Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Different Value In These Two Gambling Games?

  1. Aug 23, 2016 #1
    So it's been such a long time since I studied statistics at school/college/university that I'm rusty to say the least and hoped you guys could help me out with this.
    Something is telling me that these two games must have a different value despite being similar, but if you brilliant folks could confirm that and go in to more detail then that would be great :)

    Game 1
    If we wager £100 of our own money on roulette with an RTP (return to player) of 97.3%. We get given £10 cash.

    So in this case the game clearly has a positive EV and it is £7.30. (We expect to lose £2.70 from the first part and then get given £10 for free)

    Game 2
    If we wager £50 of our own money on roulette with an RTP of 97.3%, we then get given a bonus of £10.
    This £10 is different to before though when it was cash. This time we have to wager the bonus 5 times (£50 of play) before it will be released as cash.

    So we have to place another £50 worth of roulette bets to release the bonus. But we will never be betting with our own money, if our £10 becomes zero we just stop, no need to bet our own money since the bonus is gone. If after we've placed £50 worth of bets with the bonus £10 and it is now a bonus £20 then it is now worth £20 in cash.

    So in both scenarios we have placed £100 worth of bets and given £10 for free. But would I be wrong in saying game 2 has a higher value because of the 'stop-loss' element of the bonus? We don't have to bet with our own money if the bonus becomes worthless.

    Am I right in saying the variance of the game makes a difference, so game 2 might have a different value if we bet on numbers rather than red/black? Would I be right in saying game 2 has a different value if we lower/raise our stakes (and thus decrease/increase the variance)?

    Sorry if I'm not being clear! I appreciate any insight and a refresher in my old statistics classes!
     
  2. jcsd
  3. Aug 23, 2016 #2

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    Correct.

    If you want to maximize the expectation value (not necessarily the best thing to aim for in reality), you should play as risky as possible in game #2.
     
  4. Aug 23, 2016 #3
    Brilliant! I thought as much. Given that we know the variance of a roulette game, could you or anyone else go in to more depth about the value of game 2 if you were to play red/black?

    Also maybe more detail as to why a risky strategy is the more profitable one long term?

    Any level of maths is fine and I'm very very grateful of your response :)

    Also even hints towards methodology is great. I'd love to go away and crunch some numbers on similar pproblems.
     
  5. Aug 23, 2016 #4

    micromass

    User Avatar
    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    Maximizing expectation values shouldn't be the only thing that matter. Playing a game with expectation value 0 and variance 100 is very different than a game with expectation value -1 and variance 1.

    Teaching students to take expectation values too seriously is a deep fallacy in my opinion.
     
  6. Aug 23, 2016 #5
    Completely agree.. But in this case if we played both games with exactly the same variance (always playing red/black) they would still have slightly different expectation values.
     
  7. Aug 23, 2016 #6

    mfb

    User Avatar
    2016 Award

    Staff: Mentor

    On average, you bet less money, which reduces losses.
    Sure, but as mentioned already: expectation values are not everything.

    The St. Petersburg paradox is a common example.

    Here is one story I really like, although I don't find the original source any more: The devil tells you that you just have 1 month left to live, but offers you to gamble with your time: With 60% probability the time gets halfed, with 40% probability it gets doubled. You quickly calculate the expectation value: 1.1 times the original value. The devil also offers you get as many subsequent bets as you want, betting your previous result each time. Do you pick 0 bets? 1? 100? Let's say you picked 10000. The devil uses its perfect random number generator, and calculates your remaining lifetime: Less than a nanosecond. Bad luck? The devil lets you play again: again less than a nanosecond.
    What went wrong? Your expectation value is indeed huge (way longer than the age of the universe), but it comes from a few extremely unlikely cases. The chance that you end up with more than a second is less than 1 in a billion.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Different Value In These Two Gambling Games?
  1. Gambling Strategy (Replies: 4)

  2. Gambling mathematics (Replies: 2)

  3. Binomial gambling (Replies: 1)

Loading...