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

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The discussion revolves around calculating the expected number of draws, D, required to deplete paired tea bags from a jar containing 2*N bags. Initially, all bags are paired, and the selection process involves randomly choosing either a pair or a single bag. The range for D is established between N and 2*N-1, depending on the selection outcomes. Participants express uncertainty about the complexity of the problem, with one recalling a possible connection to a previous mathematical discussion. The conversation highlights the need for further exploration of recursion equations or a general formulation for a solution.
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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.
 
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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.
 
There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host...

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