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    $6385 on roulette wheel. Probability question


    Edit title should be $3840, not $6385
    ============


    If I ONLY bet on Red or Black, $5 each timer, what are the odds of winning ~$200 a day with $3840 reversal power?

    There are 18 black slots, 18 red slots and two green slots (0 and 00) on a US roulette wheel*. Your chances of hitting a black or red are equal, 47.4%. This translates to a 52.6% probability that the house will be scooping up all your chips on a black/red bet. The 5.26% 'spread' is known as the house edge.
    So if I take $3840 to the casino and bet ONLY on Red, and ONLY $5 each time (unless I bet the prize from last round), the chances of losing are pretty slim because of the relatively high reversal backup money.

    I had a look at the tables the other night and I didn't see any table with 9 times in a row same color.

    Am I doing the math wrong or it would be almost impossible to lose?

    Number Of Plays Bet Total Loss (If lose)
    1 $5 $5
    2 $10 $15
    3 $15 $30
    4 $30 $60
    5 $60 $120
    6 $120 $240
    7 $240 $480
    8 $480 $960
    9 $960 $1920
    10 $1920 $3850

    I read in a few forums some dealers mention it is NOT impossible to have black or red 11 times in a row.

    Tempted to check this out even though it's risky.
    Last edited by English Breakfast Tea; July 20th, 2018 at 12:51 AM.
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    Originally Posted by English Breakfast Tea
    So if I take $3840 to the casino and bet ONLY on Red, and ONLY $5 each time (unless I bet the prize from last round), the chances of losing are pretty slim because of the relatively high reversal backup money.
    Given that the chances of landing on red are 47.4%, I don't see how you could possibly deduce that your chances of losing are "pretty slim".

    Originally Posted by English Breakfast Tea
    I had a look at the tables the other night and I didn't see any table with 9 times in a row same color.
    That would be the Gambler's Falacy.

    Your odds of winning any spin are 47.4%. If you wager $5 then you can expect to win 0.474 * 5 = $2.37 on each play. That averages a net loss of $2.63 for each play.


    House always wins.
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    Originally Posted by requinix
    Given that the chances of landing on red are 47.4%, I don't see how you could possibly deduce that your chances of losing are "pretty slim"
    Look at it another way. Because I can stay in the game for 10 rounds. I just need a 1-time win to win all the loss.
    If I had only 1 shot, then 100% you're right. There' more chance than only playing once, or twice, or thrice, or 9 times?

    You have 10 shots vs you have 1 shot.

    if we look at it the way you say, the whole "probability" calculations are useless.

    And yes for sure, the house always wins 'in the long run'.

    Otherwise, there wouldn't be a business. I see how people go and put all their money on black or white.

    Tell me you see the difference I am talking about. Look me in the eyes when you tell me.

    What I get from your saying, "it won't matter if the bet is on red or black" each turn.

    What I nice is very very very dangerous is the greed. omg, you don't know how it feels when you slowly win $1000 free. Daemons take over.

    Winning with $5 is pretty slow too.
    Last edited by English Breakfast Tea; July 20th, 2018 at 03:53 AM.
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    That's the Martingale system. It's a great strategy... as long as you don't run out of money.
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    Won $400 in 1.5 hours.

    It's draining.

    So much pressure. Felt like work, but exciting.

    One time had 7 losses in a row.

    I think I ended up cashing because:

    1 - Didn't bet more than $5 unless it was house's money

    2 - Aggressively risked the win from previous hands. So won 5, 10, 20 and kept betting the win. Went up to $900 but down to $400.

    3 - Got lucky (no doubt)

    Probably won't go back.

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