Do IBM’s drug plans have a donut hole? No

It's more like a cement block wall for the medical/drug supplement plan. That plan pays $3500 maximum for the year. You get the negotiated price for your drugs but no other benefit once you hit the max. The other two drug plans have unlimited drug coverage but are significantly more expensive. The drugs only plan (no medical) also has lower copays. But it is the most expensive and you have no secondary medical insurance unless you buy a separate medigap policy.

None of the IBM drug plans are structued like a Medicare D drug plan. But they are creditable. That means they payout a comparable amount of money to recipients compared to a Medicare D plan. Creditable is really imprtant. If you don't have creditable drug insurance or no insurance at all then when you do try to buy a D policy the insurance company will charge you a lifetime penalty that is 1% of the national average premium ($33.32 in 2010) for every month you did not have it. You have to buy drug insurance within 3 months of becoming medicare eligible.

Why the penalty? The government doesn't want people to skimp or skip buying drug insurance while they are well and then jump into a policy when sick. The insurance pools need premiums from the combination of well and sick people to at least break even much less make a profit. Incidently, there is no penalty to jump in and out of IBM drug insurance from year to year. However, if IBM decides to stop providing drug insurance (what an evil thought) then you'd face a penalty.

Medicare D insurance policies do have the donut hole. These plans STOP paying out when your total drug cost (including your co-payments) is $2830 (in 2010). When you pay out a total of $4550 (they do include your co-payments in that amount) the insurance starts paying again - usually at a 95% rate. Most people never make it through the donut hole in a given year. Incidently, drugs admininstered in the hospital are covered by Medicare A and in the doctor's office by Medicare B so they do not count in this computation.