remember that when buying the chips they give me only 10%, according to what you advise me if I buy 1,000 USD they would be giving me 1,100 USD it would be to bet it in a single hand and in case of losing the game they would give me 100 refund. Do you understand that this way I have an advantage of 15%?Originally Posted by aceside
I haven’t looked at the math - frankly not even sure of what the question is. However, the second last post then clarified next post by OP.
For argument sake, let’s say advantage is actually a whopping 15%. Then, clarified that wager MUST be $1000. Can your bankroll handle digging deep to wager multiple rounds to get the return.
If your answer is yes, then fine. If the answer is no, stay away. Just because the math (may) be there, doesn’t mean it’s a good deal for everyone. Further, $1000 bets are beyond most peoples comfort zone. If it’s beyond yours - stay away.
I’ll change the scenario to one familiar to many. That lousy game of roulette (I’m fuzzy on the rules, so there is some license here) promos 105-1 payout for a normal 35-1 on 00. With no minimum, just keep betting. Let’s say minimum bet for promo is $100. For some, it’s way outside the comfort zone. I’ll keep playing, no problem. Let’s say minimum for promo is $1000.
1, 10, 30, 60, 100 spins - when does it payoff for you. Can you dig that deep if you had to. Personally, I like to sleep at night.
I don’t. Now, what’s interesting in your question is - not the average number of spins required to win, rather the average number of spins to give you a 50% chance of winning.
Extending, at a minimum of $1000 a spin, I would want to know both 50% and 100% probabilities. I don’t know the math - logic says winning is 1-37 probability and furthered suggests that 50% probability is 1-18.5 spins. My gut feel says that a 50% probability exceeds 1-18.5 - in other words, the average exceeds the median.
So, if right and let’s say that number is 24 spins - am I prepared to risk 24k for a 50% chance to win 105k. The next question, since these are independent events, can I still rely on 1-37 to win 105k, or is my new probability still only 50% to win in the next 24 spins. I tend to think the latter.
Since I don’t need the 105k (less number of spins required to win, let’s say 37 for arguments sake, am I prepared to risk big dough at $1000 a spin - I think not. At a $100 a spin, or potentially $4800 invested to win $10500 for overall potential gain of $5700 - sure, why not. At a $1 per skin - who cares.
I'm alluding to the simplest possible way to solve for your X! (How did you solve for X?)
Multiply the denominator by ln2, which is 0.69. In this case, you get 0.69 x 37 = 25.5. So, 25 spins is not quite enough to reach 50%. Make that bet, and you're a long-term loser. But 26 is enough to make you a winner.
This is the easiest possible way to make an "under/over" for how many tries it will take to give you a 50% chance to achieve an unlikely event, such as the one we're discussing.
Don
One of the most famous problems in the history of probability theory, which, in fact, led to Pascal's formulations of dice probabilities, was the dice paradox of a notorious gambler of the day, the Chevalier de Méré. He first reasoned correctly that, trying to obtain a single six with a roll of a single die, if he were given four chances to do so, he would have the edge. It's easy to see that being successful is equal to 1 - (5/6)^4 = 51.8%, so he indeed did have the edge.
Then, he made the fatal mistake of reasoning that, with two die, if he wanted to obtain a roll of two sixes, he would have the edge if he were given 6 x 4 = 24 rolls. As we say in French, Erreur! With no calculators back in the day, it wasn't so easy to show that 1 -(35/36)^24 = 49.14%. The poor Chevalier lost a fortune being on the wrong side of this bet. All he would have needed was the very handy rule of multiplying by 0.69 to see that 0.69 x 36 = 24.84 and that, therefore, he needed 25 rolls to have a positive-e.v. wager.
Don
Trial and error
Natural log of 2.
Don, what is the significance of ln2 in solving this? How was that derived?
Reminds me of sqrt 3 being the denominator in 3 phase power calculations, for which I knew the derivation at one time, but that was many years ago (which is wye (pun intended for the power engineers in the crowd) one phase of commercial 3 phase power (208V) = 120VAC). Geeze, so many parentheses...
Edited to add - I see the ln2 question was answered on the next page, before I saw it, and after this post.
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