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?
Each digit has 9/10 chance of being wrong. So (9/10)**4 or .6561. Therefore, about 66% chance.
5/9 x 4/8 x 3/7 x 2/6 = 120/3024 = 5/126 or 3.97%
pan
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
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%.
If it makes any difference, the order is not important.
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%
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