdope.com/t/easy-easy-probability-question/40107/7 "2000-11-07T14:45:10Z")
My mistake. Since order does not matter, then I believe that cabbage is right. But I need to think about it … Total number of all combinations: 10!/(6!\*4!) = 210 Total number of all wrong number: 6!/(4!\*2!) = 15 15/210 = 7.14% ---
**Author:** ![kabbes](https://avatars.discourse-cdn.com/v4/letter/k/e99b99/32.png) [@kabbes](https://boards.straightdope.com/u/kabbes)\ **Post date:** [November 7, 2000, 3:00pm UTC](https://boards.straightdope.com/t/easy-easy-probability-question/40107/8 "2000-11-07T15:00:49Z")
Doh! My sum should have been 6/10 x 5/9 x 4/8 x 3/7 = 360/5040 = 1/14 = 7.1%. Pesky 0-9 confusing me into thinking that there are only 9 numbers…mumble…grumble…couldvehadthecorrectanswerfirst… pan