1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Binomial gambling

  1. Feb 24, 2009 #1

    Let's say you start off with $500, and someone offers to give you another $500 everytime a coin is heads, or take $500 from you it's tails. You agree to play this game until a) you've either lost all your money, or b) you've made an extra $1000 (i.e. so you walk away with $1500).

    Then it will take at least 2 coin flips for you to walk away winning anything in this game, so initially considering those, we have:
    (In the following I will use +2 to signify you've one 500 twice, -1 to rep you losing a 500 and going bankrupt etc)

    hh, +2, 25% chance
    ht, +0 , 25%
    th, bankrupt after toss 1, when you stop making you -1, 25%
    tt, bankrupt after toss 1, when you stop, making you -1, 25%

    So obviously here there is a 25% chance to win in 2 moves, but you also keep on going if the combo was ht and your back to your original 500. But then the probability for you to win +2 going down this node is exactly the same anyway (since youre basically just back to square one and going to repeat yet another two moves)

    Thus the probability of you going up by +2 and walking away happy, is given by the recurssion

    P(+2)=0.25+0.25*P(+2). If you solve this you get P(+2)=33.33%

    Another way to arrive at this answer is by considering:
    Which is a geometric series with a=0.25,r=0.25
    So again P(+2)=33%


    Just wanted to check everyone agrees with this and I haven't missed anything? would really appreciate it thanks
  2. jcsd
  3. Feb 24, 2009 #2
    Second thing I wanted to check is, let's say this game changes slightly, so that you start with 500 again balance, but now you only bet 250 chunks, you again agree to play until either you lose the lot or you win 1000 (so you have to get to +4 or quite if you go -2).

    Now it takes a minimum of 4 moves for you to walk away with your 1000, so considering those initially :: (were w represents a win from getting a head, l a loss from it being a tail)

    wwww :+4
    wwwl : +2
    wwlw : +2
    wwll : +0
    wlww :+2
    wlwl :+0
    wllw : +0
    wlll :-2 (gameover at tick num 3)
    lwww :+2
    lwwl :+0
    lwlw :+0
    lwll :-2 (gameover at play num 4)
    llww :-2 (gameover at play num 2)
    llwl : -2 (gameover at play num 2)
    lllw : -2 (gameover at play num 2)
    llll :-2 (gameover at play num 2)

    The probability of any one combo occuring is 0.0625.
    So totting up all these combos we have 1 that goes +4 (prob is 0.0625), 4 that go +2 (prob is 4*0.0625=0.25), 5 that stay neutral +0 (prob of being +0 is therefore 5*0.0625=0.3125), 6 were we stop playing and have bust the account at -2.(prob of being -2 is thus 6*0.0625=0.375).

    Now following the same argument as in the prior post.
    P(+4| neutral), i.e. probability that we go up by 4 given that we obviously start neutral. Is

    P(+4| 0)=0.0625+0.25P(+4|+2)+0.3125P(+4|0).

    [//*****Now as for P(+4|2). This can be worked out in a similar way, i.e. we start of +2, and keep going, we can either have ww taking us to +4 0.25 of the time, wl/lw leaving us at +2, 0.5 of the time, or ll taking us to 0, 0.25 of the time. So to work out P(+4|2),we have:
    P(+4|2)=0.25+0.5P(+4|2), solving this algebraically gives
    0.5P(+4|2)=0.25, and thus

    Subbing this in to first equation gives

    P(+4| 0)=0.0625+(0.25*0.5)+0.3125P(+4|0).
    P(+4| 0)=0.1875+0.3125P(+4|0).
    0.6875P(+4| 0)=0.1875
    P(+4| 0)=0.272727.....

    So the probability of you going up the required +4, has actually decreased from the 33% (when doing 500 bets) to 27.7% (for 250 bets)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Similar Threads for Binomial gambling
A Newton's Generalized Binomial Theorem