I just bought 10 powerball tickets with all numbers picked by the machine.

With the 10 tickets the powerball number was 03 three times, and 13 twice.

There are 35 possible powerball numbers.

So what are the odds of having one double and one triple out of ten numbers? I know that these odds are vastly less than picking three 03’s and two 13’s. I think that this is the same as the birthday problem.

Note that having three out of three numbers the same is a whopping 1 in (35*35) chance, which is 0.081%. Does that seem very unlikely? With ten numbers instead of three, this rises to almost 10%. Having a pair on the side decreases the chance, but you specified that after you got your targets.

Playing Bridge suppose your clubs are 7-6-5-4-3-2 (six in a row). You might exclaim “What are the odds of that!?!” Another player comments “I’ve K-J-8-7-5-2 of hearts. What are the odds of that?” The odds are the same. :cool:

Experiments show that when people are asked to invent random numbers themselves, they fail, producing fewer pairs and triplets as would be found among real random numbers.

Are quick pick numbers selected only as random numbers? Are number combinations already in use a factor?

An interesting example of what septimus pointed out: I was listening to a statistician interviewed in a podcast (sorry, I can’t remember which one. Fresh Air?) who did an experiment early in each of her intro classes. She assigned one group to toss a coin a hundred times and record their results. She assigned another group to enter what they considered to be random results of the same thing, but without tossing the coin. She would then leave the room.

She said that she could almost always tell which group was random and which was “pretend” random. The “pretend” group would almost always try to enter an equal distribution of numbers, rather than the more messy result from the truly random group. And the random group was hardly ever exactly 50-50.

I realize that all number sequences are equally likely, and was actually pleased to have two 13’s randomly picked as they are unlikely to be purposely picked.

I always have trouble with probability however.

Shouldn’t the odds of picking three of the same numbers 35^3?

The Powerball websites just describe it as random. I’ve never heard of any other algorithm being used.

dauerbach, thank you for throwing your $20 on top of my pile of winnings. :smiley:

the trouble with human “random” is we try to be too random.

Examine a variety of “quick picks”. there will be quite a few with a sequential pair or several ending in the same digit (13/33, 21/41, etc.). Odds are one decade will be missing. Even likely that 3 of the 6 are in the same decade. (Our lottery is the 6 of 49.)

If a human is asked to pick random, we likely will not pick a sequential pair, or miss a decade, or over-represent others. we try too hard.

Then peruse the winning numbers - I suspect you will find the same. Decades missing or over-represented, sequential pairs, same last digits, etc.

The odds of getting 3 numbers the same out of 10 sets is not too bad, provided you don’t specify the number. After all, that’s 60 numbers, you will hav a lot of duplicates.

For the powerball - one select number -
each number can be one of 35. What one pick has, should have no bearing on the next. Random.

35^10 choices for 10 random powerballs.
There are 35 numbers. For “1” to repeat 3 times, there are 35^7 combinations (set 3 numbers to “1” and choose 7 randomly). Ditto for 2, for 3, to 35.
so there are 35^8 combinations where at least 3 numbers repeat.

Odds of that happening - 35^10/35^8=35^2 or about 1 in 1000.

But that was the first 3 positions selected - it could be any 3 - so divide by how many ways you can arrange 3 items in 10 -which is 1098=720. But the 3 are identical, so now divide by 3! or 6 ways to arrange those 3.

(35*35)/(720/6) gives about 1 in 10.2 odds.

Buy singles, keep your old tickets and test this against any random combination of tickets.
If you win the jackpot before you prove the odds, stop caring…

The odds of three consecutive quickpick powerball numbers being the same is 1/(35^2).

The odds of three consecutive quickpick powerball numbers being a particular number is 1/(35^3).

See the difference?
With the requirement that they be the same but with no restriction on what the exact number is then the first ticket can be any number. There is a 1/35 chance that the second ticket will have the same powerball number. Then there is a 1/35 chance that the third ticket will have the same powerball number. 1*(1/35)*(1/35)= 1/1225

With the requirement that all three be a certain number then the first ticket has a 1/35 chance of having that number. The same logic flows for the second and third tickets. (1/35)(1/35)(1/35)= 1/42874

The odds against getting specifically three 13’s are " 35^3 for 1 ".
But what about three 14’s or three 15’s etc.? You need to multiply your chance by 35.

Or, perhaps more simply, number #1 can be anything, a 1 in 1 chance; number #2 has a 1 in 35 chance of matching #1; number #3 a 1 in 35 chance of matching. Net chance = (1/1) * (1/35) * (1/35) = 1 in 35^2.

ETA: Or, just read Iggy’s post. :smack:

Thanks, it is now clear, but was not before.

The most frustrating thing is how clear and obvious it is when explained.

So, out of the ten tickets (60 numbers) I had only one number match. There are 59 possibilities for the first five, 35 possibilities for the second. This seems like a pathologic mirror image of good luck. What are the odds of this happening.