Ask the Wizard #355
In the movie License to Kill, James Bond plays three rounds of blackjack, playing five hands at a time. I notice he never once hit to any of them. What are the odds of that?
You may find a clip of the scene on YouTube.
The game does see a suspiciously high number of large cards, doesn't it. Here are the cards I could make out. Note that in some cases, the cards could not be seen clearly.
License to Kill -- Ranks Observed
You are indeed right, Bond does not hit any of his 15 initial hands. He splits eights once, but doesn't hit after splitting either. Here is a count of his initial action on all 15 hands:
- Double — 1
- Split — 1
- Stand — 13
Here are the basic strategy probabilities of each initial action, assuming six decks, double after split allowed, dealer peeks at the hole card, and dealer stands on soft 17.
Blackjack -- First Action Probabilities
If surrender were allowed, that probability would be 4.14%, coming out of the probability of standing.
It should be noted that Bond once stood on 16 vs. 10. The basic strategy play is to hit that, but it's very marginal. Given the flood of tens that came out that hand, Bond may have known the count was high, which would trigger a basic strategy exception, standing in that situation.
The probability of not hitting any one hand is 60.22%. The probability of not hitting 15 out of 15, assuming each hand is independent, is 0.60216215 = 0.000496253 = apx. 1 in 2015.
In The Squid Game, 16 players face a game with a glass bridge. The bridge is divided into 18 pairs of glass. In each pair, one piece of glass is tempered and can support the weight of a player. The other piece of the pair is ordinary glass and will break under the weight of a player. If a player steps on a piece of ordinary glass, he will break it and fall to his death.
The players must advance, one at a time, in a predesignated order.
Assuming random guessing at each pair of glass steps, what is the expected number of players to cross safely?
I'll put the answer and solution in spoiler tags, to allow the world to enjoy working out the answer themselves.
The probability the player 1 crosses safely is (1/2)^18 = 1/262144 = apx. 0.000004.
There are two ways the player 2 can cross safely:
- Player 1 crosses safely. In this case, player 2 can just copy his steps.
- Betwen player 1 and player 2, there is only one bad step. This could be at 18 of the possible pairs of glass. The probability of 17 good steps and one bad step is 18*(1/2)^2 = 18/262144 = 0.000069.
So, the probability player 2 crosses safely is 0.000004 + 0.000069 = 0.000072.
There are two ways the player 3 can cross safely:
- Player 2 crosses safely. In this case, player 3 can just copy his steps.
- Betwen players 1, 2, and 3, there are only two bad steps. There are combin(18,2)=153 ways to choose 2 out of 18 of the pairs of glass for the two pieces that kill players 1 and 2. The probability of 16 good steps and two bad steps is 153*(1/2)^2 = 153/262144 = 0.000584.
So, the probability player 3 crosses safely is 0.000072 + 0.000584 = 0.000656.
There are two ways the player 4 can cross safely:
- Player 3 crosses safely. In this case, player 4 can just copy his steps.
- Betwen players 1 to 4, there are only three bad steps. There are combin(18,3)=816 ways to choose 3 out of 18 of the pairs of glass for the two pieces that kill players 1 to 3. The probability of 15 good steps and 3 bad steps is 816*(1/2)^2 = 816/262144 = 0.003113.
Following with this logic, we get the following table of each player's probability.
To express the answer in closed form, it is:
This question is asked and discussed in my forum at Wizard of Vegas.
In your Ultimate X Gold, at the 3:10 point you are dealt a pat full house, threes over fours. Your multiplier for any four of kind in 2's to 4's was 9x. The multiplier for a full house was 1x. Why did you keep the pat full house, as opposed to just the three 3's and try for the four of a kind with the big multiplier?
Bob from Las Vegas
You're right, I absolutely blew it with that hand.
Recall I was playing 10-play. Thus, my win holding the full house was 10×35 = 350.
Holding the threes only, each hand had the following probabilities:
- Four of a kind — 4.26%
- Full house — 6.11%
- Three of a kind — 89.64%
Here are the wins for each hand, after the multipliers:
- Four of a kind — 1800
- Full house — 35
- Three of a kind — 15
My expected return holding just the threes would have been (4.26% * 1800) + (6.11% * 35) + (89.64% * 15) = 92.17854. That is significantly more than the 35 for the full house. So, yes, I made an embarrassing mistake with that hand.
Somebody challenged me to the following bet. I get to choose any three ranks from a standard poker deck, writing down my prediction, but keeping it hidden until the end. For example, 7-ace-2. He then offered to bet me even money that he could name at least one of my ranks if he got three guesses. What were my odds of winning?
To win, you opponent has to be incorrect with all three predictions. The first prediction has a 10/13 chance of being incorrect. The second guess has a 9/12 chance of being incorrect, because we can remove the first rank guessed as a possibility. The third guess has an 8/11 chance of being incorrect, because we can remove the first two ranks guessed as a possibility.
All three of these things would have to happen for you to win. Thus, the chances of you winning are (10/13) * (9/12) * (8/11) = 720/1716 = 41.96%.
At even money, the house edge on this bet, from your side, is 16.08% (ouch!).
This question was culled from The Book of Proposition Bets by Owen E'Shea (number 7).