What are odds?

If I have 200 people who are asked to guess a number between 1 and 1500, what are the odds someone will pick the exact number. Then, if I ask them to pick another number between 1 and 100, what are the odds they will pick both numbers correctly. While we are at it. What are the odds of them picking the second number correctly if they miss the first number.

1-1500:
Prob(W) = 1/1500
Prob(L) = 1499/1500

Probability of losing 200 in a row is (1499/1500)^200. Probability of winning at least 1 is [1 - (1499/1500)^200].

1-100:
Prob(W) = 1/100
Prob(L) = 99/100

Probability of losing 200 in a row is (99/100)^200. Probability of winning at least 1 is [1 - (99/100)^200].

Probability of a correct pick being made in the first and second is both of the probabilities multiplied by each other. If you want to know the probability of the correct picker in the first also getting the correct pick in the second, then take the first probability and multiply by 1/100.


#1: [1 - (1499/1500)^200]

#2: [1 - (99/100)^200]

One pick in both:

[1 - (1499/1500)^200]* [1 - (99/100)^200]

First picker is correct in second:

[1 - (1499/1500)^200] * [1/100]



Or something like that.

I don't think this is correct.

For (1), you would have to calculate the probability that each of the 200 people pick a different number. That's not what you calculated. This is somewhat similar to the birthday problem. I'm too tired to do the math right now, but may tackle it tomorrow.

Perhaps I'm mistaken, but it didn't seem like each person must choose a different number.

If they all chose a different number, wouldn't the odds simply be 200/1500? Then #2 would be a guaranteed success (200 people picking 100 numbers)?


Speaking of birthday problem -- how would you solve that? Say with 20 people. I'd think it'd be similar to Keno math. Instead of a 5-spot with 20 draws and 80 numbers, it'd be like a 20 spot with 20 draws and 365 numbers...then solve for the probability of hitting at least a 1 out of 20 spot. Although I'm not too confident this'd be the way to solve it.

Odds would have to take into account human mindset for instance ask some to pick a number between 25-30

Odds are 1-6 for any number
However I promise you that it's about 60 percent that they pick 27

I'm sure everyone has wanted to ask this!

Quote: RS

Perhaps I'm mistaken, but it didn't seem like each person must choose a different number.

If they all chose a different number, wouldn't the odds simply be 200/1500? Then #2 would be a guaranteed success (200 people picking 100 numbers)?

Yes, and I think your first answer is correct as well.

Quote:

Speaking of birthday problem -- how would you solve that? Say with 20 people.

If you assume there are 365 possible birthdays and all are equally likely, then let's find the probability of having no matches by going person-by-person.

The first person has some birthday. Now for the second person, there are 364 of 365 possible days where we avoid matching. If there's no match, then there are 363 choices for the third person. If still no match, 362 for the fourth. And so on.

So with N people, the probability of at least one match is

1 - (364/365)*(363/365)*(362/365)...*((366-N)/365)

Quote: MrGoldenSun

Yes, and I think your first answer is correct as well.



If you assume there are 365 possible birthdays and all are equally likely, then let's find the probability of having no matches by going person-by-person.

The first person has some birthday. Now for the second person, there are 364 of 365 possible days where we avoid matching. If there's no match, then there are 363 choices for the third person. If still no match, 362 for the fourth. And so on.

So with N people, the probability of at least one match is

1 - (364/365)*(363/365)*(362/365)...*((366-N)/365)

Seems like, when discussed in the past, the likelihood becomes near certainty around 30 or 32 people.

Quote: beachbumbabs

Seems like, when discussed in the past, the likelihood becomes near certainty around 30 or 32 people.

That's still only about 75% probability though....if I did the math right. o.O