How to calculate the odds of winning next game?

Hunter+1

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Mar 26, 2020
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The game I am referring to is highly similar to Black Jack. However, instead of cards equal to a certain value: pure integers are used.

The participant starts the game by rolling 1-100 as much as wanted. After adding each roll together, if the total passes 100 the participant loses, but if the number adds up to exactly 100, the participant wins.
If the participant hasn't lost, the host will roll until their total is greater than the participant's. If the host passes 100, the participant wins, otherwise the Host wins.

In this game if the participant only continues to roll when their total is less than 58, the odds of the participant winning each game is nearly 43%.

I simply want to know: How can you calculate the exact odds of winning each next game?

Example:
Let's say H=Host wins, and P=Participant wins.
Order of winners: PHPHHP
What are the odds of Participant winning after the following has been displayed, and how could the odds continue to be calculated?
 

Jomo

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Dec 30, 2014
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What have you tried? Can you tell us where you are stuck? This is a math help forum and it is hard to help you solve your problem if the helpers here do not know where you need help. Please post back and share your work with us.
 

Romsek

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Nov 16, 2013
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I don't see any reason that the games wouldn't be independent.

If that's the case then previous games contain no information regarding the outcome of the next game.
 

Hunter+1

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Mar 26, 2020
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The extent of my probability education was from one section of an Advanced Algebra class in high school, 5 years ago. I've always been fascinated by it,

During that segment of the course, I learned that the odds of getting heads twice in a row is no longer a 1/2 times. Heads twice in a row is 1/4, three times: 1/8, etc.

Applying that logic to 43 out of 100 (1 out of 2.33), if the participant wins once, the odds of them winning twice in a row should be calculable. right?
 

firemath

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Oct 29, 2019
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Hold up. Either I'm missing something very important (very likely) or we have a problem.

The instructions say "as much as wanted." That is a variable statement. No one can calculate another's preferences, unless you have an ideal roll for us to work with. That is, a game that is perfect. I know it says that 58 has 43%, and all that, but we can't calculate a preference.
 

Hunter+1

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Mar 26, 2020
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It's hard to wrap my mind around the odds not being calculable. especially because I've been under the understanding that you can calculate your exact odds of winning consecutive coin flips.
 

firemath

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Oct 29, 2019
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556
But how many coin flips does the person prefer? See? You never know what a person wants!

And, as Romsek said, the past cards or games do not affect the future ones.
 
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