Question: At a restaurant, how could you choose one out of three desserts with equal probability with the help of a single coin (could be biased)?
Answer: If the coin were not biased we could toss the two times and get the following events.
|HH||Choose first dessert|
|HT||Choose second dessert|
|TH||Choose third dessert|
As we have to repeat from start in case of TT, we might need more number of tosses (8/3 on average).
If the coin is biased the above four events won’t be equally likely to occur. But TH and HT would occur with equal probability. So we can toss coin four times and get the following events-
|TTHH or HHTT||Choose first dessert|
|THHT or HTTH||Choose second dessert|
|THTH or HTHT||Choose third dessert|
Another possible approach could be assigning HTT, THH to first dessert, THT, HTH to second dessert and TTH, HHT to third dessert, with other 3-toss outcomes rejected.