MHB How many draws until all paired tea bags are gone from the jar?

  • Thread starter Thread starter RWood
  • Start date Start date
RWood
Messages
4
Reaction score
0
I suspect this is not really an "advanced" probability question, but I'm not sure - haven't been near this stuff for decades.

The definition: I have a jar with 2*N tea bags in it (N>0 obviously). The beginning condition is that the teabags are joined in pairs - so there are N pairs. At each selection I select an item at random - initially that will be a pair of bags, in which case I tear one off and put the other back. On later turns I randomly select either a single bag, which would then be used, or a pair of joined bags (if there are any left), in which case I tear one off and proceed as above. What is the probability distribution - and hence the expectation - for D, the number of "drawings" required before there are no paired bags left in the jar?

It is clear that the values for D can range from N (by happening to always select paired bags) to 2*N-1.

I can see some ways of getting recursion equations, but I suspect that this problem has a simple answer resulting from a more general formulation. Any quick answers? Thanks.
 
Physics news on Phys.org
RWood said:
I suspect this is not really an "advanced" probability question, but I'm not sure - haven't been near this stuff for decades.

The definition: I have a jar with 2*N tea bags in it (N>0 obviously). The beginning condition is that the teabags are joined in pairs - so there are N pairs. At each selection I select an item at random - initially that will be a pair of bags, in which case I tear one off and put the other back. On later turns I randomly select either a single bag, which would then be used, or a pair of joined bags (if there are any left), in which case I tear one off and proceed as above. What is the probability distribution - and hence the expectation - for D, the number of "drawings" required before there are no paired bags left in the jar?

It is clear that the values for D can range from N (by happening to always select paired bags) to 2*N-1.

I can see some ways of getting recursion equations, but I suspect that this problem has a simple answer resulting from a more general formulation. Any quick answers? Thanks.

I can't give you any help with this at present, I will have to think about it. However I can say I have seen this problem somewhere before, and vaguely recall it being connected with Herman Bondi (I suspect there was a note either in Mathematics Today or the Mathematical Gazette about it, but that is no help since my filling system makes it impossible to find even if I knew which and which year..).

CB
 
Last edited:
CaptainBlack said:
I can't give you any help with this at present, I will have to think about it. However I can say I have seen this problem somewhere before, and vaguely recall it being connected with Herman Bondi (I suspect there was a note either in Mathematics Today or the Mathematical Gazette about it, but that is no help since my filling system makes it impossible to find even if I knew which and which year..).

CB

Thank you for the update, will see what develops.
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Back
Top