Probability of Passes

Math Question for the Wizard, please.

What is the probability of one shooter, making five passes. The passes may result from naturals on the come out and or points won.

Five consecutive passes? Sounds simple enough - the probability of making one pass is 244 / 495, so the probability of five in a row is (244 / 495)^5, or about 1 / 34.36.

Thank you for your answer and the math. This too is very helpful. I did not consider this possibility.

However, no, not necessarily consecutive passes. Say the shooter rolls naturals as two of the passes, along the way, and makes three points. Mixing naturals with winning points. The order probably makes a difference? But for my purposes, I am after the probability for the event to occur.

While making the three points the shooter could roll 18 more or less times. Just a hypothetical example. Shooter could achieve the three points with less rolls or more rolls. Just attempting to produce a reasonable example here, if it at all helps.

How about without the naturals, five winning points?

You are kind to indulge my questions.

Quote: ThatDonGuy

Five consecutive passes? Sounds simple enough - the probability of making one pass is 244 / 495, so the probability of five in a row is (244 / 495)^5, or about 1 / 34.36.
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Would it count if he throws craps on some come out rolls but keeps rolling?

Assuming your question was "what's the probability of 5 consecutive pass line wins", which I think it was, TDG answered correctly

The probability of 5 consecutive pass line wins with 5 points won (not necessarily distinct points) is (134/495)^5 = apprx 1 in 688

Fantastic question! No craps on the come out. Thank you for catching this.

Ace, thank you very much. You are kind to join in the game. That answer was powerful.

Pass line wins can be duplicate? Or are you interested in separate pass line number wins which would be more in line with Fire Bets.

Yes, duplicated is okay, any pass line win. Any Pass Line win, either during the come out cycle or during the point cycle, total wins five, before the seven-out and without any craps during the come out.

Example: Come out 7 / 11 / 8. Point is 8, point is made. Come out 7 / 9. Point is 9, point is made. Five wins on the pass line. Thank you for asking.

What is the probability of five random passes, on the pass line, before the seven-out, and with no craps on the come out, to include naturals and points won?

Quote: MousseTheDog

What is the probability of five random passes, on the pass line, before the seven-out, and with no craps on the come out, to include naturals and points won?
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I am pretty sure it is (244 / 495)^5, or about 1 / 34.36. I do not see why this was not acceptable the first time.

This is the probability of winning five Pass Line bets without losing any - five combinations of naturals / points made without any craps on the comeout or missed points.

It's possible to win 5 odds bets in a row without getting any 2 PL bets won in a row.

But I usually try to run a PL progression without odds.
I've tried to dabble into pressing odds but I'll be waiting a few hours for a couple of 5 point shooters.

Thank you Don for the definitive answer. Sorry, I did not ask my question thoroughly in the first post. I received offers of help from others here and in answering their question, I finally came to the heart of my question, which you just spelled out clearly. This is what I am after for a 31:1 shot, flat bet, coming out, full press. So, assuming a 35 unit buy-in, and no craps on the come out, the math says I could have a shot at one experience with five wins. If I understand the math?

Your answer was and is perfectly acceptable to me. I did not understand the expression, but I like your answer. I get 244 divided by 495, and I understand where the five comes from. However, I do not know that the special character and how it and the five function to arrive at the ratio. Which is how I stumbled upon this forum after searching several math sites.

Thanks to you all the others for providing feedback. My first time here and impressed by the kindness and comradery.

The special character means that 244/495 is raised to the power of 5. Or to put it another way that ratio is multiplied by itself four times.

Quote: MousseTheDog

Thank you Don for the definitive answer. Sorry, I did not ask my question thoroughly in the first post. I received offers of help from others here and in answering their question, I finally came to the heart of my question, which you just spelled out clearly. This is what I am after for a 31:1 shot, flat bet, coming out, full press. So, assuming a 35 unit buy-in, and no craps on the come out, the math says I could have a shot at one experience with five wins. If I understand the math?

If you make 35 PL bets, you have a 40.0501% chance of at least one winning streak of five or more. That is easily calculated with a Markov chain and can also be approximated as: 1 - P / 2^35 = 42.02%, where P is the 37th Pentanacci number.

However, if you're asking about the probability of seeing such a streak before busting a 35 unit bankroll, that probability should be essentially 100% and you can expect to see TONS of them before busting out. I can get you a better answer tomorrow.

Quote: Ace2

Quote: MousseTheDog

Thank you Don for the definitive answer. Sorry, I did not ask my question thoroughly in the first post. I received offers of help from others here and in answering their question, I finally came to the heart of my question, which you just spelled out clearly. This is what I am after for a 31:1 shot, flat bet, coming out, full press. So, assuming a 35 unit buy-in, and no craps on the come out, the math says I could have a shot at one experience with five wins. If I understand the math?

If you make 35 PL bets, you have a 40.0501% chance of at least one winning streak of five or more. That is easily calculated with a Markov chain and can also be approximated as: 1 - P / 2^35 = 42.02%, where P is the 37th Pentanacci number.

However, if you're asking about the probability of seeing such a streak before busting a 35 unit bankroll, that probability should be near 100% and you can probably expect to see TONS of them before busting out. I can get you a better answer tomorrow.
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I assume he’s trying to reverse martingale five times on pass line with no odds and with a 35 unit bankroll.