Via : "Should You Pay $250 To Play This Casino Game?"
March 28th, 2016 at 8:03:25 AM
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I saw this question in The Riddler section of FiveThirtyEight and thought it might be appropriate for this forum. Being a newbie, I'll play it safe and not copy and paste the text. Here's the link to the question.
Should You Pay $250 To Play This Casino Game?
Since there are some variables that aren't covered, like how many cards are in the deck or exactly how far each decimal can be carried out, it's somewhat confusing to me. Maybe I'm getting too far into the weeds, but I'd like to see others' thoughts.
Should You Pay $250 To Play This Casino Game?
Since there are some variables that aren't covered, like how many cards are in the deck or exactly how far each decimal can be carried out, it's somewhat confusing to me. Maybe I'm getting too far into the weeds, but I'd like to see others' thoughts.
March 28th, 2016 at 8:27:39 AM
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It's an interesting question, so I'm placing the link below that you tried to publish (your linkage ability is restricted by your new membership).
538 puzzler
538 puzzler
If the House lost every hand, they wouldn't deal the game.
March 28th, 2016 at 9:19:12 AM
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March 28th, 2016 at 9:45:52 AM
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Quote: beachbumbabsIt's an interesting question, so I'm placing the link below that you tried to publish (your linkage ability is restricted by your new membership).
538 puzzler
It is impossible not to make at least $100 back, since the first draw cannot disqualify you. Therefore the cost to play is actually $150 not $250.
The proposition is that the values are between 0 and 1, and the examples are in tenths, so we can conclude possible draw values are .1, .2, .3 ... .9
The game is actually sic bo with three dice, each with nine faces. What are the odds that the player will throw a total of 10 or less (and thus make money)? Just on the face of it, I say bank this game, don't play it. If any single die comes up 9, the player loses money. If any single die comes up 8, the player only makes money if the other two dice are a pair of 1's (a 3*((1 in 9)*(1 in 81) occurence).
Total combinations = 9^3 = 729
Total combinations that total 10 or less = C
HE = 1-(C/729)= pretty steep.
The proposition is that the values are between 0 and 1, and the examples are in tenths, so we can conclude possible draw values are .1, .2, .3 ... .9
The game is actually sic bo with three dice, each with nine faces. What are the odds that the player will throw a total of 10 or less (and thus make money)? Just on the face of it, I say bank this game, don't play it. If any single die comes up 9, the player loses money. If any single die comes up 8, the player only makes money if the other two dice are a pair of 1's (a 3*((1 in 9)*(1 in 81) occurence).
Total combinations = 9^3 = 729
Total combinations that total 10 or less = C
HE = 1-(C/729)= pretty steep.
Simplicity is the ultimate sophistication - Leonardo da Vinci
March 28th, 2016 at 9:57:08 AM
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From the context of the question there exists a random number generator pulling numbers between 0 and 1 (.1,... .9, etc). It will add the numbers together and whenever the number is OVER 1, it stops. This is the key factor in the wording of the riddle, in my opinion. The value of the result must be GREATER THAN 1 for the game to stop. For every draw you earn $100. If this is the case the mean of these numbers would be .5, and the average number of draws to meet the requirements would be 3, because the result value of 1 requires another draw to get GREATER THAN 1. Thus, on average, you would require 3 draws, making $300. There is a $250 entry fee, so your Expected Value each time you play the game would be $50.
Playing it correctly means you've already won.
March 28th, 2016 at 10:12:23 AM
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I'm closing this thread because there is another recent thread devoted to the same question.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
