What is the expected number of marks on a single bingo card to form various common winning patterns?

anonymous

Here is the average number of marks needed on a card for common winning patterns:

• Single bingo — 13.60808351
• Double bingo — 16.37193746
• Triple bingo — 18.02284989
• Single hardway — 15.29273554
• Double hardway — 18.09327842
• Triple hardway — 19.79294406
• Six pack — 14.62449358
• Nine pack — 18.97212394

In the previous Ask the Wizard column, you were asked about the expected number of rolls to achieve a total of 12 with two dice twice in a row. On a related note, I see somebody on your forum is claiming to have witnessed 18 consecutive yo's (total of 11) at the craps table. What is the expected rolls required for that to happen?

anonymous

41660902667961039785742

Here is my solution (PDF).

This question is asked and discussed in my forum at Wizard of Vegas.

The exact answer found with the aid of WizCalc.

Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.

It’s \$1 per flip and they are really bored, so they decide to do one million flips. At the end of the session, the loser will write a check to the winner for the final balance. What is the expected value of the check amount?

Ace2

797.88456080286535587989211986876373695171726 232986931533185165934131585179860367700250466 781461387286060511772527036537102198390911167 448599242546125101541269054116544099863512903 269161506119450728546416733918695654340599837 28381269120656178667772134093073...

The general formula for the answer is the sqrt(variance * (2/pi)).

The variance in this case is 1,000,000. So, the expected absolute difference between actual and expected results is sqrt(1,000,000 × (2/pi)) =~ 797.88456080286535587989211986876373695171726 232986931533185165934131585179860367700250466 781461387286060511772527036537102198390911167 448599242546125101541269054116544099863512903 269161506119450728546416733918695654340599837 28381269120656178667772134093073.

I ask a related question in Ask the Wizard #358, which will help show where I get the sqrt(2/pi) term.

This question was asked and discussed in forum at Wizard of Vegas.