Assuming no house edge, what would be the average time of the crash in a Crash game?

anonymous

For the benefit of other readers, Crash is a popular game at Internet casinos, especially ones based on Crypto currency. The idea is the player makes a bet and then a plane or rocket takes off from the ground. Every fraction of a second, the altitude increases by a small percentage. The player may eject at any time and will win according to the altitude of the object. However, with every fraction of a second the object my rise, it may also crash. If the object crashes before the player ejects, the player loses.

That said, the average altitude attained, assuming no house edge, is 1/ln(2) =~ 1.442695041.

What is the probability of getting in royal flush in both Omaha and Texas Hold 'em?

zjbird

For the benefit of other readers, here are the cards that may be used in both games.

• Texas Hold 'Em -- Player may make the best poker hand consisting of two hole cards and five community cards. The player may use any five of these seven cards.
• Omaha -- Player must make use of exactly two out of four hole cards and exactly three out of five community cards.

Using Excel formulas, here is the probability in each game.

• Texas Hold 'Em -- 4*COMBIN(47,2)/COMBIN(52,7) =~ 1 in 30,940
• Omaha -- 4*COMBIN(5,2)*COMBIN(47,2)*COMBIN(45,2)/(COMBIN(52,4)*COMBIN(48,5)) =~ 1 in 10,829

What is your advice on strategy in the game show Lucky 13?

anonymous

For the benefit of other readers, here are the rules.

1. The player will be asked 13 true or false questions.
2. Afterward, the player will be asked to choose a range of how many he thinks he got right. If he is within that range, he will win according to the table below.
3. There is also a \$25,000 bonus if the player correctly guesses exactly how many he got right.
4. The host, Shaq, can offer a surrender value at any time.

Photo source: ABC Press

 Range Correct Win 13 \$1,000,000 10 to 12 \$100,000 7 to 9 \$25,000 4 to 6 \$15,000 1 to 3 \$5,000

The following table shows the strategy for maximizing expected value. For example, if the player thinks he will get 9.5 right, he should choose the 10 to 12 range. Although 9.5 is not within that range, the expected value is greater than the 7 to 9 range, because the win is 4x as much.

 Estimated Correct Optimal Range 10.62 to 13 \$13 7.66 to 10.62 10 to 12 5.83 to 7.66 7 to 9 2.54 to 5.83 4 to 6 0 to 2.54 1 to 3

As to the surrender offers, every time I have watched the show the offers were very stingy, about half of expected value. I recommend just saying "no" to Shaq.

In this YouTube video, Matt Parker alleges that a horizontal bingo is more likely to win that a vertical one. Is this correct?

anonymous

In his video, Matt is validating one element of this question as raised in the article The Bingo Paradox by Arthur Benjamin, Joseph Kisenwether and Ben Weiss.

Assuming a simplified game with no free square and no diagonal wins, they both make the case that if balls are drawn until either (1) at least one ball in each column or (2) five balls in one column, the following are the probabilities of which will happen first.

• At least one ball in each column (possible horizontal bingo) = 0.751779.
• Five balls drawn in any one column (possible vertical bingo) = 0.248221.

I have done all the math and I agree with this. However, just because a bingo is possible, it doesn't mean one will be called in the average game.

Before I get into that, with just one bingo card, the probability the first bingo will be horizontal or vertical is exactly 50/50.

However, this is not the case with a finite number of cards and only the winning cards count.

For purposes of my analysis, I assumed that every possible bingo card was in play exactly one. That is a total of 6,076,911,214,672,420,000,000,000,000 cards.

If multiple cards formed a bingo at the same time, each win was counted equally.

Assuming a vertical bingo was possible first, I show the probability of any given card winning is 1 in 3003 =~ 0.000333. Assuming a horizontal bingo was possible first, I show the probability of any given card winning is 1 in 8294 =~ 0.000121.

The following table puts together both the first possible bingo type and the probability any given cards gets a bingo. The right column shows that the probability a bingo will be horizontal is 52.3%, given that a bingo is called.

 Bingo Probability Possible Probability Random Card Wins Product Ratio Horizontal 0.751779 0.000121 0.000091 0.523040 Vertical 0.248221 0.000333 0.000083 0.476960 Total 1.000000 1.000000

I know my 52.3% differs from the 2 to 1 ratio quoted in the paper, but that is based on a simulation of a 1,000 card game and I don't know how they count multiple bingos at the same time.

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