Different Value In These Two Gambling Games?

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)?

If I am right, then what is the method behind working out the value of game 2? Given that it's quite easy for us to work out the variance for a game of roulette.

Sorry if I'm not being clear! I appreciate any insight and a refresher in my old statistics classes!

EV(Game 1) = (100*.973) + 10 = 107.30, so after $100 in action you stand to make $7.30, for a 7.3% advantage.

I don't follow your logic for the 2nd game. You must make $50 in wagers... that's going to result in (50*.973), for a return of $48.65, and an expected loss of $1.35. At this point your bankroll grows by $10, but you can't cash out unless you clear the bonus with another $50 in wagers. You're stating something like "only bet with that $10 otherwise don't dip" but once that bonus is in your account, it's essentially your money already.

To be free and clear to cash out you must again lose another expected $1.35 making $50 in wagers. At this point, according to you, you would have expected to lose $2.70, but your $10 bonus is somehow now a $20 bonus which is $20 cash? Do you have to play AGAIN to release the extra $10 bonus you got to bump you up to $20? Where did this $20 bonus come from?

Money management, such as a stop loss, does not affect the house edge of a game. This is why games can't be beated simply by using money management. The BONUS is what changes the house edge because it changes your expected value of the outcome of the play. You more than likely would be best fit to commit to the bonus and bet the other $50 regardless if you lose the bonus $10 on the first spin... because at the end of the rainbow you get another bonus $10 whilst your expected loss is only $1.35.

Okay, maybe I didn't explain the bonus properly. You get given £10 which is a bonus balance, at no point do you need to actually risk your own funds now since you have been granted the £10 bonus already.

Whereas in game 1 you got given the £10 bonus only after the £100 wagering.

The bonus can only become real funds once you've wagered another £50, but when you win it gets added to your bonus balance, so it's possible that upon completion of the £50 wagering that bonus £10 could now be £30 if you got lucky with your bets.

I'm not really talking about money management. I'm talking more about the difference between these two bonuses that I'm sure change the EV depending on the variance of the game you're playing (in game 2 anyway, game 1 the EV is unaffected by the variance)

Game 1 differs from game 2 in the way that in game 1 you have to wager £100 of your own money, it can go up and down, only after that you get your £10. In game 2 you get it earlier, and can stop risking any of your own funds.. If your bonus £10 goes down to £0 before the wagering is complete, you just stop since you don't want to play with your own money any more.

Hope this explains what I'm saying a little more! :) And thanks for your reply.

Edit: Here's one thing that makes me think the second game has to be more profitable.. Consider these two games:

Game 3 - We have to wager £100 on a 90% payout game and then we will be given a £10 cash bonus. EV is zero.

Game 4 - The casino just give us a £10 bonus that has to be wagered 10 times on the same 90% payout game (£100 wagering). If you worked it out with the logic above, you would say this bonus has an EV of zero. [same sort of rules on the bonus, it's like a play balance, starts at £10 but can go up and down depending on your luck. Once wagering is complete you get to keep whatever the bonus balance now is (could be zero though)]

How can this be true? If the casino gave out this bonus to 1 million people they would definitely be out of pocket since the player never had to risk his own money. So game 4 must have a value. If we know the variance of the game, how do we go about working out it's value? (Let's assume it's a really basic roulette style game).

If I am reading it right, in game 2, after betting £50, you get £10 worth of bonus; bets with bonus money are paid off with more bonus money, and if you have any bonus money left after 5 bonus bets of £10, it gets converted to cash.
There are four possible different results:
If you ever have more losses than wins with bonus money, you will lose all of it.
If you have 3 wins and 2 losses (and don't have more losses than wins at any point), you will have £20 (the initial 10, plus 30 for the wins, minus 20 for the losses)
If you have 4 wins and 1 loss, you will have £40
If you have 5 wins, you will have £60

It turns out that, of the 32 different sets of results of five bets, 1 has 5 wins, 4 have 4 wins, and 4 have 3 wins without having more losses than wins.
The expected value of the bonus = (20 x 4 x (18/37)³ x (19/37)²) + (40 x 4 x (18/37)⁴ x 19/37) + (60 x 1 x (18/37)⁵) = about 8.666.
When combined with the expected loss of 1.35 from the initial 50 bet, this is an EV of 7.316, which is slightly better than the 7.3 for game 1.

You've understood it exactly right! I knew game 2 had to be more profitable!

So I imagine it gets more profitable the more risky you play the game? So higher stakes would increase the EV? So if you did 2 bets of £25 instead. (in fact, ignore that.. I forgot I said it was a £10 bonus)
But if I lowered the stakes to £5 spins that would slightly decrease the EV?

Also I'm guessing it gets even more profitable (although higher chance of losing), if you play your £10 bet like you just explained then, on a number instead of red/black?
This would change the logic slightly from your previous argument am I right? Since now you could win one bet and lose four, and still come out with a profit.. As long as the win was before the losses.

So by my calculations.. If we play numbers instead of red/black we increase our EV from £7.316 to £8.35.

We are retaining £9.70 of the £10 bonus and then obviously we expect to lose £1.35 from the original £50 wagering. Does this sound right?