B Different Value In These Two Gambling Games?

[Korg]

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 into 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!

 

[mfb]

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.

 

[Korg]

Brilliant! I thought as much. Given that we know the variance of a roulette game, could you or anyone else go into 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.

 

[micromass]

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.

 

[Korg]

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.

 

[mfb]

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

On average, you bet less money, which reduces losses.

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.

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 let's 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.