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Expected value for a Lucky 7 game

  1. Jan 22, 2008 #1
    Around 3 years back, I had a stall setup at my school's fun fair. It was a simple gambling setup. You put your money on either of 3 things: i] Above 7, ii] Lucky 7 or iii] Below 7.

    And then, you roll the dice [2 of 'em]. If you put on either below 7 or above 7 and the dice sums up to below 7 or above [resp.].. you get twice your money. On 'Lucky 7', you get thrice your money.

    Here obviously it depends on how the player chooses his bet that determines the expected value. Let's assume our player is quite fair.. he picks up either of the 3 categories with equal likelihood. So, what is the expected value for each $1 he bets?
  2. jcsd
  3. Jan 22, 2008 #2
    They are 36 possible outcomes. 7 is the most likely number coming up 6/36=1/6 of the time. Thus the expected value of the bet is:

    [tex] \frac{5}{12}x2+\frac{5}{12}x2 +\frac{1}{6}x3 = \frac{26}{12} [/tex]

    Which, of course, can not make any sense. The better idea from the standpoint of the bank is to pay double on only one of the two possibilities not a seven, and to return the bet on the 7. This then gives:

    [tex] \frac{5}{12}x2+\frac{5}{12}x0 +\frac{1}{6}x1 = \frac{12}{12}=1. [/tex]

    Which is a perfectly fair game. (This could have been seen without the details since the chances of getting less than 7 is exactly the same as getting more than 7.)

    Of course the bank would not approve of that either, so we probably need further restrictions. The way the 7 comes up is: 2 ways for each case, (1,5),(2,5),(3,4), so some restriction like not paying on (5,1) would be added. Then the 7 returns the money 4/36 = 1/9. The new case then yields:

    [tex] \frac{5}{12}x2+\frac{5}{12}x0 +\frac{1}{9}x1 = \frac{34}{36}=\frac{17}{18}. [/tex]

    This says that for evey $18 handled in bets the bank expects to pay out $17. Even so the bank may not even want to handle even that.
    Last edited: Jan 22, 2008
  4. Jan 22, 2008 #3


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    This analysis doesn't make any sense; what is this supposed to be the expected value of? It's certainly not the expected value of any wager available to the player.
    Last edited: Jan 22, 2008
  5. Jan 22, 2008 #4
    I see that that is not the problem. The player is to pick which of the three cases he prefers. Acording to the requirements specified that he plays all different cases equally. Thus after placing three bets the return would be:

    (5/12)2B + (5/12)2B +(1/6)3B = 26B/12, which we divide by 3 giving (13/18)B, or a loss of (5/18) about 28% loss per bet. Which means he expect to receive about 72 cents for each dollar he wages.
    Last edited: Jan 22, 2008
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