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I Math question regarding the game of craps

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  1. Mar 7, 2016 #1
    Not sure if anyone plays craps here, but I think one of the probabilities is wrong.The C and E bet which is a single roll on craps (2,3,12, four total possible combinations for 2 dice which has 36 total possible combinations) or on 11 (yo) which has two combinations for a total of six combinations. Your bet is split between the two bets so this bet must an even number, split between craps and 11.

    If 11 rolls you have 15:1 payout on half your bet and the other on craps lose. If craps (2,3,12) is rolled you have a 7:1 payout on half your money and lose the half on 11. For example if you bet $2 on the C and E the dealer will put $1 on each. If craps is rolled you get $7 - $1 you lose on 11 so total profit is $6, you can take the $1 on craps back if you wish. If an eleven is rolled you get 15:1 on your $1 not $2, because the bet is split so you would profit $14 dollars $15 - $1 you had on C (craps).

    Because you lose half your money when this bet wins and lose everything when a number other than craps or 11 is rolled, shouldn't the house edge be better than 11.11% (as in worse for the player higher house edge than 11.11%)? If you bet on the 11 alone you get 15:1 payout when the true odds are 17:1 for an 11.11% house edge. How can this be the same if you are making a bet on the C and E where you lose half your bet even if you win? Are the odds on Wikipedia right?
     
    Last edited: Mar 7, 2016
  2. jcsd
  3. Mar 7, 2016 #2
    Hi eiyaz:

    CORRECTION
    My expectation calculation is that the house averages taking 11.111. . . cents out of every dollar bet.

    Wikipedia says that the odds given is 7:1 odds for the 11 bet, and 3:1 odds on the craps bet.
    If 11, better wins 3. 3 x 4 rolls = 12
    If craps, better wins 7. 7 x 2 rolls = 14
    If other, better loses 1. -1 x 30 rolls = -30
    Net = -4
    4/36 = 1/9 = 0.111...

    I am not sure what you mean by "odds" here.
    Wikipedia says "True Odds = 5:1".
    I assume that this means that the better wins something on 6 rolls and loses 1 on 30 rolls.

    Wikipedia says "House Edge = 11.11%".

    Regards,
    Buzz
     
    Last edited: Mar 7, 2016
  4. Mar 7, 2016 #3
    Thanks, just to confirm Wikipedia's article on Craps (https://en.wikipedia.org/wiki/Craps#Bet_odds_and_summary) says 11.11% this is wrong correct this actual number should be 13.8%?

    Wait you have six combinations (1-1, 1-2, 2-1, 6-6, 5-6, 6-5,) so its 6/36 I think, but the thing is you lose half your bet so craps is (1-1, 1-2, 2-1, 6-6) and get 7:1 payout on half your money so if you bet $2 you win 7:1 on $1 and lose the $1 on 11 (yo) so you get $6 net profit ($7 - $1 on the 11) you can still take the $1 on craps back for a total revenue of $7. If 11 (5-6, 6-5) rolls you get 15:1 on $1 and lose the $1 on craps (1-1, 1-2, 2-1, 6-6) so net profit is $14 plus you can take the $1 on 11 back for a total revenue of $15. If any other number besides craps or 11 comes you lose your whole bet on (both bets), this is a single roll bet. Is the 11.11% correct? I was thinking it should be between 11.11% and 13.88% because you still lose half your bet when you win.
     
    Last edited: Mar 7, 2016
  5. Mar 7, 2016 #4
    That's true odds and its 17:1 not 16:1 and 8:1.
    True Odds for Craps (1-1, 1-2, 2-1, 6-6) is 8:1 payout is only 7:1
    True Odds for 11 (5-6, 6-5) is 17:1 payout is only 15:1

    However, if you bet C and E you are betting both, so you are only paid 15:1 or 7:1 (depending on 11 or craps) on half your bet. If you bet on 11 alone (not C and E) you get 15:1 payout but you also get your whole investment back so the house edge is 11.11%, a bet on the C and E has more possibilities yes, but is the odds the same 11.11% when you only get half your original bet back?
     
  6. Mar 7, 2016 #5
  7. Mar 7, 2016 #6
    Hi @eiyaz:

    Sorry about my mistakes in earlier versions of my previous post. Please look at the corrected version.

    Regards,
    Buzz
     
  8. Mar 7, 2016 #7
    Does that include the fact you lose half your bet on C or E when you win? I was under the impression that the calculations out there did not include this fact, I am not sure though
     
    Last edited: Mar 7, 2016
  9. Mar 7, 2016 #8
    Hi @eiyaz:

    No. The odds given
    "Wikipedia says that the odds given is 7:1 odds for the 11 bet, and 3:1 odds on the craps bet."
    assumes a single bet on the CE combination. This would actually correspond to making a $0.50 bet on each of
    (a) the 11, and (b) craps. The (a) bet would pay 16:1 and the (b) bet 8:1.
    The $0.50 bet on 11 would return a gain of $8 minus the dollar bet = 7:1.
    The $0.50 bet on craps would return a gain of $4 minus the dollar bet = 3:1.

    Regards,
    Buzz
     
  10. Mar 7, 2016 #9
    That is a calculation for the total bet. If you were to bet $2 and you hit craps you get $6 in net profit plus half your original bet back for 3:1 on $2 and a total revenue of $7. If you were to bet $2 and 11 (yo) came you would get 7:1 on $2 for $14 of net profit but you would only get back $1 for 11 not $2 for a total revenue of $15 since the craps (C) bet lost. Wikipedia did not mention you do not get back your whole bet back just half.
     
  11. Mar 7, 2016 #10
    Hi eiyaz:

    I am not sure what you mean by "the total bet".
    (1) Bet $0.50 on craps and $0.50 on 11.
    (2) Bet $1 on the EC combination.

    Please choose one of the above.

    Regards,
    Buzz
     
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