Math(s) Question: Probability scenario and formula

[69dodge]

I think you misunderstood me. You are more likely to pick a whole pill than a half pill because it is bigger.

Yes, I did misunderstand. I thought you meant that it didn't matter whether or not you were more likely to pick a whole piil.

I know, how about this: whenever you compare a full pill and a half pill, you are twice as likely to pick the full pill. In any given situation you are likely to have twice as many half pills as full pills (because of the above). Thus, it cancels out and you end up with 50/50.

I'll think about this some more.

 

[ddt]

Nope. The probability of drawing a whole pill is not (n-1)/n on the second day as half pills are half the size. I am about to edit in the answer for any given set up as well

Edit: it won't let me edit earlier posts. The full answer is this:

On any day with x pills and y half pills, the chance of picking a pill is 2x/n where n is 2x+y
If this is day 0, then for any day d, the probability of picking a pill is (2x/n)-d(2x/(n^2-n))
Feel free to write that in a nicer way
If your jar has 32 half pills, and 73 pills, then the chance of picking a full pill on day 11 is 79.24% marvellous!

Okay, so you assume it's twice as likely to pick a whole pill than a half pill, all other things being equal. Your earlier wording suggested you assumed both were equally likely. I worked from the latter assumption.

I can see that such a linear formula turns up, but just checking some instances doesn't cut it. Math requires proof. (Hint: there's a latex BBcode tag).

As to edits: there's a 2 hour edit window after you submitted a post.

 

[Loss Leader]

Okay, here's what I've got. Assume chance of picking half or whole is equal.

5 pills total

X = half pill
O = whole pill

Day 1 - O O O O O

Day 2 - O O O O X

Day 3 - O O O X X
----or O O O O

Day 4 - O O X X X
----or O O O X

Day 5 - O X X X X
----or O O X X
----or O O O

Day 6 - X X X X X
----or O X X X
----or O O X

Day 7 - X X X X
----or O X X
----or O O

Day 8 - X X X
----or O X

Day 9 - X X
----or O

Day 10 - X

(the word "or" does not imply equal likelihood.)

 

[sol invictus]

On average, the chance of picking a half pill is equal to the chance of picking a whole pill. Size does not matter, as by the time you've finished the jar, you will have picked the same amount of half pills as you have picked whole pills.

That is incorrect. It's true that the average chance is the same - in fact it's simply 1/2, since by the time you've finished the bottle you've picked N wholes and N halves out of a total of 2N draws. But that obviously isn't enough information to determine the probabilities on a given day (c.f. my first post).

Thus if you have n pills in a jar. The chance of you getting a whole pill is simply 1-d/(2n-1) where d is the day number (starting at day 0).

That formula is correct only in a very special case (the one where you're twice as likely to pick a whole pill as a half pill when the numbers are equal), as I see from reading the comments below has been understood. Incidentally that same (linear) result in that case follows trivially from the differential equations I wrote modified for that case.

But the case of equal probability seems to be significantly more complicated. And of course the real problem is much more complicated, since it really depends on the physics of the pills in the bottle and the way it is handled.

 

[phelix]

Yup. I hacked away at it for a bit and conceded that you are correct sol. There is no simple formula for this it seems. Bit of a pity.

 

[blobru]

This is a nice way to set it up: sort of a graph theory approach (were you to draw it as nodes and edges; then weight and count up the edges, as different chances to get there, to track the probability for each node's combination; see Markov chain^WP)...

Okay, here's what I've got. Assume chance of picking half or whole is equal.

5 pills total

X = half pill
O = whole pill

Day 1 - O O O O O

Day 2 - O O O O X [P=1]

Day 3 - O O O X X [P=4/5=.8]
----or O O O O [P=1/5=.2]

Day 4 - O O X X X [P=.8*3/5=.48]
----or O O O X [P=.2+.8*2/5=.52]

Day 5 - O X X X X [P=.48*2/5=.192]
----or O O X X [P=.48*3/5+.52*3/4=.678]
----or O O O [P=.52*1/4=.13]

Day 6 - X X X X X [P=.192*1/5=.0384]
----or O X X X [P=.192*4/5+.678*2/4=.4926]
----or O O X [P=.678*2/4+.13=.469]

Day 7 - X X X X [P=.0384+.4926*1/4=.16155]
----or O X X [P=.4926*3/4+.469*2/3=.68211[U]6[/U]]
----or O O [P=.469*1/3=.156[U]3[/U]]

Day 8 - X X X [P=.16155+.68211[U]6[/U]*1/3=.38892]
----or O X [P=.68211[U]6[/U]*2/3+.1563=.61107]

Day 9 - X X [P=.3889[U]2[/U]+.61107*1/2=.694461]
----or O [P=.6110[U]7[/U]*1/2=.305538]

Day 10 - X [P=.69446[U]1[/U]+.305538=1 ]

(the word "or" does not imply equal likelihood.) [that's for sure! ]

[ed: to insert calculations for Probabilities] Amazing, even for the relatively simple case of five pills, how gnarly the probabilities become (maybe should have left it all in fractions in hindsight; in any case, a good illustration how quickly complexity can arise from simple starting conditions and rules; and what a boon computers are for calculations of this type).

 

[Loss Leader]

[ed: to insert calculations for Probabilities] Amazing, even for the relatively simple case of five pills, how gnarly the probabilities become (maybe should have left it all in fractions in hindsight; in any case, a good illustration how quickly complexity can arise from simple starting conditions and rules; and what a boon computers are for calculations of this type).

Thanks for doing the math. I knew that some outcomes were far more likely than others, but beyond the third day, I couldn't figure out how to express them. It also threw me that some outcomes could be the same, even from different starting positions.

For instance, when the condition arises that you either have two whole pills or one pill and two half pills, you can get to one whole pill and one half pill from either state.