Ask The Wizard #300
The king has a barrel full of wine.
On Monday night, a servant steals three cups from the barrel and replaces them with three cups of water.
On Tuesday night another servant steals three cups from the now diluted barrel of wine and replaces them three cups of water.
On Wednesday night yet another servant steals three cups from the now-more diluted barrel of wine and replaces them three cups of water.
On Thursday morning the barrel is 50% wine and 50% water.
How much wine was initially in the barrel?
On Monday night, a servant steals three cups from the barrel and replaces them with three cups of water.
On Tuesday night another servant steals three cups from the now diluted barrel of wine and replaces them three cups of water.
On Wednesday night yet another servant steals three cups from the now-more diluted barrel of wine and replaces them three cups of water.
On Thursday morning the barrel is 50% wine and 50% water.
How much wine was initially in the barrel?
Here is the answer and solution (PDF).
For discussion about this problem, please visit my forum at Wizard of Vegas.
For discussion about this problem, please visit my forum at Wizard of Vegas.
There is a biased coin with 60% chance of landing on heads. It is flipped until either there are two heads or two tails in a row. What is the chances two heads comes first?
Here is the answer and solution (PDF).
For discussion about this problem, please visit my forum at Wizard of Vegas.
For discussion about this problem, please visit my forum at Wizard of Vegas.
I know of a blackjack machine where, through a bug, the player can get back a losing insurance bet. The rules are eight decks, blackjack pays 3 to 2, dealer hits a soft 17, double after split allowed, no surrender, and no re-splitting. What is the house edge of this game if I get back every losing insurance bet?
This bug is equivalent to getting an extra unit every time the dealer has a blackjack with an ace up. That will happen 2.37% of the time. The house edge would be 0.72% without this bug. With it, the player advantage is 2.37% - 0.72% = 1.66%.
For further discussion on this question, please see my forum at Wizard of Vegas.
For further discussion on this question, please see my forum at Wizard of Vegas.
I saw that that YouTube video where you allegedly solved the Rubik's Cube on a unicycle. However, I submit for your consideration that you faked it. I think you used two Rubik's Cubes and had a solved one in the pocket of your hoodie. When your back was turned to the camera, you switched them. Fraud!
I'm never offended by a healthy degree of skepticism. To satisfy those skeptics, I created this follow-up video. Enjoy.
In your page on video poker programming methodology, you mention that you don't need to analyze all 2,598,960 possible starting hands, but only the 134,459 distinct hands and then weight each one appropriately. My question is how many distinct seven-card hands are there?
Before I answer, I'd like to remind everybody that the number of ways to choose k out of n items, with replacement, is combin(n+k-1,k) = (n+k-1)!/((n-1)!×k!).
That said, here are the following types of seven-card hands and the number of distinct ways to make each:
That said, here are the following types of seven-card hands and the number of distinct ways to make each:
- 7 cards of one suit: combin(13,7)=1,176.
- 6 cards of one suit and 1 of another: COMBIN(13,6)×13 = 22,308.
- 5 cards of one suit and 2 of another: COMBIN(13,5)×combin(13,2) = 100,386.
- 5 cards of one suit and 1 each of another two: COMBIN(13,5)×combin(13+2-1,2) = 117,117.
- 4 cards of one suit and 3 of another: COMBIN(13,4)×combin(13,3) = 204,490.
- 4 cards of one suit, 2 of a second, and 1 of third: COMBIN(13,4)×combin(13,2)×13 = 725,010.
- 4 cards of one suit and one each of another 3: COMBIN(13,4)×combin(13+3-1,3)×13 = 325,325.
- 3 cards of two different suits and one card of a third suit: 13×((COMBIN(13,3)×(COMBIN(13,3)-1)/2+COMBIN(13,3))) = 533,533.
- 3 cards of one suit and two cards each of two other suits: COMBIN(13,3)×(COMBIN(13,2)×(COMBIN(13,2)+1)/2) = 881,166.
- 3 cards of one suit, 2 cards of a second, and one card each of the two other suits: COMBIN(13,3)×COMBIN(13,2)×COMBIN(13+2-1,2) = 2,030,028.
- 2 cards each of three suits and 1 from the fourth: ((COMBIN(13,2)×(COMBIN(13,2)+1)×(COMBIN(13,2)+2)/6) = 1,068,080.
The sum of those combinations is 6,009,159. Compared to the combin(52,7)= 133,784,560 ways to pick 7 cards out of 52, that is a 95.5% reduction in hands analyzed.
For more discussion on this question, please see my forum at Wizard of Vegas.
If I had a bowl of 200 Skittles and 3 would kill me, what are the chances I would die if I took a handful of 12 and ate all of them. Assume that it takes only one poison Skittle to kill me.
The answer is 1-combin(197,12)/combin(200,12) = 17.02%.
For more discussion on this question, please visit my forum at Wizard of Vegas.